Minimum Time to Make Rope Colorful

10 Prerequisite LeetCode Problems

“1578. Minimum Time to Make Rope Colorful” involves the use of strings, arrays, and implementing an optimal strategy for removing elements. Practice problems that involve similar concepts such as greedy algorithms, string manipulation, and sorting.

Clarification Questions

Here are some questions to ask about this problem:

Are there any restrictions on the number of colors available for the balloons?
What should be the course of action if all balloons are of the same color?
Is it guaranteed that the time needed to remove a balloon is always a positive integer?
What happens if there are multiple ways to achieve the goal with the same minimum time? Should we just return the minimum time or all possible solutions?
Is it guaranteed that the ‘colors’ string and the ’neededTime’ array will always have the same length?
In the ‘colors’ string, will each unique character always represent a unique color?
Are we assuming that removing any balloon will not affect the positions of the other balloons?

Problem Analysis and Key Insights

From the analysis of the problem, several key insights emerge:

The problem involves array manipulation and string manipulation, and is likely to be solved using a combination of these techniques.
The goal of the problem is to minimize the total time taken to remove balloons, under the constraint that no two adjacent balloons have the same color. This makes it an optimization problem.
The ‘colors’ string and the ’neededTime’ array have a one-to-one correspondence, implying that each color has an associated removal time.
The time complexity of the solution is likely to be critical, as the number of balloons (n) can be as large as 105.
We’re looking for contiguous subarrays of balloons with the same color, indicating that we might need to use a sliding window or two-pointer technique.
Balloon removal does not depend on the removal of other balloons, suggesting that we can make a greedy choice at each step to achieve the optimal solution.
The problem does not specify what to do when there are multiple ways to achieve the minimum time. This suggests that there are multiple correct answers, and any one of them could be the solution.

Problem Boundary

The scope of the problem “Alice’s Colorful Balloons” involves understanding and applying concepts from string manipulation, array manipulation, and optimization techniques.

The primary objective is to determine the minimum time Bob needs to remove enough balloons such that no two consecutive balloons are of the same color. Here are the specific points that define the scope of this problem:

The problem must be solved by using the given input, which includes an array of integers and a string of characters.
The solution must consider the time it takes to remove each balloon, as specified in the ’neededTime’ array.
The solution must ensure that no two consecutive balloons of the same color remain after the operation.
The problem statement does not indicate a requirement for a specific order in the removal of the balloons. Therefore, the balloons can be removed in any order as long as the final condition of no two consecutive balloons of the same color is met.
The constraints limit the size of the array/string to 105, meaning the solution should be efficient and avoid excessive computation.
The final solution must return the minimum time required to make the string of balloons meet Alice’s requirements.

Establishing the boundary of a problem is all about understanding its limits and constraints. Here’s how we can establish the boundaries of the “Alice’s Colorful Balloons” problem:

Input Constraints: The string of colors and the array of time values should have the same length, which is between 1 and 105, as specified in the problem statement. All elements in the neededTime array are positive and up to 104. The colors string only contains lowercase English letters.
Processing Constraints: The solution should not result in two consecutive balloons of the same color. The goal is to minimize the time needed to achieve this, meaning we need to optimize the balloon removal process.
Output Constraints: The solution should return the minimum time needed to make the rope colorful. If the string is already colorful, the solution should return 0, indicating no time is needed for balloon removal.
Time Complexity Constraints: The constraints limit the size of the array/string to 105, which indicates that the solution should be optimized to avoid a time complexity of O(n^2) or worse.

These boundaries help in understanding the problem’s requirements and limitations, ensuring the devised solution fits within these constraints.

Problem Classification

This falls under ‘Arrays and Strings’. It involves working with an array of strings and corresponding integers, and finding the minimum sum of certain integer values, such that the condition of no two consecutive same color balloons is fulfilled.

The ‘What’ components are as follows:

An array of characters or colors represented as strings.
An array of integers representing the time needed to remove a balloon of a certain color.
A condition that no two consecutive balloons can have the same color.
The task to find the minimum total time to remove some balloons to fulfill the condition.

Considering these aspects, this problem can be classified as a ‘Greedy Algorithm’ problem. In this case, a solution will involve making the locally optimal choice at each step (i.e., choosing the balloon to remove that requires the least time) in the hope that these local choices will lead to a globally optimal solution, which is to make the entire rope colorful with the minimum total time. The elements of ‘Arrays’, ‘Strings’, and ‘Greedy Algorithms’ all come into play in this problem classification.

Distilling the Problem to Its Core Elements

Fundamental Concept: This problem fundamentally relies on the concept of array manipulation and string processing in tandem with optimization for achieving a minimal cost. It also utilizes the principle of greediness, as we want to spend as little time as possible to achieve the goal.
Simple Description: Imagine you have a string of beads in different colors. If two same-colored beads are next to each other, you want to remove one. Each bead takes a different amount of time to remove, and you want to spend the least amount of time possible. Your task is to figure out how to do this.
Core Problem: The core problem here is to identify and remove the minimal number of balloons (characters from the string) in the least amount of time such that no two adjacent balloons have the same color.
Key Components:
- Identifying pairs of adjacent balloons of the same color.
- Determining the time it would take to remove each balloon in a pair.
- Deciding which balloon to remove to achieve the minimal total time.
Minimal Operations:
- Scan through the string to identify adjacent balloons of the same color.
- For each pair of same-colored balloons, decide which one to remove based on the needed time.
- Sum up the time needed to remove the chosen balloons.
- Return the total time.

Visual Model of the Problem

To visualize the problem statement, imagine a line of balloons, each represented by a circle of a certain color. Each balloon is connected to the next, creating a sort of chain. You could also represent the time required to remove each balloon as a number inside or next to the balloon.

Let’s use an example for better understanding:

Consider colors = "aabac" and neededTime = [1,2,3,4,5].

Visualize it like this:

a(1)-a(2)-b(3)-a(4)-c(5)

Here, the letters represent the color of the balloon, and the number in the brackets represents the time taken to remove that balloon.

Then, Bob removes the balloons to avoid having two consecutive balloons of the same color, ideally removing those that require the least amount of time.

For example, Bob could remove the balloon a(1) and a(4) from the chain:

a(2)-b(3)-c(5)

This leaves no consecutive balloons with the same color, and Bob spent 1 + 4 = 5 units of time, which in this case is the minimum time possible.

Through this visualization, the problem of determining which balloons to remove to minimize the total removal time is clearly depicted.

Problem Restatement

Alice owns a series of balloons, each one having a specific color and placed in a row. Each balloon’s color is represented by a character in the given string, “colors”. Alice desires that no two neighboring balloons share the same color.

Bob is tasked with removing any balloons as necessary to achieve this colorful variation. Each balloon has a removal time associated with it, provided in the array “neededTime”, where the time corresponds to the same index as the balloon in the string.

The goal of the problem is to figure out the smallest total time Bob would need to make sure no two adjacent balloons have the same color. If the string of balloons already meets Alice’s conditions, Bob doesn’t have to remove any balloons, and the minimum time is then 0.

The constraints specify the number of balloons, the range of their removal times, and that the color of each balloon is represented by a lowercase English letter.

Abstract Representation of the Problem

We have a sequence of elements, each associated with a distinct identifier (the color) and a cost value (time needed for removal). The sequence must be transformed so that no two adjacent elements share the same identifier.

The transformation process involves removing elements from the sequence. Each removal incurs a cost, which is specific to the element being removed. The aim is to achieve the required transformation with the minimal total cost.

The problem’s constraints define the sequence length, the range of cost values, and state that identifiers are unique symbols from a predefined set.

Terminology

In the context of this problem, there are a few key terms to understand:

Balloon (or Element): The item that forms the sequence. Each balloon is unique and associated with two properties: color and time needed for removal.
Color (or Identifier): An attribute that is used to differentiate between balloons. It can be considered similar to a key or ID in other contexts.
NeededTime (or Cost): This denotes the cost of removing a balloon from the sequence. In other contexts, this might be referred to as a weight, value, or penalty.
Consecutive (or Adjacent): This term refers to elements that are next to each other in the sequence. In this problem, we want to avoid having consecutive elements with the same color.
Rope (or Sequence): A rope refers to the sequence or array of balloons. In other contexts, this might be called a list, array, or collection.

Understanding these terms is crucial to correctly interpreting and solving the problem.

Problem Simplification and Explanation

Imagine you’re at a party and you see a row of balloons tied to a rope. Each balloon has a color and each color represents the unique identity of the balloon. Now, you want this rope of balloons to be as colorful as possible - no two balloons that are tied next to each other should have the same color.

However, removing a balloon from the rope takes some effort (or time, in our problem). The time needed to remove each balloon is different.

You want to make the rope of balloons as colorful as possible but also want to spend as little time as possible in removing the balloons.

This problem asks us to figure out the minimum amount of time it will take to ensure that no two adjacent balloons have the same color.

A simpler way to look at this problem is to imagine you are editing a word document and you want to ensure that no two consecutive letters are the same. Deleting each letter takes a certain amount of time and your goal is to finish editing in the minimum time possible.

Key concepts involved:

Sequence: The sequence of balloons or letters in the document.
Unique Elements: The need for each element in the sequence to be unique.
Cost: The time needed to remove an element from the sequence.

The core problem is to minimize the cost (time) of making the sequence (rope of balloons or document) unique (colorful or having different letters).

Constraints

Let’s examine the problem for specific characteristics:

Sequence of Colors: The balloons are arranged in a sequence and we’re trying to remove certain elements (balloons) from this sequence to satisfy a certain condition (no two consecutive balloons of the same color). The sequential nature of this problem hints at the use of dynamic programming or greedy strategies, which exploit the property of subproblems or local optimality respectively.
Removal Cost: The time taken to remove each balloon is given. The goal is to minimize the total time taken to satisfy the condition. This aspect of the problem resembles a minimization problem which could be solved by a variety of strategies, including greedy algorithms, where at each step we make the decision that looks best at the moment to get the overall minimum time.
Non-Unique Elements: Consecutive balloons can have the same color and we want to avoid that. This means that we can exploit this property and target the removal of balloons that are surrounded by balloons of the same color.
Character Strings: Since the problem is dealing with string and array, it naturally lends itself to strategies involving string manipulation and array indexing. Techniques like two-pointer approach, prefix-sum, and sliding-window could potentially be used.

These specific conditions and characteristics provide us with clues about potential strategies for an efficient solution, though the exact details would depend on the precise structure of the problem and any additional constraints.

Upon analyzing the constraints, the following key insights can be derived:

Length Constraints: The number of balloons (length of ‘colors’ and ’neededTime’) can go up to 10^5. This is a common indication in algorithmic problems that our solution needs to be more efficient than O(n^2), as n^2 where n is 10^5 would result in 10^10 operations, which is typically too large to complete in a reasonable amount of time.
Character Constraints: ‘colors’ contains only lowercase English letters. This means there are a maximum of 26 distinct colors.
Time Constraint: The time needed to remove a balloon ranges from 1 to 10^4, so we’re not dealing with negative numbers or zero, which simplifies calculations.
Same Size Arrays: The ‘colors’ and ’neededTime’ arrays are of the same size, which means each balloon (character in the string) has a corresponding removal time. This direct mapping could potentially simplify our logic and coding efforts.

Understanding these constraints helps us rule out certain types of solutions that are unlikely to meet the problem’s requirements, and point us towards potential solutions that will be efficient enough.

Case Analysis

Let’s consider a few examples that could aid in a better understanding of the problem and cover different scenarios.

1. Case of all same colors (“monochrome”)

Input: colors = “aaaaa”, neededTime = [1,2,3,4,5]

In this case, all the balloons are of the same color. To satisfy Alice’s condition of not having two consecutive balloons of the same color, Bob has to remove all but one of the balloons. Ideally, he would remove the balloons in order of decreasing time, as he wants to minimize the total time spent. So, the minimum time would be the sum of all times except for the minimum one. For this case, the output would be 1+2+3+4 = 10.

2. Case of all different colors (“rainbow”)

Input: colors = “abcde”, neededTime = [1,2,3,4,5]

In this case, none of the balloons are of the same color, so Bob doesn’t need to remove any balloons. The minimum time would be 0, regardless of the values in ’neededTime’.

3. Case of repeated patterns (“striped”)

Input: colors = “ababab”, neededTime = [1,2,1,2,1,2]

Here, although there are two colors, none of them are consecutive, so Bob doesn’t need to remove any balloons, and the minimum time is 0.

4. Case of one color dominating (“dominant”)

Input: colors = “abaaa”, neededTime = [1,2,3,4,5]

In this scenario, color ‘a’ dominates the string. Bob has to remove two ‘a’s to satisfy Alice’s requirement. Following the logic of removing the balloons that take the most time first, Bob would remove the last two ‘a’s, and the minimum time would be 4+5 = 9.

These examples highlight different aspects of the problem and cover scenarios where the solution might have to perform different operations based on the input. They also cover edge cases, such as having all the balloons of the same color or all different colors.

Visualizing these cases could involve using diagrams or sketches that represent the sequences of colors and their associated removal times. Let’s illustrate the cases:

1. Case of all same colors (“monochrome”)

We have a line of balloons all in the same color. To make them colorful, we have to remove all but one, preferably those with the highest removal times:

aaaaa --> a 12345 --> 1

2. Case of all different colors (“rainbow”)

All the balloons are of different colors. No removal is necessary:

abcde --> abcde 12345 --> 12345

3. Case of repeated patterns (“striped”)

The colors alternate, so no balloons need to be removed:

ababab --> ababab 121212 --> 121212

4. Case of one color dominating (“dominant”)

One color is more prevalent. Remove the balloons with the highest removal times:

abaaa --> aba 12345 --> 123

Each balloon can be visualized as a pair of color and removal time. The operations we perform (removal of balloons) change these sequences towards our goal (no consecutive balloons of the same color). The visualization also highlights our strategy of minimizing the total time by removing the balloons with the highest removal times first.

Key insights derived from the analysis of different cases include:

No need for removal in diverse cases: In cases where all the balloons are of different colors, there is no need for any removal operation. The rope is already colorful.
Optimization in homogenous cases: In cases where all the balloons are of the same color, all but one balloon needs to be removed. However, since our goal is to minimize the total time of removal, we should target balloons that require the least time to remove.
Selectivity in dominant color cases: If one color is dominant but others are present, selectively removing balloons of the dominant color can achieve a colorful rope. Here, the decision should prioritize removing balloons that require less time, aiming for an efficient solution.
Pattern presence can help: In cases where colors alternate or follow a certain pattern, no balloons need to be removed. The pattern itself ensures that no two consecutive balloons are of the same color.

These insights should guide the development of an efficient strategy for solving the problem. They highlight the need to consider the characteristics of the given balloon sequence, the time required for removal, and the ultimate goal of achieving a colorful rope with minimum time expenditure.

Identification of Applicable Theoretical Concepts

The given problem revolves around string manipulation, and finding the minimum sum. Both these concepts hint towards the application of Dynamic Programming and Greedy Algorithms.

Here’s how these concepts fit into our problem:

Greedy Algorithm: In the context of this problem, the greedy approach can be used to determine which balloons to remove to get the least total removal time. Greedy algorithms build up a solution piece by piece, always choosing the next piece that offers the most immediate benefit - in this case, removing the balloon that takes the least time to remove when there’s a choice.
Dynamic Programming: Dynamic programming can be used to keep track of the minimum time required to make the rope colorful up to a certain balloon. The state of each balloon can depend on the previous ones, which fits the problem into the overlapping subproblems property of dynamic programming. With dynamic programming, we could create a solution that considers different possibilities and stores past results for future use, thus increasing efficiency.
Sliding Window: The sliding window concept could be applied here for removing sequences of same-colored balloons. This can be particularly useful in handling large inputs and improving the runtime efficiency of our solution.
Prefix Sum: The prefix sum could be used to quickly calculate the total time required to remove a range of balloons. It can reduce the time complexity of obtaining the sum of a subarray from O(n) to O(1), which can be a significant improvement for large inputs.

By identifying and applying these concepts, we can break down the problem into simpler, manageable parts, and solve it in a more efficient manner.

Simple Explanation

Let’s imagine a scenario where you have a long line of different colored beads on a string. Your friend, Bob, likes colorful things and doesn’t like when the same color appears next to each other. He thinks it’s more interesting when the colors alternate.

So, Bob decides to remove some beads to make sure no two beads of the same color are next to each other. But, there’s a catch. Each bead is tied to the string with a special knot, and the time to untie each knot varies. Some knots take longer to untie than others.

Your goal is to help Bob figure out the fastest way to remove the beads so that no two beads of the same color are side-by-side on the string, and so he spends the least amount of time untying knots.

This problem is like a puzzle where you’re trying to figure out the best order to do things to get the result you want as quickly as possible.

Problem Breakdown and Solution Methodology

Given this problem, a detailed approach could be as follows:

Initial Inspection: Look at the string of colors. If there are no two consecutive balloons of the same color, then no work is needed, and the time will be 0. For example, in the case of ‘abc’ with any given neededTime array, the answer will be 0.
Identify Repetitions: For strings where the same color balloon appears consecutively, identify the indices where we would need to remove a balloon to ensure no two same-colored balloons are side-by-side. This is like finding pairs of the same color in a bead necklace and marking one bead from each pair to be removed.
Removal Time Calculation: Once these indices are identified, we look at the neededTime array to see how much time it would take to remove each balloon at those indices. This step is akin to calculating how long it would take to untie each marked bead’s knot.
Find the Minimum: The goal is to minimize the total time taken. So, for each color repetition, we choose the balloon that takes the least time to remove. This is like deciding which bead to remove based on the time it takes to untie its knot.
Sum up the Time: Sum up the minimum times required to remove a balloon for each color repetition. This will give the minimum time needed to make the rope colorful. This is like adding up all the time it would take to untie all the chosen beads’ knots.

Let’s visualize this approach using an example:

Consider the example: colors = ‘aabaa’, neededTime = [1,2,3,4,1]

Here, we have ‘a’ repeated at index 0, 1 and ‘a’ repeated at index 3, 4. We need to remove one ‘a’ from each repetition.

From the first repetition, we can either remove the balloon at index 0 which takes 1 second or the balloon at index 1 which takes 2 seconds. We choose the one that takes lesser time, i.e., at index 0.

From the second repetition, we can either remove the balloon at index 3 which takes 4 seconds or the balloon at index 4 which takes 1 second. Again, we choose the one that takes lesser time, i.e., at index 4.

Adding up these times, 1 + 1, gives us the total minimum time of 2 seconds to make the rope colorful.

This approach will change based on the problem parameters. For example, if the colors string is longer or the neededTime for removal of each balloon is higher, the total minimum time to make the rope colorful would increase. If there are more color repetitions, there would be more balloons to remove which would increase the total minimum time. But the methodology of finding the minimum time remains the same.

Inference of Problem-Solving Approach from the Problem Statement

Here are some key terms or concepts in this problem:

String (colors): This is the input that represents the colors of the balloons. Since it’s a string, we can use standard string manipulation methods and properties, like indexing and length.
Array (neededTime): This is another input that gives the time needed to remove each balloon. An array is an ordered data structure that can be accessed via indices. This corresponds directly to the string colors and we can utilize this to associate each character in the string with its corresponding removal time.
Minimum Time: The problem is asking for a minimum time, implying that we’ll have to find the lowest possible sum of the neededTime array elements under certain conditions. This guides us towards using optimization techniques.
Consecutive Characters: The problem states that Alice doesn’t want two consecutive balloons to have the same color. This influences our approach in that we have to inspect consecutive characters in the string and their associated times in the neededTime array.
Removal of Balloons: Bob can make the rope colorful by removing balloons. This aspect of the problem implies that we are not just looking for the minimum value in the neededTime array, but the minimum values associated with each group of repeating characters in the colors string.

Taking all these key terms into consideration guides us towards an approach where we iterate through the string, identifying groups of consecutive same-colored balloons, and within each group, select the balloon with the minimum removal time to be removed. The sum of these minimum times will then give us the solution to the problem.

Creating a table can definitely help recognize the properties of this problem. Here’s how we can structure such a table:

Index	Color	Needed Time	Minimum Time in Group	Action
0	a	1	1	Remove
1	a	2	1	Keep
2	b	3	3	Remove
3	a	4	1	Remove
4	a	1	1	Keep

This table is based on the “aabaa” example from the problem. In the table:

The “Index” column simply tracks the position of each balloon.
The “Color” and “Needed Time” columns come directly from the problem input.
The “Minimum Time in Group” column shows the minimum needed time in each group of consecutive same-colored balloons. For example, in the group of “a” at indices 0 and 1, the minimum time is 1. We compute this value as we iterate through the colors string.
The “Action” column is a decision we make based on the “Minimum Time in Group”. In each group, we choose to “Remove” the balloon with the smallest “Needed Time”, and “Keep” the others.

To visualize this process, you could draw a diagram representing the balloons on the rope. Each balloon could be a circle filled with its color and showing its needed time. Draw an arrow from the balloon you decide to remove in each group. This visual representation can help illustrate the concept of removing a single balloon from each group to minimize the total removal time.

While the problem statement doesn’t directly specify that we can use techniques such as Greedy, Dynamic Programming (DP), Prefix Sum, or Sliding Window, we can infer the possible use of these techniques based on the nature of the problem and the type of solution we’re trying to find.

Greedy Approach: The problem is asking for a minimum time, which is a classic indication that a Greedy approach may be applicable. Greedy algorithms work by making the locally optimal choice at each stage, with the hope that these local choices will lead to a global optimum. In this case, removing the balloon that takes the least time in each group of same-colored balloons is a Greedy choice.
Dynamic Programming (DP): DP can be useful when the problem has an optimal substructure and overlapping subproblems, which might be the case here. As we’re trying to minimize the time to remove balloons, we could potentially build a solution by making decisions based on smaller, simpler subproblems (like choosing which balloon to remove in each group).
Prefix Sum: The Prefix Sum technique is often useful for quickly calculating the sums of elements in a range of an array, which could be useful here as we are looking at the sum of the times to remove balloons.
Sliding Window: A Sliding Window approach could potentially be used to keep track of the current group of same-colored balloons, making it easy to update our choice of which balloon to remove as we slide the window along the rope.

These techniques are all common strategies for solving optimization problems like this one. However, whether they can be applied effectively would depend on the specific details and constraints of the problem, and determining that would require a deeper analysis or actual implementation.

Simple Explanation of the Proof

The crux of the proof for this greedy algorithm lies in the choice we make when we encounter two consecutive balloons of the same color. Given two such balloons, we always opt to remove the balloon that takes less time to remove.

The reason we can make this choice is based on the principle of local optimality - a key idea in greedy algorithms. The local optimality principle suggests that by making the best choice at each step, we can arrive at a global optimum solution.

In this problem, our local optimum choice is removing the balloon with the lesser removal time, which we believe will lead to the globally optimal solution of minimizing the total time for removing balloons.

Now, let’s argue why this local optimum leads to a global optimum.

Let’s say we have two consecutive balloons i and j of the same color, and we’re deciding which to remove. Without loss of generality, let’s say balloon i takes less or equal time to remove than balloon j. So, we decide to remove balloon i. After removal, the colors sequence becomes colorful at position i since the balloon at position i-1 and the balloon at position j have different colors (because balloon i and balloon j have the same color).

Now, let’s suppose we made a different choice and removed balloon j instead. We would have spent more or equal time and also the sequence would have been colorful at position i (balloon at position i-1 and balloon at position i have different colors). But the total time taken would have been more or equal, so the previous choice was a better one.

This shows that our greedy strategy of always removing the balloon with the lesser removal time when confronted with two consecutive balloons of the same color ensures we are minimizing the total time, which is our objective.

Therefore, the greedy algorithm correctly solves this problem.

Let’s approach this in a stepwise manner:

High-level approach: The primary task is to ensure that no two adjacent balloons have the same color. To accomplish this, we need to identify and remove the minimum time-consuming balloons.
Stepwise Refinement:
- Step 1: Start from the first balloon and traverse through the string of colors.
- Step 2: Check if the current balloon color is the same as the previous one. If it isn’t, move to the next balloon.
- Step 3: If the colors match, we need to decide which balloon to remove. Here, we opt for the balloon that takes less time to remove. Remove that balloon and update the total time taken so far. Continue with the next balloon.
Parts of the problem that can be solved independently: Each choice to remove a balloon can be made independently based on the local information (the colors of the adjacent balloons and their respective removal times).
Repeatable patterns: A recurring pattern in our solution is the comparison of two adjacent balloons and decision making based on their colors and removal times. This is a pattern that is repeated throughout the entire traversal of the balloon string. The use of greedy strategy here means that we are always making the locally optimal choice at each step with the hope that these local optimal choices will lead to a globally optimal solution. In this case, the locally optimal choice is always to remove the balloon that takes less time when we encounter two adjacent balloons of the same color.

Solution Approach and Analysis

Let’s get into the details of solving this problem. We will be using the greedy algorithm.

Step 1: Initialize a variable, say total_time, to 0. This will keep track of the total time needed to make the rope colorful.
Step 2: Start traversing the given string colors from the first balloon. Keep a variable, say prev, that will store the time needed to remove the previous balloon. Initialize prev with the time needed to remove the first balloon.
Step 3: For every balloon from the second one, check if its color is the same as the previous balloon. If it’s not, just move on to the next balloon.
Step 4: If the current and previous balloons have the same color, you have two choices - remove the previous balloon or the current one. Since we want to minimize the total time, we choose the one with the lesser time. We add this time to total_time, and set prev to the time of the balloon we didn’t remove.
Step 5: Repeat steps 3 and 4 until all balloons have been checked.
Step 6: At the end, the value of total_time will be the minimum time Bob needs to make the rope colorful.

Let’s understand this with an example:

Suppose colors = "abaac" and neededTime = [1, 2, 3, 4, 5].

We start with the first balloon. total_time = 0, prev = 1.

Then we move to the second balloon. Its color is different from the first one, so we just move on. prev is now 2.

We arrive at the third balloon. Its color is the same as the second one, so we have to remove one of them. The second balloon takes 2 seconds to remove and the third one takes 3 seconds. So, we remove the second one. total_time = total_time + 2 = 2. prev is now 3.

The fourth balloon’s color is different from the third one, so we just move on. prev is now 4.

The fifth balloon’s color is the same as the fourth one. The fourth balloon takes 4 seconds to remove and the fifth one takes 5 seconds. We remove the fourth one. total_time = total_time + 4 = 6. prev is now 5.

Finally, we’ve checked all the balloons. The rope is now colorful and the total time taken is 6 seconds.

Any change in the problem’s parameters, like the order of colors or the times needed to remove the balloons, can change the solution. The solution relies heavily on these parameters.

This approach ensures that we minimize the time to make the rope colorful by always choosing to remove the balloon that takes less time whenever we encounter two consecutive balloons of the same color.

Identify Invariant

An invariant in the context of algorithm design is a condition that can be relied upon to be true during the execution of an algorithm. For this problem, our invariant is:

“At any point of time, no two consecutive balloons in the remaining sequence have the same color.”

Initially, the sequence may have consecutive balloons of the same color. We handle this in our algorithm by iteratively removing balloons. After each removal operation, our invariant is maintained by removing one balloon from each pair of consecutive balloons with the same color.

This invariant holds true right from the start where no balloons are removed, during the execution of the algorithm where balloons are being removed, and at the end where no two consecutive balloons of the same color are left.

Having such an invariant helps in reasoning about the algorithm’s correctness and efficiency. It allows us to focus on maintaining this condition throughout the algorithm and assures us that if this invariant is preserved at every step, our final output will meet the problem’s requirements.

Identify Loop Invariant

In the context of this problem, a suitable loop invariant would be:

“At the start of each iteration of the main loop, the balloons in the sequence up to the current index do not have two consecutive balloons of the same color.”

This loop invariant works under the assumption that we’re iterating over the sequence of balloons from left to right and making decisions about which balloons to remove.

Before the loop starts (i.e., when no balloons have been removed yet), the invariant is trivially true because we haven’t made any decisions about removals yet.

During each iteration of the loop, we maintain the invariant by deciding whether to remove the current balloon based on the color of the previous balloon in the sequence.

By the end of the loop, when we’ve made a decision for every balloon in the sequence, the invariant ensures that there are no two consecutive balloons of the same color in the final sequence.

The correctness of the loop is therefore tied to maintaining this loop invariant at each iteration, which, when upheld, ensures the correctness of the entire algorithm.

In the context of this problem, the invariant and the loop invariant can be seen as the same. The invariant is a condition that remains unchanged throughout the execution of the algorithm, while a loop invariant is a condition that remains unchanged during each iteration of a loop specifically.

Here, we are discussing an algorithm that uses a loop to iterate over the balloons. The invariant or condition that we want to maintain—no two consecutive balloons of the same color—is assessed and enforced within this loop. Hence, in this specific context, the invariant becomes a loop invariant because the condition is being checked and maintained within the loop of the algorithm.

In general, not all invariants are loop invariants (as there might be invariants outside loops), but in this specific problem, our main invariant is indeed a loop invariant.

Thought Process

Failed to solve using the above steps in this article. Even after several iterations. Official solution:

Your outlined algorithm makes sense and correctly applies the greedy approach. The algorithm breaks down as follows:

Initialize two variables: totalTime and currMaxTime, setting both to 0. totalTime will keep track of the total removal time, while currMaxTime will track the maximum removal time in the current group.
Iterate over each balloon. Each balloon i has a color colors[i] and a removal time neededTime[i].
If the current balloon is the first balloon of a new group (its color is different from the previous balloon’s color), reset currMaxTime to 0.
Increase totalTime by the smaller of neededTime[i] and currMaxTime. This step is the crux of the greedy strategy: always choosing the balloon with the smaller removal time to remove.
Update currMaxTime as the larger of neededTime[i] and currMaxTime. This ensures that currMaxTime always represents the maximum removal time in the current group.
Finally, return totalTime, which now holds the minimum total removal time.

This approach ensures that, for each group of balloons of the same color, the balloon with the maximum removal time is retained and all others are removed. As a result, the total removal time is minimized, which is the objective of the problem.

Here’s the Python code that corresponds to this algorithm:

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class Solution:
    def minCost(self, colors: str, neededTime: List[int]) -> int:
        n = len(colors)
        colors = ' ' + colors  # add a space at the beginning to handle the first balloon
        totalTime, currMaxTime = 0, 0
        for i in range(1, n+1):
            if colors[i] != colors[i-1]:
                currMaxTime = 0
            totalTime += min(neededTime[i-1], currMaxTime)
            currMaxTime = max(neededTime[i-1], currMaxTime)
        return totalTime

Establishing Preconditions and Postconditions

Parameters:
- This method takes two parameters: colors and neededTime.
- colors is a string and neededTime is a list of integers.
- colors represents the colors of the balloons. Each character in the string represents the color of a balloon. The neededTime list represents the removal time for each balloon. The ith integer in neededTime corresponds to the ith character in colors.
Preconditions:
- Before this method is called, it’s assumed that the input parameters colors and neededTime are of equal length.
- colors should only contain lower case alphabetic characters. Each element in neededTime is a positive integer.
- The program doesn’t need to be in any specific state before this method is called.
Method Functionality:
- The method is expected to calculate the minimum total removal time for the balloons. It iteratively processes each balloon and makes a decision based on the greedy approach as discussed earlier.
- The method doesn’t change the state of the program or the inputs.
Postconditions:
- After the method has been called, the return value indicates the minimum total time to remove the balloons.
- The state of the program or the values of the parameters remain unchanged after the method has been called. The method doesn’t have any side effects.
- The return value represents the minimum total time to remove the balloons such that no two adjacent balloons have the same color.
Error Handling:
- The method doesn’t explicitly handle errors. If the preconditions are not met (for example, if colors and neededTime are of unequal length), a runtime error may occur.
- It’s the responsibility of the caller to ensure that the preconditions are met before calling the method.

Problem Decomposition

Problem Understanding:
- The problem involves a string of characters, where each character represents a balloon of a specific color. There’s an associated list of integers, each representing the time needed to remove a balloon. The task is to find the minimum total time to remove all balloons, such that no two adjacent balloons have the same color.
Initial Breakdown:
- The major stages of this problem are: group the balloons by their colors, keep at most one balloon from each group, and calculate the total time for removing the balloons.
Subproblem Refinement:
- Grouping the balloons by colors: For each balloon, check if its color is the same as the previous one. If it is, it’s in the same group as the previous balloon; otherwise, it starts a new group.
- Keeping one balloon from each group: For each group, keep the balloon that requires the longest time to remove.
- Calculating total time: Add up the removal times of all the kept balloons.
Task Identification:
- Identifying the color groups is a repeatable task, as we need to do it for each balloon.
Task Abstraction:
- The task of identifying color groups is abstracted to a level where it’s clear, reusable, and still makes sense in the context of the problem.
Method Naming:
- We can name the method that performs the task “groupByColor”.
Subproblem Interactions:
- The subproblems need to be performed in a specific order. We first need to identify the color groups. After that, we can decide which balloons to keep and which ones to remove. Once we have the list of balloons to keep, we can calculate the total removal time. The task of identifying color groups depends on the input data (colors of balloons). The task of deciding which balloons to keep depends on the results of the first task (color groups). The calculation of total removal time depends on the results of the second task (balloons to keep).

From Brute Force to Optimal Solution

A brute force solution for this problem generates all possible arrangements of balloons such that no two adjacent balloons have the same color, calculating the removal time for each arrangement, and then finding the minimum removal time among all arrangements.

Brute Force Approach:
We can start by generating all possible arrangements of balloons, which is a permutation problem. We generate all permutations and for each permutation, check if it follows the rule that no two adjacent balloons have the same color. If it does, we calculate the removal time. We keep track of the minimum removal time across all valid permutations.
This brute force approach, however, is highly inefficient. It has a time complexity of O(n!) where n is the length of the string, since we are generating all permutations. In addition, checking whether each permutation is valid and calculating the removal time also takes time, leading to a high computational cost.
Optimization Process:
Realizing that generating all permutations is inefficient, we turn to a key insight: we don’t need to consider all permutations. We only need to consider the groups of same-colored balloons.
We can group the balloons by their colors. Within each group, we keep the balloon that requires the longest time to remove, and remove the rest. The total time required is the sum of the removal times of the balloons we removed.
This greedy approach works because it always ensures we remove the balloons with smaller removal times within each group of same-colored balloons, thus minimizing the total removal time.
Impact on Time and Space Complexity:
With this optimization, our time complexity reduces to O(n), where n is the length of the string. This is because we only iterate over the string once to group the balloons and calculate the total removal time. The space complexity is also O(n), as we keep track of the maximum removal time for each group.
In conclusion, this optimized solution is significantly more efficient than the brute force approach.

Code Explanation and Design Decisions

Initial Parameters:
- The input to our function is a string colors representing the colors of balloons and a list neededTime representing the removal time of each balloon. These parameters define the problem space and dictate the nature of our solution.
Primary Loop:
- Our primary loop iterates over each balloon in the colors string. Each iteration represents a step in the process of minimizing the total removal time, where we decide whether to keep or remove a balloon based on its color and removal time.
Conditions within the Loop:
- Within the loop, we have a condition that checks if the current balloon is the first balloon of a color group. This condition allows us to manage the grouping of balloons by color and ensure we’re making decisions on a group-by-group basis, as per the problem constraints.
Parameter Modifications:
- Inside the loop, we modify totalTime and currMaxTime. totalTime represents the total removal time we’ve accumulated so far and is updated by adding the minimum of currMaxTime and the removal time of the current balloon. currMaxTime represents the maximum removal time among balloons of the current color group and is updated as the maximum of its current value and the removal time of the current balloon. These modifications help us track our progress towards the solution.
Invariant:
- Throughout the code, we maintain an invariant that totalTime is always the minimum possible total removal time we can achieve for the balloons we’ve processed so far. This invariant aligns with our objective of minimizing the total removal time.
Final Output:
- Our final output is totalTime, which represents the minimum total removal time for all balloons. It meets the problem’s requirement by providing the least amount of time needed to remove all the balloons such that no two adjacent balloons have the same color.

Coding Constructs

Problem-Solving Strategies:
- This code is utilizing a Greedy algorithm approach, which makes the locally optimal choice at each stage with the hope of finding a global optimum. The goal is to minimize the total time required to remove all the balloons.
Purpose of the Code:
- If I were to explain the purpose of this code to a non-programmer, I’d say it’s like planning a sequence of actions to pop a series of balloons where each balloon requires a specific time to pop, and the rule is that you cannot pop two adjacent balloons of the same color. The aim is to find the best sequence to pop all balloons in the least total time.
Logical Constructs:
- The code uses a loop to iterate over the elements in a list, conditions to compare values and decide actions, and variables to store and update the current state of the problem.
Algorithmic Approach:
- In plain English, the code goes through each balloon, looking at its color and the time it takes to pop it. If it’s the first balloon of a particular color group, it will store its popping time. For the next balloon of the same color, it will decide whether to pop the current balloon or the previous balloon based on which one takes less time. This way, it ensures that no two balloons of the same color are adjacent.
Key Steps or Operations:
- The code performs a few key steps: (i) Identifying when a new color group starts (ii) Selecting which balloon to pop within a color group based on the popping time (iii) Summing the popping times of selected balloons. These steps allow the code to implement the algorithm and calculate the minimum total time.
Algorithmic Patterns or Strategies:
- The primary algorithmic pattern used in this code is the Greedy strategy, where we make the locally optimal choice at each step. Also, there is an element of dynamic programming as we store the maximum popping time for each color group, which helps us make future decisions.

Language Agnostic Coding Drills

Concept Identification:
- Variable assignment: It’s a basic building block of any programming language. You assign values to variables so you can refer to and manipulate them later.
- List iteration: This is a fundamental concept of looping over elements of a list. It’s necessary to traverse the data structure and apply operations on each item.
- Conditional statements: They allow the code to make decisions. Here, they’re used to check the conditions related to the balloon’s color and popping time.
- Comparison operators: Used in the conditions to compare the popping times of the balloons.
- Mathematical operations: Specifically addition and the use of min and max functions. They help calculate the minimum total popping time.
Difficulty Classification:
- Variable assignment: Easy. It’s a fundamental concept that is typically taught at the beginning of any programming course.
- List iteration: Easy-Medium. While a basic concept, using it effectively requires understanding the data structure and the concept of loops.
- Conditional statements: Medium. It requires understanding the problem’s logic and making decisions based on it.
- Comparison operators: Easy-Medium. Comparing values is straightforward, but correctly applying comparisons in the problem context can be challenging.
- Mathematical operations: Medium. Using addition is easy, but using min and max effectively in this problem requires more understanding.
Problem-Solving Approach:
- Variable assignment: Begin by creating variables to store the total popping time and the maximum popping time for the current color group. This allows tracking the current state of the problem.
- List iteration: Iterate over the given lists of colors and popping times. This allows us to examine each balloon one by one.
- Conditional statements: Use conditions to check if we’ve encountered a new color group. If yes, reset the max popping time for the new group. Otherwise, decide which balloon to pop based on their popping times.
- Comparison operators: Apply comparisons to decide which balloon to pop - the one with the lesser popping time. Also, use comparisons to update the max popping time.
- Mathematical operations: Use addition to update the total popping time. Use min and max to make decisions on which balloon to pop and to update the max popping time for the current group.

By understanding these concepts and how they contribute to the solution, you can effectively break down the problem, implement each part separately, and then combine them to form the complete solution.

Targeted Drills in Python

Coding Drills:

Variable assignment: This drill aims to demonstrate variable assignment in Python. Here we create two variables: total_time and max_time.
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total_time = 0 max_time = 0

List iteration: This drill illustrates how to loop through two lists concurrently using the zip function in Python. The two lists are colors and needed_time.

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colors = ['a', 'b', 'a', 'a', 'c']
needed_time = [1, 2, 3, 4, 5]
for color, time in zip(colors, needed_time):
    print(color, time)

Conditional statements: This drill demonstrates how to use if-else conditions in Python. In this case, it checks if the color of the current balloon is the same as the color of the previous balloon.
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prev_color = 'a' curr_color = 'b' if prev_color == curr_color: print("Same color") else: print("Different color")
Comparison operators: This drill illustrates the use of the min and max functions in Python. These functions are used to compare two numbers and return the smallest or largest number, respectively.
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time1 = 3 time2 = 5 print(min(time1, time2)) # returns 3 print(max(time1, time2)) # returns 5
Mathematical operations: This drill shows how to perform basic mathematical operations like addition in Python.
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total_time = 2 curr_time = 3 total_time += curr_time print(total_time) # prints 5

Problem-Specific Concepts:

The above drills cover both the general and problem-specific concepts required for this problem. Particularly, the use of min and max functions and if-else conditions are tailored to the problem’s requirement of choosing between two balloons to pop.

Integrating the Drills:

After understanding these drills, you can start to assemble them in the following way:

Initiate variables total_time and max_time.
Begin iteration over the lists colors and needed_time using the zip function.
Within the iteration, first check if the color of the current balloon is different from the previous one. If it is, update total_time by adding max_time to it and reset max_time to 0.
Then, increment total_time by the smaller one among the current needed_time and max_time, and update max_time as the larger one among the current needed_time and max_time.
Continue this until you’ve iterated over all the balloons. The total_time will be your minimum removal time, which is the solution to your problem.

By sequentially combining these drills, you can build the complete solution to the balloon popping problem.