Broken Calculator

The calculator can either multiply its current value by 2 or subtract 1 from its current value. We’re given an initial start value and a target value, and we need to find the minimum number of operations to get from the start value to the target value.

One way to approach this problem is by working backwards from the target, and using a greedy algorithm. If the current value is greater than the start value, we need to consider two scenarios:

  1. If it’s even, we’ll divide by 2, since the target value has been multiplied to reach this point.
  2. If it’s odd, we’ll add 1, as the target value must have been decremented by 1 before being multiplied.

By repeating this process, we’ll eventually reach the start value, and the number of operations will be the number of steps taken.

Here’s the code:

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class Solution:
    def brokenCalc(self, startValue: int, target: int) -> int:
        operations = 0
        while target > startValue:
            if target % 2 == 0:
                target //= 2
            else:
                target += 1
            operations += 1

        return operations + (startValue - target)

This code first takes care of the scenario where the target value is greater than the start value by repeatedly dividing by 2 if even or incrementing by 1 if odd. Once the target value is no greater than the start value, the remaining difference between the start value and the modified target value is simply the number of decrement operations needed, and it’s added to the total operations count.

Identifying Problem Isomorphism

“Broken Calculator” can be mapped to “2 Keys Keyboard”.

In “Broken Calculator”, you start with a number X, and you want to get to a number Y using only two operations: doubling the current number, or decrementing the current number by 1. The task is to find the minimum number of operations to get from X to Y.

In “2 Keys Keyboard”, you start with a single ‘A’, and you want to get to n ‘A’s using only two operations: copying all the ‘A’s you currently have, or pasting the ‘A’s you last copied. The task is to find the minimum number of operations to get n ‘A’s.

The similarity between these problems is in the operations available, which are doubling the current state or adding one unit to it, and the goal, which is to reach a specific state in the minimum number of operations. In both problems, you are manipulating a number (either an actual number or the number of ‘A’s) and trying to reach a target.

“2 Keys Keyboard” is simpler due to the context being a bit more straightforward.

10 Prerequisite LeetCode Problems

For “991. Broken Calculator”, the following are a good preparation:

  1. “279. Perfect Squares” - Practices dynamic programming and understanding of minimizing operations.

  2. “322. Coin Change” - Similar concept of finding minimum operations, but with coins. Enhances understanding of dynamic programming.

  3. “365. Water and Jug Problem” - This problem uses a similar strategy of reducing a problem to simpler subproblems.

  4. “621. Task Scheduler” - Helps in understanding the problem-solving aspect related to operations and tasks.

  5. “134. Gas Station” - Practices solving problems from a circular perspective, like in the given problem where we could double or decrement the number.

  6. “752. Open the Lock” - This problem uses a similar approach of getting to the target in minimum operations.

  7. “127. Word Ladder” - It has a similar strategy where we need to reach a target word from a starting word using minimum changes.

  8. “518. Coin Change 2” - Like Coin Change, but also counts the number of combinations which might help understand problem-solving strategies.

  9. “416. Partition Equal Subset Sum” - Practices dynamic programming and understanding of operations to achieve a target.

  10. “45. Jump Game II” - It requires similar optimization to find the smallest number of jumps (or operations) to reach the end (or target).

These cover dynamic programming, problem reduction, and operation minimization, which are essential for solving the “991. Broken Calculator” problem.

Problem Classification

The problem falls under the “Mathematics and Simulation” domain. It involves basic arithmetic operations (multiplication and subtraction) and minimization.

What

  1. startValue: The initial value displayed on the broken calculator.
  2. target: The desired value you aim to display on the calculator.
  3. Operations: You can either double the current value or subtract one from it.
  4. Minimum Number of Operations: The objective is to find the minimum number of operations needed to reach from startValue to target.

This problem can be further classified as an “Optimization” problem. Specifically, it’s a “Minimization” problem where you’re required to find the minimum number of operations needed to transform startValue into target using the given operations.

Clarification Questions

  1. Are there any constraints on how many times each operation can be used?
  2. Is it possible for startValue to be greater than target?
  3. Do we need to account for any cases where it is impossible to reach the target from startValue?
  4. Are all operations reversible? For instance, can we divide by 2 instead of multiplying, or add 1 instead of subtracting?
  5. Can multiple operations be performed simultaneously, or do we have to complete one before starting another?
  6. Is there any time constraint on performing these operations?
  7. Can we perform the same operation back-to-back multiple times?
  8. Are we allowed to exceed the target temporarily and then come back to it?
  9. Are we interested only in the count of the minimum operations or do we also need to know the exact sequence of operations?
  10. Do we assume that the startValue and target are always integers, or could they be floating-point numbers?

Identifying Problem Isomorphism

To find an isomorphism for this problem, you can think of it as a path-finding problem. Imagine that you are at a location on a map, and this location is labeled with your startValue. Your destination on the same map is marked with your target. Each possible move from your current location corresponds to an operation on the calculator—either “multiply by 2” or “subtract 1.”

  1. Multiply by 2: Think of this operation as taking a big leap in a direction. This will approximately double your distance from the starting point.

  2. Subtract 1: Think of this operation as taking a single step. This operation moves you a fixed distance closer to your destination.

Your goal is to reach the destination (target value) in the fewest number of steps (operations). This makes the problem similar to finding the shortest path in a graph, where each node represents a value the calculator can display, and each edge represents an operation.

By framing the problem this way, you can leverage insights and algorithms from graph theory and path-finding to solve the original problem.

Problem Analysis and Key Insights

  1. Bounded Operations: You have only two operations available - multiply by 2 or subtract 1. This limits the ways in which you can manipulate the startValue.

  2. Directionality: You’re moving from a startValue to a target. The problem is directional, so choices made early can affect the ease or difficulty of reaching the final target value.

  3. Integer Constraints: All values, including the startValue and target, are integers. This means you can’t reach fractional or decimal values during your operations, which simplifies the search space.

  4. Minimal Steps: The objective is to minimize the number of operations, so you’ll need to find the most efficient path from startValue to target.

  5. Upper Limit: The constraints specify an upper limit of 10^9 for both the startValue and target, which suggests that your solution needs to be highly efficient to solve the problem within these constraints.

These insights guide you toward looking for the most efficient algorithmic approach, likely leveraging techniques from optimization and path-finding problems.

Problem Boundary

The scope of this problem is limited to:

  1. Integer values for both startValue and target, ranging from 1 to 10^9.

  2. Two types of operations: multiplying by 2 or subtracting 1.

  3. Finding the minimum number of operations needed to transform the startValue into the target.

  4. A single startValue and a single target for each query, without considering sequences of different values or multiple queries in one go.

The problem focuses purely on optimization under these constraints. It doesn’t involve any additional factors like time-based changes, variations in available operations, or probabilistic elements.

To establish the boundary of this problem, consider the following factors:

  1. Input Range: Both startValue and target are confined to integer values between 1 and 10^9. This sets a numerical boundary.

  2. Operations Allowed: Only two operations are permitted - either multiply the display number by 2 or subtract 1 from it. No other operations can be considered.

  3. Objective: The focus is to minimize the number of operations. This is not a question of finding all possible ways to reach the target but to find the one requiring the least number of steps.

  4. Output: The output is a single integer, representing the minimum number of operations needed. No other types of output, like the sequence of operations, are required.

  5. Constraints: The problem explicitly states the numerical constraints for startValue and target. No other variables or scenarios are indicated, making it a self-contained problem.

By considering these factors, you set a clear boundary around what is relevant for solving this problem and what can be ignored.

Distilling the Problem to Its Core Elements

  1. Fundamental Concept: The fundamental concept in this problem is “Dynamic Programming” or “Greedy Algorithm,” where you need to make optimal decisions at each step to minimize the total number of operations.

  2. Simplest Description: You have a calculator that shows a number. You can either double this number or subtract one from it. What’s the fewest number of moves to turn a starting number into a target number?

  3. Core Problem: The core problem is to find the minimum number of operations (either multiply by 2 or subtract 1) needed to convert startValue to target.

  4. Key Components:

    • Starting number (startValue)
    • Target number (target)
    • Permitted operations (multiply by 2, subtract 1)
    • Minimum operations (objective)

  5. Minimal Set of Operations: At each step, you have to decide between two operations:

    • Multiply by 2: To quickly approach the target if it’s far away
    • Subtract 1: To fine-tune your value to reach the target

By breaking down the problem this way, you can better focus on each component to find a solution that minimizes the operations needed.

Visual Model of the Problem

  1. State Transition Diagram: You can visualize this problem by drawing a state transition diagram where each node represents a possible calculator display value and each edge represents an operation (either multiply by 2 or subtract 1).

  2. Tree Structure: You could also visualize the problem as a tree, where the root is the startValue and each branch is a possible operation. The leaves would be potential values that can be achieved after a series of operations. Here, the goal is to reach the leaf that represents the target value, using the fewest operations.

  3. Number Line: Place startValue and target on a number line. Show the jumps when you multiply by 2 and the small steps when you subtract 1. This visual can help understand the distance between startValue and target and how each operation moves you closer or farther away.

  4. Flow Chart: Create a flow chart for decision-making. At each point, represent the two options (multiply by 2 or subtract 1) and the conditions that would make one preferable over the other. This can encapsulate your algorithmic approach to solving the problem.

  5. Table: A table could be used to keep track of the minimum operations needed to reach specific values. Each row could represent an achievable number, and the corresponding cell would hold the fewest number of operations needed to reach that number from startValue.

By visualizing the problem in one or more of these ways, you can better understand the task at hand and think more clearly about how to solve it.

Problem Restatement

You have a calculator that starts with a given number, called startValue. The calculator is broken and can only perform two types of operations: either multiply the current display value by 2 or subtract 1 from it. Your goal is to reach a specified number, called target, using the fewest number of these operations. Both the startValue and target are integers between 1 and 1,000,000,000. You need to find out the minimum number of operations needed to go from startValue to target.

Abstract Representation of the Problem

In an abstract sense, you have an initial state S and a goal state G. You can transition between states using a set of operations, in this case, multiplication by 2 or subtraction by 1. The task is to find the shortest path from S to G in this state space, where the path is defined as a sequence of operations.

This can be framed as a shortest-path problem in a directed graph. Each node in the graph represents a possible state (or value) of the calculator, and each directed edge represents an operation that transitions you from one state to another. The weight of each edge is 1, representing a single operation. Your goal is to find the shortest path from node S to node G.

The constraints bound the values for S and G to between 1 and 1,000,000,000, which also bounds the size of your state space. The objective is to minimize the number of operations, or in graph terms, to find the path with the minimum sum of edge weights.

Terminology

  1. State Space: In the context of this problem, the state space includes all the possible values the calculator can display. Each state is a node in a directed graph.

  2. Directed Graph: A set of nodes connected by edges, where each edge has a direction, indicating a one-way relationship. In this problem, each operation (multiply by 2 or subtract 1) is a directed edge from one state to another.

  3. Shortest Path: The sequence of operations (edges in the graph) that takes you from the initial state to the target state in the fewest number of steps. In this problem, finding the shortest path essentially means finding the minimum number of operations needed to reach the target from the start value.

  4. Edge Weight: The “cost” associated with an edge in a graph. In this problem, each operation has a weight of 1, as each operation counts as one step.

  5. Constraints: These are the limitations set on the problem, like the range of possible values for startValue and target. Constraints can impact the strategies you might employ to find a solution.

  6. Optimization Problem: A problem where you need to find the best solution according to some criteria. Here, you’re optimizing the number of steps taken to reach the target value from the start value.

Understanding these terms is crucial for conceptualizing the problem as a shortest-path problem in a directed graph and for discussing potential solutions in a clear and precise manner.

Problem Simplification and Explanation

Let’s break it down:

  1. Key Concepts:

    • Start Value: The number you begin with on the calculator display.
    • Target Value: The number you aim to get on the display.
    • Operations: Actions you can take - either double the number or subtract 1 from it.
    • Minimum Steps: The least number of actions needed to reach the target value from the start value.

  2. How They Interact:

    • You begin at the start value.
    • You use the allowed operations (double or subtract 1) to change the number on the display.
    • Your aim is to reach the target value in the minimum number of steps (operations).

  3. Metaphor: Imagine you’re in a maze with rooms labeled with numbers. You start in a room labeled with your “Start Value.” Your goal is to find the room labeled with your “Target Value.”

    • From any room, you have two paths:
      1. A path that takes you to a room with double the current room’s number.
      2. A path that takes you to a room with the current room’s number minus one.

    Your challenge is to find the shortest route to your target room, taking as few steps as possible.

Breaking the problem down this way helps you focus on what’s essential: navigating from the start value to the target value using only the given operations while minimizing the number of steps taken.

Constraints

  1. Operations Are Limited: You can either multiply by 2 or subtract 1. These limited choices can simplify the decision-making process at each step.

  2. Numerical Ranges: The start and target values range from 1 to (10^9). This large range suggests that multiplication (doubling) could get us closer to the target quickly, especially for larger numbers.

  3. Upper Limit on Steps: The maximum number of steps you can take is limited by the constraints. For example, doubling the smallest start value (1) ten times will exceed (10^9). So, theoretically, you won’t need more than ten “double” operations. Similar logic can apply to “subtract 1” operations.

  4. Start and End Values:

    • If the start value is already larger than or equal to the target, then you only need to perform “subtract 1” operations.
    • If the start value is significantly smaller than the target, doubling early on might be more beneficial.

  5. Odd or Even: The target being odd or even can be a point of exploitation. If it’s even, it might be more efficient to double your way there. If it’s odd, you might consider getting to an even number close to the target and then using “subtract 1”.

  6. Incremental Progress: Each step either brings you closer to the target or potentially overshoots it. Overshooting could be useful because it might get you closer to a number that’s easier to manipulate down to the target.

  7. Multiple Paths: Since each operation changes the number in a very specific way, it’s possible to consider multiple paths and select the one that minimizes the number of steps.

By recognizing these characteristics, you can develop a more targeted and efficient approach to solving the problem.

  1. Finite Range: Both the startValue and target are between 1 and (10^9). This sets a definite limit on the space you need to explore.

  2. Limited Operations: You have only two operations to consider: multiplying by 2 or subtracting 1. This simplifies the decision tree you have to navigate.

  3. Upper Bound on Steps: Given the constraints, you can deduce an upper bound on the number of steps needed. For instance, multiplying 1 ten times will get you beyond (10^9), so you don’t need more than 10 multiplication operations.

  4. Significance of 1: The lowest limit for startValue and target is 1, which means you have to consider scenarios where reducing the startValue to 1 is beneficial or where you have to reach 1 from a larger startValue.

  5. Large Number Handling: The upper limit of (10^9) forces you to think about efficiency since straightforward methods might not be fast enough for large numbers.

  6. Possibility of Backtracking: Given the nature of operations and constraints, you might sometimes benefit from overshooting the target and then working your way down.

  7. No Negative Numbers: Since you can only subtract 1 and the lower limit is 1, negative numbers are not a concern, simplifying the problem.

  8. Importance of Even/Odd: Given that you can multiply by 2, reaching an even number could be more efficient than aiming directly for an odd target.

Recognizing these aspects can guide you toward an efficient solution by helping you make educated choices at each step.

Case Analysis

Additional Examples and Test Cases

1. Minimal Input (Edge Case)

  • Input: startValue = 1, target = 1
  • Output: 0
  • Analysis: The startValue and target are the same. No operations are needed. This tests the lower boundary condition.

2. Reachable in One Step

  • Input: startValue = 4, target = 8
  • Output: 1
  • Analysis: You can reach the target in one “multiply by 2” operation. This case examines scenarios where minimal effort is needed.

3. Reduction Needed (Edge Case)

  • Input: startValue = (10^9), target = 1
  • Output: (10^9 - 1)
  • Analysis: The startValue is at its maximum and target is at its minimum. You can only use the “subtract 1” operation to reach the target. This tests the upper boundary condition.

4. Mixed Operations

  • Input: startValue = 6, target = 14
  • Output: 3
  • Analysis: Use double, decrement, and double: {6 -> 12 -> 11 -> 22 -> 14}. The focus here is on mixing both types of operations.

5. Reverse Approach

  • Input: startValue = 20, target = 41
  • Output: 4
  • Analysis: Sometimes it might be efficient to go beyond the target and then come back. Here, you could go {20 -> 40 -> 80 -> 79 -> 41}.

6. Odd Target

  • Input: startValue = 2, target = 7
  • Output: 4
  • Analysis: When the target is odd, you’ll likely need to use “subtract 1” at least once. This case checks for odd targets.

7. Even Target

  • Input: startValue = 3, target = 12
  • Output: 3
  • Analysis: With even targets, a sequence of “multiply by 2” might be more efficient: {3 -> 6 -> 12}.

8. Decrement Then Multiply

  • Input: startValue = 9, target = 16
  • Output: 2
  • Analysis: Sometimes it’s more efficient to decrement before multiplying. Here, you go {9 -> 8 -> 16}.

Edge Cases:

  • Minimal input with startValue and target at their lower limit.
  • Reduction needed with startValue at its upper limit and target at its lower limit.

These examples cover a variety of scenarios, including but not limited to edge conditions, odd/even targets, and the necessity for mixed operations. This should provide a robust foundation for problem-solving.

Visualizing these cases can be done through state-transition diagrams. Imagine each state as a node in a graph, and each operation (“multiply by 2” or “subtract 1”) as an arrow that takes you from one state to another. Let’s discuss a few cases to see how this works.

  1. Minimal Input (Edge Case)

    • A single node labeled “1” (both startValue and target), indicating no transition needed.

  2. Reachable in One Step

    • Two nodes: “4” and “8”, with an arrow from “4” to “8” labeled “x2”.

  3. Reduction Needed (Edge Case)

    • Two nodes: (10^9) and “1”, with (10^9 - 1) arrows from (10^9) to “1”, each labeled “-1”.

  4. Mixed Operations

    • Nodes: “6”, “12”, “11”, “22”, “14”. Arrows would represent transitions based on operations. From “6” to “12” (x2), “12” to “11” (-1), “11” to “22” (x2), and “22” to “14” (-1).

  5. Reverse Approach

    • Nodes: “20”, “40”, “80”, “79”, “41”. Arrows from “20” to “40” (x2), “40” to “80” (x2), “80” to “79” (-1), “79” to “41” (-1).

  6. Odd Target

    • Nodes: “2”, “4”, “3”, “6”, “7”. Arrows from “2” to “4” (x2), “4” to “3” (-1), “3” to “6” (x2), “6” to “7” (-1).

  7. Even Target

    • Nodes: “3”, “6”, “12”. Arrows from “3” to “6” (x2), “6” to “12” (x2).

  8. Decrement Then Multiply

    • Nodes: “9”, “8”, “16”. Arrows from “9” to “8” (-1), “8” to “16” (x2).

Each diagram would illustrate how many steps are needed to reach from startValue to target, thereby providing a good visualization of the operations involved.

From analyzing different cases, we can gather the following key insights:

  1. Minimal Movement: In the edge case where startValue and target are the same, no operations are needed.

  2. Single Operation: Some cases can be resolved in one step, typically when target is a multiple of 2 or target + 1 is a multiple of 2.

  3. Avoid Unnecessary Multiplication: When the target is less than startValue, multiplying will only take you farther away.

  4. Multiple Routes: In many cases, there are multiple ways to get to the target, but some are more efficient.

  5. Working Backwards: Sometimes, it’s more efficient to consider operations that transform the target back to the startValue.

  6. Odd and Even Targets: Depending on whether the target is odd or even, different strategies might be more efficient. Odd targets often require at least one decrement operation.

  7. Power of Two: If either startValue or target is a power of 2, and the other is close to it, you might reach the target more efficiently by multiplying or dividing by 2 multiple times.

  8. Upper Limit: The constraint of (1 \leq startValue, target \leq 10^9) indicates that simple looping might not be efficient for extreme cases.

These insights will inform the algorithmic approach to finding an efficient solution.

Identification of Applicable Theoretical Concepts

Several mathematical and algorithmic concepts could make the problem more manageable:

  1. Greedy Algorithms: Greedy approaches often work well when you have a set of choices and you need to make an optimal choice at each step.

  2. Dynamic Programming: The problem has overlapping subproblems. Calculating the minimum steps for smaller targets could help in finding the minimum steps for larger targets.

  3. Bit Manipulation: As one of the operations involves multiplying by 2, which is equivalent to left-shifting a bit, bit-level operations could be useful for efficiency.

  4. Recursion: The problem can be broken down into smaller instances of the same problem. Recursion may provide a simple way to implement a solution.

  5. Backtracking: The problem could involve making a series of choices, and backtracking can help in exploring these choices to find the minimum number of steps.

  6. Queue-based BFS: Since we’re looking for the minimum number of operations, breadth-first search (BFS) can help us explore all possible states level-by-level until we reach our target.

  7. Mathematical Patterns: Observing that multiplying by 2 is geometric progression, and subtraction is arithmetic, could provide insights for a more efficient algorithm.

  8. State-space Search: This is essentially a state-space problem where the goal is to go from one state (startValue) to another (target) with the least number of operations.

  9. Memoization: Storing previously computed states can prevent redundant calculations, speeding up a recursive solution.

  10. Complexity Analysis: Understanding the time and space complexity of each operation can help in choosing the most efficient approach.

By applying one or more of these concepts, the problem-solving process will likely be more efficient and effective.

Simple Explanation

You have a magic calculator that starts with a number you choose. The calculator can only do two things: double the number on the screen or subtract 1 from it. Your goal is to turn your starting number into a target number by using the fewest magic spells—oops, I mean operations—possible.

Imagine you’re baking cookies and you start with one cookie dough ball. You can either double the dough to make more cookies or take a small piece off. You want to end up with exactly the right amount of dough to make a specific number of cookies, and you want to do it as fast as possible. What’s the quickest way to get there?

Problem Breakdown and Solution Methodology

To solve this problem, we can use an approach called “Breadth-First Search” (BFS). Think of BFS as a treasure hunt. You’re starting at a point (the initial number on the calculator), and you have a map that shows different paths you can take (either multiply by 2 or subtract 1). Each move you make is like taking one step closer to the treasure (the target number).

Here’s the breakdown:

  1. Initialize the Queue:

    • Start with a queue that has one element: your starting number (startValue).

  2. Set up the ‘Visited’ List:

    • Create a list to keep track of numbers you’ve already calculated, so you don’t end up going in circles.

  3. Start the Hunt:

    • While the queue is not empty, take the first number out.

  4. Check the Map:

    • From that number, you have two paths: either double it or subtract one. Add these to the queue if they haven’t been visited yet.

  5. Count the Steps:

    • Every time you make a move, add one to the number of steps needed to reach each new number. Store this information so you can retrieve it later.

  6. Found the Treasure?

    • If you encounter the target number (target), the number of steps taken is your answer.

  7. Rinse and Repeat:

    • Keep doing this until you find the target or have visited all possible numbers within the constraints.

  8. Result:

    • Return the number of steps required to reach the target.

Impact of Changing Parameters:

  1. Higher startValue:

    • A higher startValue could either make it easier to reach the target or have no effect at all, depending on how close it is to the target.

  2. Higher target:

    • A higher target generally needs more operations, especially if it’s much larger than the startValue.

Example Case:

  • startValue = 5, target = 8
    1. Start with 5. Queue: [5], Steps: 0
    2. Double (10) and subtract 1 (4). Queue: [10, 4], Steps: 1
    3. Pop 10, no match. Double (20) and subtract 1 (9). Queue: [4, 20, 9], Steps: 2
    4. Pop 4, no match. Double (8) and subtract 1 (3). Queue: [20, 9, 8, 3], Steps: 2
    5. Pop 8, match! Total Steps: 2

This way, you can find the minimum number of operations needed to transform the startValue into the target.

Inference of Problem-Solving Approach from the Problem Statement

  1. startValue and target:

    • These are your starting point and destination, dictating the problem’s boundaries. Knowing them helps you initialize your algorithm and set the end condition for your search.

  2. Minimum Number of Operations:

    • This requirement guides us toward optimizing the algorithm to find the smallest number of steps, which makes Breadth-First Search a suitable choice.

  3. Multiply by 2 or Subtract 1:

    • These are your possible moves. Understanding these options informs how you’ll expand your search space during each iteration.

  4. Constraints (1 <= startValue, target <= 10^9):

    • These define the limits for your inputs, and help in considering the time complexity of your solution.

  5. Breadth-First Search (BFS):

    • Not explicitly stated in the problem, but a crucial concept for solving it. Knowing how BFS works enables us to systematically explore all possible moves to reach the target.

  6. Queue:

    • This is a data structure that fits naturally with BFS. It holds numbers as they wait for processing, ensuring that we explore possibilities in the order in which they are discovered, thus ensuring the minimum number of steps.

  7. Visited List:

    • This keeps track of numbers we’ve already considered, so we don’t process them again, which is essential for making the algorithm efficient.

Each keyword or concept informs a part of the solution, from how to initialize variables and data structures to how to structure the algorithm for efficiency and accuracy.

  1. State Transition Table:

    • Use a table to represent how the state (the number displayed on the calculator) changes based on the actions “multiply by 2” or “subtract 1”. This helps you visualize the possible next steps from any given state.

  2. Decision Tree Diagram:

    • Draw a tree where each node represents a state and each edge represents an operation. This will help you visualize the breadth-first search (BFS) strategy. You start from the root (startValue) and expand each node until you find the target value.

  3. Queue Representation:

    • You could use a horizontal list to mimic the queue in BFS, showing elements being added to one end and removed from the other. This helps to visualize the FIFO (First-In, First-Out) nature of the queue used in BFS.

  4. Visited List:

    • Keep a separate list/table of visited states. Every time you visit a state, mark it. This table will help you visualize the set of states you’ve already processed.

  5. Operations Counter:

    • Keep a table or column next to each state to represent the number of operations taken to reach that state from the startValue. This will help you keep track of the minimum steps required to reach each state.

  6. Constraints Table:

    • A simple table can list the constraints like the range of the startValue and target. This can serve as a quick reference while thinking through the algorithm.

Each of these visual aids gives you a different lens to understand the problem’s properties and constraints, making it easier to plan your algorithm effectively.

How did you infer from the problem statement that this problem can be solved using ?

Simple Explanation of the Proof

I’m having trouble understanding the proof of this algorithm. Could you explain it in a way that’s easy to understand?

Stepwise Refinement

  1. Stepwise Refinement of Approach:

    • Step 1: Initialize the variables

      • Set up the initial state with startValue, target value, and a counter for operations.

    • Step 2: Set up data structures

      • Prepare a queue for BFS and a set to track visited states.

    • Step 3: Begin BFS Loop

      • Start a loop that continues as long as the queue is not empty.

    • Step 4: Process each state

      • Dequeue the front state, apply the operations (“multiply by 2” and “subtract 1”), and enqueue the new states.

    • Step 5: Check Constraints

      • Verify if each new state is within the given constraints before proceeding.

    • Step 6: Update Counter

      • Every time a new state is enqueued, update the operations counter.

    • Step 7: Check for Target

      • If any state matches the target, exit the loop and return the counter.

  2. Granular, Actionable Steps:

    • Initialize variables: Declare startValue, target, and counter to zero.
    • Prepare Data Structures: Create an empty queue and an empty set.
    • BFS Loop: Use a while loop to process states from the queue.
    • Dequeue State: Remove the front state from the queue for processing.
    • Generate New States: Multiply by 2 and subtract 1 to generate new states.
    • Check Constraints: Ensure new states are within the 1 to 10^9 range.
    • Enqueue and Mark Visited: If a new state is valid and not visited, enqueue it and mark as visited.
    • Update Counter: For each newly enqueued state, increment the operations counter.
    • Check for Target: If the target is reached, return the counter.

  3. Independently Solvable Parts:

    • Checking constraints can be a separate function.
    • State generation (multiply by 2 or subtract 1) can also be modularized.

  4. Repeatable Patterns:

    • The BFS loop follows a repeatable pattern where you dequeue a state, generate new states, check for constraints, and enqueue valid states.
    • The operations counter is updated in a consistent manner, offering another repeatable pattern.

By breaking the problem down into these steps and components, we can focus on each part independently, making the problem easier to tackle.

Solution Approach and Analysis

Let’s break down the problem into smaller, manageable steps:

Approach:

  1. Initialization:

    • Start with startValue on your “calculator screen.”
    • Create a counter to keep track of the number of operations.
    • Initialize a queue to hold states to explore.

  2. State Exploration:

    • Think of this step as a maze-solving task where each room has two doors: one that doubles your number and another that decreases it by one.
    • Add the startValue into the queue.

  3. Breadth-First Search (BFS) Loop:

    • Loop through the queue. For each state:
      • If it matches the target, you’re done. Return the counter.
      • Otherwise, generate the next states by multiplying by 2 and subtracting 1.

  4. Constraints and Validity Checks:

    • For each new state, check if it’s within the boundaries (1 to 10^9).

  5. Enqueue New States:

    • Add these new, valid states back into the queue for further exploration.

  6. Increment Counter:

    • For every new level of states added into the queue, increment your operations counter by 1.

  7. Repeat Until Target Reached:

    • Continue this process until you reach the target state or run out of states to explore.

Considerations for Parameter Changes:

  • Changes in startValue and target:

    • A higher startValue might reach the target faster, but it’s not always guaranteed.
    • A large target will likely require more operations, thus increasing computation time.

  • Constraint Changes:

    • If the upper limit changes from 10^9, you’ll need to adjust your constraint checks.

Example Case:

Let’s consider startValue = 2, target = 3.

  1. Initialize startValue = 2, counter = 0, and queue = [2].
  2. Enter the BFS loop.
    • Dequeue 2 from the queue.
    • Generate new states: 4 (double) and 1 (subtract 1).
  3. Check constraints: Both 4 and 1 are within 1 to 10^9.
  4. Enqueue new states: Add 4 and 1 to the queue.
  5. Increment counter: counter = 1.
  6. Next iteration: Dequeue 4.
    • Check for target: 4 is not 3.
    • Generate new states: 8 and 3.
  7. Check constraints: Both are valid.
  8. Enqueue new states: Add 8 and 3 to the queue.
  9. Increment counter: counter = 2.
  10. Next iteration: Dequeue 3.
    • It matches the target. Exit and return counter = 2.

This way, each step contributes to the overall solution, and you can reach your answer efficiently.

Identify Invariant

The invariant in this problem is that you can always perform one of two operations on the current value displayed on the “calculator”: either multiply it by 2 or subtract 1 from it. This property holds true for any integer value on the display as long as it’s within the given constraints (1 to (10^9)).

An invariant serves as a rule or property that remains unchanged as you go through different states of the algorithm. Here, no matter what the value on the display is, as long as it’s within the constraints, you always have the two operations available. This invariant guides the solution approach, particularly in the state exploration step, where each state branches into two new states based on these operations.

Identify Loop Invariant

What is the loop invariant in this problem?

Is invariant and loop invariant the same for this problem?

Identify Recursion Invariant

Is there an invariant during recursion in this problem?

Is invariant and invariant during recursion the same for this problem?

Thought Process

Thought Process and Steps

  1. Understanding the Problem: The first step is to understand what the problem asks for. You have a calculator that starts with a given value and you can either multiply by 2 or subtract 1. The goal is to get to the target value with the minimum number of steps.

  2. Identify Constraints and Invariants: The next step is to look at the constraints and invariants. The invariant, as mentioned earlier, is that you have two operations available at any state: multiply by 2 or subtract 1.

  3. State Representation: Each state of the problem can be represented by the current value on the calculator. From each state, you can move to two new states by applying one of the two operations.

  4. Approach Direction: Since the aim is to minimize the number of steps, a Breadth-First Search (BFS) algorithm seems suitable. BFS explores all states level-by-level, ensuring that you find the minimum steps needed to reach the target.

  5. State Exploration: Starting from the initial value, apply BFS to explore each state until you reach the target value. Keep track of visited states to avoid redundant exploration.

  6. Count Steps: As you go through the states using BFS, count the steps taken to get to each state from the initial state. When you reach the target, that count will be the minimum number of steps needed.

  7. Implementation: Translate these steps into code.

Code

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from collections import deque

def min_operations(startValue, target):
    visited = set()  # Keep track of visited states
    queue = deque([(startValue, 0)])  # The queue for BFS, each element is a tuple (current_value, steps_taken)

    while queue:
        current_value, steps = queue.popleft()

        # If we've reached the target, return the steps taken
        if current_value == target:
            return steps

        # Mark the current state as visited
        visited.add(current_value)

        # Generate next states and add them to the queue if not visited
        for next_value in [current_value * 2, current_value - 1]:
            if next_value not in visited and 1 <= next_value <= 10**9:
                queue.append((next_value, steps + 1))
                visited.add(next_value)

# Test the function
print(min_operations(2, 3))  # Output should be 2
print(min_operations(5, 8))  # Output should be 2
print(min_operations(3, 10))  # Output should be 3

Cues and Insights

  • The problem statement subtly hints at a state-based approach because you move from one state (calculator value) to another through operations.
  • Constraints like “1 <= startValue, target <= 10^9” are hints that you should avoid solutions that would take too much time or memory.
  • The objective to minimize steps naturally leads us to consider BFS, which is excellent for finding shortest paths in an unweighted graph. Here, each state can be considered as a graph node.

By following this structured approach, you’re guided to an effective and efficient solution.

Memory Limit Exceeded

To make it more memory-efficient, we should start from the target and go backwards to the startValue because the target is limited to (10^9), but the startValue can be multiplied by 2 repeatedly and get very large.

Here’s a Python code snippet with the more optimized approach:

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class Solution:
    def brokenCalc(self, startValue: int, target: int) -> int:
        steps = 0
        while target > startValue:
            steps += 1
            if target % 2 == 1:
                target += 1
            else:
                target //= 2
        return steps + (startValue - target)

# Test the class and function
sol = Solution()
print(sol.brokenCalc(2, 3))  # Output should be 2
print(sol.brokenCalc(5, 8))  # Output should be 2
print(sol.brokenCalc(3, 10))  # Output should be 3

In this version, we start from the target and try to get back to the startValue. This limits the search space and ensures that we are only dealing with integers, thus avoiding memory issues.

Establishing Preconditions and Postconditions

  1. Parameters:

    • Inputs: startValue, target
    • Types: Both are integers
    • Representation: startValue represents the initial value on the calculator’s display. target is the value we aim to reach using the calculator.

  2. Preconditions:

    • State: None specific for the program.
    • Constraints: (1 \leq \text{startValue}, \text{target} \leq 10^9)
    • Specific State: No specific state requirements for the program.

  3. Method Functionality:

    • Expected to: Return the minimum number of operations needed to display target on the calculator starting from startValue.
    • Interaction: The method reads the startValue and target, then performs operations to find the minimum steps needed.

  4. Postconditions:

    • State: No change in the state of the program or the values of the parameters.
    • Return Value: The returned integer represents the minimum number of operations needed.
    • Side Effects: None

  5. Error Handling:

    • Response: The constraints are guaranteed by the problem statement, so there’s no built-in error handling for invalid inputs.
    • Exception/Special Value: None, as the constraints ensure valid input.

This method is a standalone function within a class and does not interact with or change any state within the program. It performs its task in a “pure” manner, reading inputs, and returning an output.

Problem Decomposition

  1. Problem Understanding:

    • In my own words, the problem is about a calculator that can either double its current display value or subtract 1 from it. Starting with a given number on the display, the goal is to reach a target number with the fewest operations.

  2. Initial Breakdown:

    • Major Parts:
      1. Read the start and target values.
      2. Perform operations to get from start to target.
      3. Count the number of operations.
      4. Return the count.

  3. Subproblem Refinement:

    • Read the start and target values:
      1. Validate constraints.
    • Perform operations:
      1. Choose between doubling or subtracting 1.
      2. Apply the chosen operation.
      3. Update the current value.
    • Count the number of operations:
      1. Initialize a counter.
      2. Increment the counter after each operation.

  4. Task Identification:

    • Choosing between doubling and subtracting could be a repeated task if we loop or recurse until we reach the target.

  5. Task Abstraction:

    • Each task is abstracted to a level where it’s clear but still meaningful. For instance, “choose between doubling or subtracting 1” is abstract enough yet clear in this context.

  6. Method Naming:

    • readInputValues: for reading start and target values.
    • chooseOperation: for selecting between the two operations.
    • applyOperation: for performing the selected operation.
    • updateCurrentValue: for updating the value after the operation.
    • initializeCounter: for setting up the counter.
    • incrementCounter: for increasing the counter by one.

  7. Subproblem Interactions:

    • The reading of values happens once and is independent.
    • The choice, application, and updating of operations are interdependent and iterate until the target is reached.
    • The counter needs to be initialized once, then incremented within the loop or recursion.
    • The final count is returned after the loop or recursion ends, thus dependent on all the above steps.

From Brute Force to Optimal Solution

Brute Force Solution

The brute-force way to solve this problem is to use breadth-first search (BFS) to explore all possible sequences of operations starting from startValue, until we reach target.

Here’s a simple Python code snippet to illustrate the brute-force solution:

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from collections import deque

def brokenCalc(startValue, target):
    visited = set()
    queue = deque([(startValue, 0)])  # Pair of (current_value, num_operations)

    while queue:
        value, steps = queue.popleft()

        if value == target:
            return steps

        visited.add(value)

        if value * 2 not in visited:
            queue.append((value * 2, steps + 1))
        
        if value - 1 > 0 and value - 1 not in visited:
            queue.append((value - 1, steps + 1))

print(brokenCalc(2, 9))  # Output should be the minimum number of steps to get from 2 to 9

Inefficiencies in Brute Force:

  1. Time Complexity: In the worst case, this will have a time complexity of O(2^N), which is extremely slow.
  2. Space Complexity: We’re also storing visited numbers in a set, which adds space complexity of O(N).

Optimization Steps

  1. Reverse the Operation: Instead of going from startValue to target, go in reverse from target to startValue. This way, you limit the operations to only division and addition, both of which reduce the number of possible next steps.

    • Reasoning: Going in reverse will limit the breadth of our search at each level, thus reducing the overall number of operations.

  2. Greedy Choice: Whenever we’re at an even number while going in reverse, it’s always more efficient to divide by 2, instead of adding 1.

    • Reasoning: Division by 2 reduces the number faster, getting us closer to startValue.

Here’s the optimized code:

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def brokenCalc(startValue, target):
    steps = 0
    while target > startValue:
        if target % 2 == 0:
            target //= 2
        else:
            target += 1
        steps += 1
    return steps + startValue - target

print(brokenCalc(2, 9))  # Output should be the minimum number of steps to get from 2 to 9

Complexity Analysis:

  1. Time Complexity: O(log N) - each operation at most halves the target, so the number of steps is logarithmic.
  2. Space Complexity: O(1) - constant extra space for variables.

By making these optimizations, we have significantly improved both the time and space complexity of the solution.

Code Explanation and Design Decisions

  1. Initial Parameters:

    • startValue: The integer value where the calculation begins.
    • target: The integer value we want to reach.

    These parameters define the starting and ending points of our transformation process. We need to apply a series of operations to get from startValue to target in the minimum number of steps.

  2. Primary Loop: The primary loop iterates as long as target > startValue. Each iteration represents one operation—either division by 2 or addition by 1—on target. We work backward, modifying target to approach startValue. This approach narrows down the options, making the solution efficient.

  3. Conditions within the Loop:

    • if target % 2 == 0: checks if target is even.

    If target is even, we divide it by 2; otherwise, we increment it by 1. The logic is to reduce the target value as quickly as possible, which is more efficient when we divide.

  4. Parameter Modifications:

    • target //= 2 or target += 1: These operations either halve target or increment it by 1.
    • steps += 1: This keeps track of the number of operations performed.

    These changes are essential because they represent the actual operations needed to transform target to startValue.

  5. Invariant: The invariant is that we always approach the startValue from target by using the least number of steps. After the loop, we add the absolute difference between startValue and the modified target to steps. This sum represents the minimum steps needed to achieve the transformation.

  6. Final Output: The final output is the value of steps, which gives us the minimum number of operations to get from startValue to target. This satisfies the problem’s primary objective—finding this transformation in the minimum number of steps.

By understanding these aspects, we can gain deeper insights into not only what the code is doing but why it’s doing it that way. This ensures that our approach is both efficient and aligned with the problem’s requirements.

Coding Constructs

  1. High-Level Strategies: The code uses a greedy approach, working backward from the target to the startValue. By focusing on reducing target to startValue as quickly as possible, the solution finds the minimum number of operations needed for the transformation.

  2. Explaining to a Non-Programmer: Imagine you have to turn a starting number into a target number by either adding 1 or dividing by 2. This code finds the fewest steps to make that happen, like a shortcut-finding map but for numbers.

  3. Logical Elements:

    • Iteration: A loop that continues until target is less than or equal to startValue.
    • Conditional Branching: If-else statements to decide whether to divide or add 1.
    • Counter: A variable to keep track of the number of steps.

  4. Algorithmic Approach in Plain English: Start with the target number. If it’s even, cut it in half; if it’s odd, add 1 to it. Keep doing this until you reach the starting number or go below it. Count these steps. Finally, if you went below the starting number, add the difference to the step count.

  5. Key Steps:

    • Check if target is even or odd.
    • Either divide target by 2 or add 1 to it based on the check.
    • Increment a step counter.
    • Compare the modified target with startValue.

    These steps are crucial for narrowing down the target to startValue efficiently.

  6. Algorithmic Patterns:

    • Greedy Strategy: Opting for the operation that most quickly reduces target.
    • Loop with Conditional: To iterate and decide on the operation at each step.
    • Counter: To keep track of the number of operations done.

By dissecting the code in this way, we gain a deeper understanding of the problem-solving strategies it employs.

Language Agnostic Coding Drills

  1. Distinct Concepts in Code:

    1. Variables and Initialization: The need to hold and initialize data.
    2. Conditional Statements: Checking conditions to decide the flow of logic.
    3. Looping Mechanism: Running the same piece of code multiple times.
    4. Arithmetic Operations: Performing calculations like division or addition.
    5. Counter Mechanics: Keeping a tally of actions.
    6. Comparisons: Checking whether one value is less than, equal to, or greater than another.
    7. Return Statement: Sending the result back.

  2. List of Concepts in Order of Increasing Difficulty:

    1. Variables and Initialization: Basic storage and retrieval of data.
    2. Arithmetic Operations: Basic calculations; almost everyone knows how to divide or add.
    3. Comparisons: Another simple operation, but it requires understanding the values you are comparing.
    4. Conditional Statements: Requires logical thinking to set up conditions correctly.
    5. Counter Mechanics: Involves both variable initialization and arithmetic operations.
    6. Looping Mechanism: Combines conditionals and counter mechanics, can get complex.
    7. Return Statement: Simple but needs an understanding of the overall flow and what needs to be returned.

  3. Problem-Solving Approach:

    1. Initialize Variables: Start by setting up variables for startValue, target, and a counter for the number of operations.
    2. Setup Loop: Next, create a loop that continues running until the target is not less than the startValue.
    3. Condition Check: Within the loop, use a conditional statement to decide whether to halve the target or increment it.
      • If target is even, it’s efficient to halve it.
      • If target is odd, increment it by 1 to make it even in the next iteration.
    4. Update Counter: After each operation, increment the counter by 1.
    5. Perform Comparisons: Continually compare target and startValue to know when to exit the loop.
    6. Adjust Counter: If target is less than startValue, increment the counter by their difference to reach the startValue.
    7. Return Value: Finally, return the counter, which now holds the minimum number of steps to transform startValue to target.

Each drill contributes to the problem-solving process, eventually leading to the minimum number of operations needed for the transformation.

Targeted Drills in Python

General Coding Drills

  1. Variables and Initialization

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    my_var = 10
print("The value of my_var is:", my_var)

  2. Arithmetic Operations

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    a = 5
b = 2
result = a + b
print("The sum of a and b is:", result)

  3. Comparisons

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    if a > b:
    print("a is greater than b")

  4. Conditional Statements

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    if a > b:
    print("a is greater")
else:
    print("b is greater or equal")

  5. Counter Mechanics

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    counter = 0
counter += 1
print("Counter:", counter)

  6. Looping Mechanism

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    for i in range(3):
    print("This is loop iteration", i)

  7. Return Statement

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    def return_example():
    return "Hello, World!"

result = return_example()
print(result)

Problem-Specific Coding Drills

  1. Loop Until Condition Met

    • This is crucial because our problem involves looping until startValue reaches target.
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    startValue = 5
target = 12
while startValue < target:
    startValue += 1
    print("Current startValue:", startValue)

Assembling the Drills

  1. Initialize Variables: Start by declaring startValue, target, and counter using Variables and Initialization drill.

  2. Set Up Loop: Utilize the “Loop Until Condition Met” drill to set up the main loop.

  3. Conditional Check and Arithmetic Operations: Within the loop, insert the Conditional Statements drill. Utilize Arithmetic Operations to halve or increment target.

  4. Update Counter: Use Counter Mechanics drill inside the loop to keep track of operations.

  5. Perform Comparisons: Use the Comparisons drill to exit the loop when the conditions are met.

  6. Adjust Counter: After exiting the loop, if needed, adjust the counter using Arithmetic Operations and Counter Mechanics drills.

  7. Return Counter: At the end, use the Return Statement drill to return the counter.

By assembling these drills in the outlined order, you will create a solution for the problem. Each drill serves as a building block in crafting the final solution.

Q&A

I thought this is a DP question and i spanned two trees one with X-1 and other with X*2 and return the minimum of both ,which reached Y fast. When But i failed to understand its a greedy. How to decide if it is a greedy or DP?

Deciding between Dynamic Programming (DP) and Greedy algorithms often depends on the properties of the problem. Here are some factors to consider:

  1. Optimal Substructure: Both Greedy and DP require the problem to have optimal substructure, meaning the problem can be broken down into smaller, simpler subproblems which can be solved independently.

  2. Overlapping Subproblems: If a problem has overlapping subproblems, meaning the same subproblems are solved multiple times, DP is usually a good choice. Greedy doesn’t handle overlapping subproblems well.

  3. Choice at Each Step: In Greedy algorithms, we make a choice that looks the best at that moment, and we never reconsider. If your problem allows you to make such choices and still reach the global optimum, it might be solvable using Greedy.

  4. Global vs. Local Optimum: If a local optimum guarantees a global optimum, it’s often a good indicator for a Greedy approach. If not, you’d likely need to use DP to explore more possibilities.

  5. Memory Constraints: DP uses extra memory to save subproblem solutions, which could be a limitation for some problems. Greedy algorithms generally use less memory.

  6. Time Complexity: If solving all subproblems is needed and each can contribute to the final answer in some way, DP is often more efficient. Greedy is generally faster but might not explore all possibilities.

  7. Invariant: An invariant can be a good indicator. If there’s something that remains constant whether you solve the problem in small chunks (greedy) or by considering all possibilities (DP), understanding that can guide your approach.

In your case, you used a tree-based approach that had elements of both Greedy and DP. Understanding the problem’s constraints and requirements can give you insight into which method to use. You initially opted for DP because you were exploring all paths to find the minimum one, which is typical of DP. However, Greedy could work here because taking the maximum allowable step at each point guarantees reaching the minimum steps at the end, satisfying the ’local optimum to global optimum’ condition of Greedy algorithms.

To summarize, distinguishing between Greedy and DP often comes down to understanding the problem’s structure and constraints. If you find that local optima lead to a global optimum and there are no overlapping subproblems, Greedy is often the right choice. If not, DP is usually more appropriate.

1 <= startValue, target <= 10^9 constraint screams it is not DP. But I’ve seen a problem that had such a constraint and had to be solved using DP with some trick.

Indeed, constraints like (1 \leq \text{startValue}, \text{target} \leq 10^9) usually signal that a straightforward Dynamic Programming (DP) approach could be inefficient due to time and memory constraints. However, that’s not an absolute rule; there are instances where DP can still be applied with certain optimizations or tricks.

  1. State Compression: In some problems, even though the input range is large, the actual number of states that need to be computed could be significantly smaller. This can be exploited.

  2. Memoization with Pruning: Sometimes, not all states in the DP table are needed for finding the solution. Memoization with aggressive pruning can help reduce time complexity.

  3. Dimensionality Reduction: The trick might involve reducing the problem to a smaller subset that still captures the essence of the problem.

  4. Pattern Recognition: In certain problems, especially with sequences or arrays, a pattern may emerge that allows you to solve the problem with a DP approach without having to visit all states.

  5. Fast Matrix Exponentiation: Some DP problems can be converted into matrix multiplication problems, and then solved efficiently using fast matrix exponentiation.

  6. Convex Hull Trick or Data Structures: Some DP problems have an inherent mathematical structure that allows them to be solved more efficiently using advanced data structures or mathematical tricks.

  7. Cycle Detection: If there’s a cycle in the states, you can detect it and then jump ahead, essentially leapfrogging through the DP states to arrive at the answer more quickly.

It’s crucial to consider the specific characteristics and constraints of the problem at hand. Sometimes, a seemingly DP-unfriendly constraint may become manageable when combined with the right insight or optimization.

So, when you see constraints that suggest DP might not be applicable, think about these tricks and see if any could fit into the problem’s structure. Also, you might find it beneficial to verify whether the problem satisfies the characteristics typically solved by Greedy algorithms, like local optima leading to a global optimum, as this could guide you towards the correct approach.

in the DP approach when it will go with case -1 , it will create endless loop of only substracting one from itself as it will be moving away from target so DP is not a good option

You’ve pointed out a significant issue with a naive DP approach for this problem: the potential for an endless loop in the (X - 1) case. This would happen because subtracting 1 repeatedly would move the value further away from the target, without contributing to a valid solution path. This is especially problematic because DP would try to explore all possible states and would get stuck in this infinite loop of decreasing values.

In contrast, a Greedy approach would intelligently choose which operation to perform at each step, aiming to reduce the gap between the start value and the target as quickly as possible. The greedy choice in this context would involve decisions that lead to a quicker convergence to the target value, avoiding the traps like the endless loop in the (X - 1) case.

So, this problem showcases why understanding the nature of the problem is crucial. It helps you avoid pitfalls associated with certain algorithmic approaches and guides you toward a more efficient solution.

A lot of people are having trouble getting their heads around why are we moving from target to startValue. So here I try to tell you the way I got to the solution.

Multiplication is an independent operation irrespective of whether a number is even or odd, so it is a possibility that multiplication by 2 would overshoot our target by a large value and then to tone it down we would have to apply a large number of subtraction operations(as for this case only subtraction would be possible) to reach the target, which is an overkill. So the greedy approach aims at dividing the number by 2 only when it is possible, which was not the case with multiplication as we could have multiplied at any instance of time. As the target becomes less than the startValue then we simply do addition operations to reach the target value.

Your explanation captures the essence of why moving from the target to the startValue is a sound approach for this problem. The primary reason this strategy works is because it avoids the pitfalls of overshooting the target, which can happen easily when multiplying by 2 from the starting point.

By working backwards from the target to the start value, you exploit the constraint that division by 2 can only happen when the number is even. This naturally limits the operations you can perform and ensures a more directed path to the solution, eliminating the need for a large number of subtraction operations to adjust for overshooting.

Your focus on “reversing” the operations, or working backwards, is indeed a clever way to navigate through the problem constraints. It effectively utilizes the fact that division by 2 is a “conditional” operation, whereas multiplication by 2 and subtraction by 1 are “unconditional” when moving forward from the start value. This insight provides a powerful guiding principle for making greedy choices that lead to an efficient solution.

everyone in the discussion is having the same code in which they are changing the value of target and not the value of start. can anyone explain why everyone is doing so

Changing the value of the target rather than the start value is a result of the reverse engineering strategy that has been widely recognized as efficient for this problem. Here’s why this approach is popular:

  1. Conditional Operations: Dividing by 2 is conditional on the number being even. When you work forwards, you have to consider whether to multiply by 2 or subtract 1, and choosing multiplication can lead to overshooting the target. Working backwards, you can only divide when the number is even, making the choices more constrained and thus easier to manage.

  2. Greedy Choice: By working backwards and dividing by 2 whenever possible, you make a greedy choice that ensures you minimize the number of steps needed to reach the start value. This is because division is the inverse of multiplication, and it’s faster to get to a lower number (our start value) by dividing than by subtracting.

  3. Reduced Complexity: Working from target to start value simplifies the problem. Once the target is less than the start value, you can simply count the difference as additional steps required. This helps in reaching a solution more directly.

  4. Avoids Overshooting: As you’ve also mentioned, multiplying by 2 has the risk of overshooting the target, necessitating a series of -1 operations. Working backwards inherently avoids this issue.

  5. Unifies the Approach: Everyone is using this strategy because it’s efficient and simplifies the problem. Once a good method is found, it often becomes the standard approach to solving that particular problem.

By applying these insights, the community has generally agreed upon a method that is both efficient and intuitive for solving this problem.

I started off in that way and got stuck for a while. I guess the realization that made me switch is that you can only divide by 2 for even numbers, but multiplying by 2, you can do that for even or odd numbers so going backward seems more restricted than going forward.

Exactly, you’ve nailed a key insight. When working backwards from the target to the start value, the division operation is more restrictive. This restriction is beneficial because it narrows down your choices, making the solution path more straightforward. In contrast, when you work forwards, the multiplication operation can be applied to any number, whether it’s even or odd. This openness adds complexity because it raises the question of when to multiply and when to subtract 1.

The backwards approach capitalizes on this inherent restriction, thereby simplifying the decision-making process. It’s easier to make the “greedy” choice of dividing whenever you can, as it is more likely to result in a minimal set of operations to reach the start value. This is a good example of how constraints in a problem can actually aid in finding an efficient solution.

I tried BFS approach starting from root, picked element from queue, performed multiply and subtract, push new value to queue. Repeated the process till we reach target value, While pushing new element to queue keep updating steps count.

But ended up with TLE for worst case startValue=1, target=10000000

So BFS won’t work here.. Though its actually to find out shortest route from source to target.

Your observation is correct. While BFS is generally a good approach for finding the shortest path, it’s not efficient enough for this particular problem due to the large input space. The constraints (1 \leq \text{startValue, target} \leq 10^9) indicate that a brute-force BFS could lead to an excessive number of states, leading to a Time Limit Exceeded (TLE) error.

In this problem, a greedy approach or optimized BFS would be much more efficient. Your original thought of using BFS isn’t entirely off the mark, but it needs to be fine-tuned for this specific problem’s constraints.

The greedy approach, which works backwards from the target, restricts the decision-making, thereby cutting down on the number of states you have to explore. This significantly reduces the time complexity and avoids the TLE issue you encountered.

This problem reminded me a lot of a problem from the 2005 AMC 8 competition (American Math Competition for 8th grade).

This was problem 24 from the 2005 AMC 8 competition:

“A certain calculator has only two keys [+1] and [x2]. When you press one of the keys, the calculator automatically displays the result. For instance, if the calculator originally displayed “9” and you pressed [+1], it would display “10.” If you then pressed [x2], it would display “20.” Starting with the display “1,” what is the fewest number of keystrokes you would need to reach “200”?”

What would you say is the answer?

One of the simplest solutions to this problem is also a “work backards” greedy aproach.

Yes, the AMC 8 problem you’ve described closely resembles the problem we’ve been discussing. Both involve optimizing a sequence of operations to reach a target number.

For the AMC 8 problem, you would also start with the number 200 and work backwards. Since multiplication by 2 is only possible for even numbers, you divide by 2 when you can. If you can’t divide by 2, then you subtract 1. Here’s how it works:

  • 200 is even, divide by 2: 100 (1 step)
  • 100 is even, divide by 2: 50 (1 step)
  • 50 is even, divide by 2: 25 (1 step)
  • 25 is odd, subtract 1: 24 (1 step)
  • 24 is even, divide by 2: 12 (1 step)
  • 12 is even, divide by 2: 6 (1 step)
  • 6 is even, divide by 2: 3 (1 step)
  • 3 is odd, subtract 1: 2 (1 step)
  • 2 is even, divide by 2: 1 (1 step)

So in total, you need 9 steps to go from 1 to 200.

The “work backward” approach is a type of greedy strategy. It cuts down the decision tree significantly, making it easier to reach an optimal solution quickly. The strategy is particularly effective when you can identify specific restrictions that help you decide the next move, like being able to divide by 2 only when the number is even.

Vasya has found a strange device. On the front panel of a device there are: a red button, a blue button and a display showing some positive integer. After clicking the red button, device multiplies the displayed number by two. After clicking the blue button, device subtracts one from the number on the display. If at some point the number stops being positive, the device breaks down. The display can show arbitrarily large numbers. Initially, the display shows number n.

Bob wants to get number m on the display. What minimum number of clicks he has to make in order to achieve this result?

Input The first and the only line of the input contains two distinct integers n and m (1 ≤ n, m ≤ 104), separated by a space .

Output Print a single number — the minimum number of times one needs to push the button required to get the number m out of number n.

Examples input 4 6 output 2 input 10 1 output 9 Note In the first example you need to push the blue button once, and then push the red button once.

In the second example, doubling the number is unnecessary, so we need to push the blue button nine times.

The problem you’ve described is quite similar to the ones we’ve discussed earlier. Here the task is to convert an initial number n to a target number m using two operations: multiplication by 2 (red button) and subtraction by 1 (blue button). You have to find the minimum number of steps to do this.

Problem Understanding

We need to get to the target number m starting from n using the least number of button presses. The operations are multiply by 2 or subtract 1.

Initial Breakdown

  1. Handle the case when n >= m. In this case, you just subtract n - m to reach m.
  2. Handle the case when n < m. In this situation, you can either multiply n by 2 or subtract 1 to get closer to m.

Subproblem Refinement

For the case when n < m:

  1. If m is even, divide it by 2.
  2. If m is odd, add 1 to make it even and then divide by 2.

Task Identification

  1. Button Press: A function to perform a button operation.
  2. Decision Maker: A function to decide which button to press.

Task Abstraction

The button press can be abstracted into a function that takes an integer and a button type (‘red’ or ‘blue’) and returns the new number.

Method Naming

  1. pressButton(int number, string buttonType): int
  2. decideButton(int target): string

Subproblem Interactions

  1. For n >= m, we just subtract. For n < m, we keep using decideButton until we reach n.

Brute-force

Starting from n, use a recursive function that explores both possibilities at each step (multiply by 2 or subtract 1) until it reaches m. Keep track of the minimum number of steps.

Inefficiencies

This would take exponential time. For each number, we explore two possibilities, leading to a time complexity of O(2^N).

Optimization

Instead, start from m and work your way back to n. If m is even, divide by 2. If m is odd, add 1 to it to make it even, and then divide by 2. Count the steps.

Time and Space Complexity

The optimized approach would have a time complexity of O(log m) and a space complexity of O(1).

Here, the greedy approach is more efficient than DP due to the constraints and the nature of the problem. If m is even, dividing it by 2 is the most efficient way to get closer to n. If it’s odd, we make it even by adding 1 and then divide by 2. This ensures that we get to n in the least number of steps.

Similar Problems

Here are 10 problems that use similar concepts:

  1. Reverse Integer

    • Similarity: Looping and arithmetic operations are key in both problems. Both require careful handling of integers.

  2. Two Sum

    • Similarity: It requires looping through an array and involves conditional checks similar to the problem discussed.

  3. Palindrome Number

    • Similarity: Like our original problem, this one also requires reversing numbers and checking conditions.

  4. Remove Duplicates from Sorted Array

    • Similarity: This problem involves iteration and condition-checking, much like our original problem.

  5. Maximum Subarray

    • Similarity: Both problems require iteration over a range and involve arithmetic operations and condition checks.

  6. Longest Common Prefix

    • Similarity: This problem involves iterating through strings and checking conditions which is similar to iterating and checking conditions in our original problem.

  7. Valid Parentheses

    • Similarity: The key part is to iterate through a sequence and maintain a kind of counter, in this case, a stack, similar to our original problem.

  8. Search Insert Position

    • Similarity: This problem involves looping and condition-checking to find or insert a value, similar to our loop and conditions to reach the target.

  9. Count and Say

    • Similarity: Both problems require string manipulation and looping through sequences.

  10. Remove Element

    • Similarity: It requires iterating through an array, making checks, and modifying the array.

These require skills used in solving the original problem—be it looping, condition-checking, arithmetic operations, or maintaining counters.