Find Minimum in Rotated Sorted Array II

The problem is to find the minimum element in a sorted and rotated array that may contain duplicates. A binary search approach can efficiently solve this problem by reducing the search space.

Here’s the Python code:

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class Solution:
    def findMin(self, nums: List[int]) -> int:
        left, right = 0, len(nums) - 1

        while left < right:
            mid = left + (right - left) // 2

            # If the middle element is greater than the last element, the minimum is in the right half
            if nums[mid] > nums[right]:
                left = mid + 1
            # If the middle element is equal to the last element, we can't determine which half the minimum is in,
            # so we reduce the search space by 1
            elif nums[mid] == nums[right]:
                right -= 1
            # If the middle element is less than or equal to the last element, the minimum is in the left half
            else:
                right = mid

        return nums[left]

Explanation:

  1. Initialize pointers: left and right point to the start and end of the array.
  2. Binary Search: In each iteration of the loop, compare the middle element mid with the last element right.
    • If mid is greater, the minimum must be to the right of mid.
    • If mid is less, the minimum must be at mid or to its left.
    • If mid is equal to right, we can’t determine which half the minimum is in, so we decrease the search space by 1.
  3. Return the Minimum: Once the loop ends, left will point to the minimum element.

This approach accounts for duplicates and minimizes the number of operations needed, thereby meeting the problem’s constraints.

10 Prerequisite LeetCode Problems

“Find Minimum in Rotated Sorted Array II” is a binary search problem, but with the complexity of handling duplicates in the array.

Here are 10 problems as a preparation:

  1. Find Minimum in Rotated Sorted Array (LeetCode #153): This problem is a simplified version of the problem at hand, where no duplicates are present in the array.

  2. Binary Search (LeetCode #704): This problem will give you the fundamental understanding of binary search algorithm, which is crucial for solving the original problem.

  3. Search in Rotated Sorted Array (LeetCode #33): This problem helps you understand how to search in a rotated sorted array, which is similar to finding the minimum in such array.

  4. Two Sum II - Input array is sorted (LeetCode #167): This problem helps you understand how to handle sorted array and use two-pointer method.

  5. First Bad Version (LeetCode #278): This problem is another good introduction to binary search algorithm.

  6. Search Insert Position (LeetCode #35): This problem provides practice for dealing with sorted arrays and binary search.

  7. Find Smallest Letter Greater Than Target (LeetCode #744): This problem can help you to understand how to find a specific condition in a sorted array using binary search.

  8. Guess Number Higher or Lower (LeetCode #374): This problem gives you practice with the concept of binary search.

  9. Single Element in a Sorted Array (LeetCode #540): This problem involves binary search and handling duplicates in a sorted array.

  10. Find Peak Element (LeetCode #162): This problem can provide you with practice on finding specific elements in sorted arrays, similar to finding the minimum.

These cover how to handle binary search, especially in the context of rotated sorted arrays, which is necessary for solving the “Find Minimum in Rotated Sorted Array II” problem.

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    def findMin(self, nums: List[int]) -> int:
        l, r = 0, len(nums)-1
        while l < r:
            mid = (l+r)//2
            if nums[mid] == nums[r]:
                if nums[mid] == nums[l]:
                    l = l+1 # or r = r-1
                else:
                    r = mid
            elif nums[mid] > nums[r]:
                l = mid+1
            else:
                r = mid
        return nums[l]

Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,4,4,5,6,7] might become:

[4,5,6,7,0,1,4] if it was rotated 4 times. [0,1,4,4,5,6,7] if it was rotated 7 times. Notice that rotating an array [a[0], a[1], a[2], …, a[n-1]] 1 time results in the array [a[n-1], a[0], a[1], a[2], …, a[n-2]].

Follow up: This problem is similar to Find Minimum in Rotated Sorted Array, but nums may contain duplicates. Would this affect the runtime complexity? How and why?

Problem Classification

The task is to find the smallest number in a list that was sorted and then possibly rotated. It’s about analyzing a list to find the smallest value, so it’s a List Analysis Problem.

This is an array/list problem in the domain of algorithms and data structures.

The ‘What’ components are:

  • Integer array as input
  • Array is sorted but rotated
  • Find minimum element in array
  • Array may contain duplicates
  • Require overall operation efficiency

Based on this, I would categorize this problem as:

  • Search problem, since we need to find a target value (the minimum)
  • More specifically, a minimum finding problem
  • Array problem, as it involves manipulating/searching an array structure
  • Can also be considered a variation of the binary search problem, since the input array is sorted

The core of the problem is finding the minimum value in a sorted, rotated array efficiently, made more challenging by the potential duplicates. It requires leveraging the sorted structure while handling rotations and duplicates.

I would classify this as a minimum finding search problem on a sorted rotated array with duplicates. It constrained binary search in an array.

Clarification Questions

Here are some clarification questions I would ask about this problem of finding the minimum in a rotated sorted array:

  • What are the time and space complexity requirements? Should we optimize for faster running time or lower memory usage?

  • Can we assume the input array will always have at least one element?

  • Can there be duplicate elements across the rotation boundary? For example, [3, 1, 1, 3].

  • Are there any constraints on the number of rotations the input array can have?

  • Can we assume there will always be a single minimum element? Or could the array be something like [1, 1, 1, 1]?

  • Can the array have both positive and negative numbers?

  • Do the duplicate elements have to be adjacent to each other, or can they be interspersed?

  • What should the output be if the input array is empty?

  • Should we return the index of the minimum or the element itself?

  • Can we mutate or rearrange the input array while processing?

  • Is the rotation amount or pivot element known/given beforehand?

Getting clarity on these aspects would allow me to make better assumptions and design decisions regarding efficiency, edge cases, mutation rules, and handling of duplicates.

Identifying Problem Isomorphism

An exact isomorphism does not exist for this problem of finding the minimum in a rotated sorted array. However, here is an approximate mapping to a related problem:

The problem Minimum in Rotated Sorted Array can be approximately mapped to Search in Rotated Sorted Array.

The key similarities are:

  • Both involve searching a sorted array that has been rotated
  • The rotation divides the array into two sorted halves
  • We can leverage the sorted property within each half

The main differences are:

  • This problem asks to find just the minimum element
  • Search in Rotated Sorted Array looks for a specific target element

So while not exact isomorphs, the core technique of handling rotations by dividing the array into sorted halves is very similar.

Of the two, I would consider Minimum in Rotated Sorted Array to be simpler since we only need to find one element, the minimum, rather than handling all possible target elements.

The approximate mapping highlights the rotational divide and conquer approach in common, even though the specifics of the search goal differ.

Problem Analysis and Key Insights

Here are some key insights gained from analyzing this problem of finding the minimum in a rotated sorted array:

  • The array being sorted (though rotated) means we can leverage the ordering and comparison relationships between elements to guide the search.

  • The rotation divides the array into two sorted halves. One of the halves contains the minimum element.

  • Duplicates add ambiguity as equal elements do not clearly distinguish the halves.

  • We likely want some variation of binary search that handles the rotation.

  • The minimum could be the first or last element if no rotation. Boundary cases need handling.

  • No specific time/space complexity given, so multiple approaches are possible depending on optimization goal.

  • The problem is self-contained - no dependencies or interactions outside the array.

In summary, the core insights are that the sorted rotation structure allows variants of binary search, but handling duplicates adds complexity. No specific efficiency requirements suggests multiple algorithm options are likely viable.

Problem Boundary

Here are some key points about the scope of this problem of finding the minimum in a rotated sorted array:

  • The input is an integer array of unspecified size, but reasonably bounded between 1-5000 elements.

  • Elements are integers ranging from -5000 to 5000.

  • The array is sorted but rotated 1 or more times. The rotation point is unknown.

  • Duplicates are allowed anywhere in the array, even spanning the rotation boundary.

  • Only a single result is needed - the minimum element value.

  • No other operations on the array are required.

  • There are no constraints specified on runtime or memory complexity.

  • The array is provided - no need to handle generating test cases or input validation.

  • No additional context or dependencies outside the array. Self-contained problem.

The scope is narrow - focused just on efficiently finding the minimum value in the provided rotated sorted array with duplicates. The problem is well-specified and self-contained.

Here are some ways we can establish boundaries for the problem of finding the minimum in a rotated sorted array:

Input Space

  • Integer array
  • Length from 1 to 5000
  • Values from -5000 to 5000
  • Sorted in ascending order
  • Rotated 1 or more times
  • Duplicates allowed
  • Rotation pivot not provided

Output Space

  • Integer minimum value
  • In range -5000 to 5000

Algorithm

  • Find minimum only
  • No other operations needed
  • No modification of input array

Performance

  • No specified time/space complexity bounds
  • But optimize as much as possible

Functionality

  • Handle all valid input arrays
  • Return minimum value, not its index

Incorrect Inputs

  • Empty array
  • Not sorted
  • Not rotated
  • Invalid out of bounds elements

Defining these clear boundaries on the I/O format, required functionality, performance criteria, and incorrect inputs provides a solid problem specification.

Distilling the Problem to Its Core Elements

This problem of finding the minimum in a rotated sorted array is fundamentally based on the concept of binary search and efficient array traversal.

In simplest terms, I would describe it as:

“Finding the smallest number in an array that is sorted in a circular way.”

The core problem is identifying the minimum value in an array with an unknown pivot point dividing it into two sorted halves. We can simplify it to:

“Find min in rotated sorted array.”

The key components are:

  • Sorted array
  • Unknown rotation point
  • Duplicates allowed
  • Find just the min value

The minimal operations we need are:

  • Search algorithm to traverse array
  • Logic to handle unknown rotation
  • Duplicate handling
  • Find and return min value

At its essence, this leverages binary search algorithms adapted for the rotational division. The core is efficiently traversing the sorted array despite complications of the rotation and duplicates.

Visual Model of the Problem

Here are some ways to visualize the problem statement for finding the minimum in a rotated sorted array:

  • Draw the array as a horizontal line of numbers and use arrows to indicate it is rotated. Label one end as start and the other as end to show unknown pivot.

  • Animate the rotation process on the visualized array to show how an originally sorted array is rotated.

  • Use ascending/descending heights to represent the sorted order within each half. Show how the minimum breaks the height pattern.

  • Draw duplicates as numbers stacked vertically. Illustrate how they create ambiguity.

  • Highlight the minimum value once found after searching the array.

  • Visualize binary search and modified search patterns on the array.

  • Contrast with normal sorted array to emphasize impact of rotation.

  • For large arrays, show entire array but highlight and zoom in on pertinent section.

  • Use color coding and annotations to enhance intuitiveness.

These types of visuals could help better understand how an unknown rotation divides the sorted array into two halves, complicates search, and requires handling duplicates. Visualization brings clarity.

Problem Restatement

Here’s how I would paraphrase the problem statement in my own words:

We are given an array of integers that is sorted in ascending order, but has been rotated by an unknown number of positions. This divides the array into two parts - each individually sorted, but the relationship between the parts is unknown.

Our goal is to find the minimum value in the rotated array. The array may also contain duplicate values.

We want to optimize for efficiency and solve this in as few operations or steps as possible. No specific limitations are provided for memory usage or runtime complexity.

The input array is reasonably sized, containing between 1 and 5000 elements. The values are integers ranging from -5000 to 5000.

We simply need to find and return the minimum integer value. There are no other operations required on the array.

Let me know if I’m missing or unclear on any important details based on this paraphrasing. Getting the essence of the problem aids in orienting our solution approach.

Abstract Representation of the Problem

Here is one way to formulate an abstract representation of this problem:

Let’s consider a function f operating on an abstract ordered data structure D containing unique elements (no duplicates).

D has the following properties:

  • The elements are sorted in ascending order based on value
  • A pivot point splits D into two contiguous parts
  • Each part is sorted internally
  • The relationship between the parts is unknown

We want to define f such that:

  • It takes D as input
  • It returns the element with the minimum value in D
  • It operates in as efficient a manner as possible

By abstracting away the specifics of the integer array and rotation, we can represent the core problem as:

  • Operating on an ordered data structure D with an unknown internal pivot
  • Finding the minimum element value efficiently
  • Optimizing the operation based on the structure of D

This framing focuses on the underlying ordered structure, unknown pivot, and efficiency goals without domain-specific details.

Terminology

Here are some key terms and concepts important for understanding this problem of finding the minimum in a rotated sorted array:

  • Sorted array - Array where elements are arranged in ascending/descending order. Allows binary search.

  • Rotation - Shifting elements uniformly to the left/right. Divides sorted array into two parts.

  • Pivot element - The element separating the two sorted halves. Unknown here.

  • Binary search - Efficient search algorithm on sorted arrays. Can adapt for rotation.

  • Duplicates - Identical elements. Obscure rotation point and binary search logic.

  • Time complexity - Runtime as function of input size. Should optimize operations.

  • Space complexity - Memory usage as function of input size. Optimization criteria.

  • Divide and conquer - Breaking problem into subproblems, key strategy for handling rotation.

The core goals are leveraging the sorted structure through variations of binary search, handling the rotation divider, and optimizing search efficiency despite duplicates - thus understanding these algorithmic concepts is key.

Problem Simplification and Explanation

Here’s one way I could explain this problem in simpler terms:

Let’s compare the rotated array to an alphabetically sorted book with its pages rotated (analogy). The book pages go A to Z, but someone rotated the book contents so page A may now be somewhere in the middle.

The key concepts are:

  • The array is like the book pages - sorted alphabetically but rotated.

  • The rotation divides the array/book into two sorted halves (A-M, N-Z).

  • We don’t know where the division between halves occurs - the “pivot page”.

  • Duplicates are like having multiple pages with the same letter. They make the division ambiguous.

  • We want to efficiently find the first page/minimum letter by searching intelligently through the two halves.

  • Rather than searching page-by-page linearly, we can leverage binary search by guessing middle pages and honing in on start.

So essentially we need to adapt binary search to handle the unknown rotation point and duplicate elements obscuring the division.

Let me know if this metaphor makes sense or if any part needs more clarification!

Constraints

Here are some characteristics of this problem that we can potentially leverage to optimize finding the minimum in the rotated sorted array:

  • The array size is reasonably small at 5000 elements. This allows simpler brute force approaches without needing heavy optimization.

  • Array values are integers within a moderate range of -5000 to 5000. We don’t have to worry about overflow or high precision.

  • Only the minimum value is needed, not its index or other statistics. This simplifies the output.

  • No modification of input array is required. We don’t have to preserve state.

  • Duplicates are allowed but their positioning is unspecified. We can make simplifying assumptions.

  • No other constraints are specified, so we have freedom to tradeoff time versus space if needed.

  • The rotated halves remain sorted, allowing exploitation of binary search.

Overall, the small discrete input space, lack of complex constraints, and freedom to optimize for time or space allow us to use simpler algorithms and data structures like binary search that take advantage of the sorted subarrays.

Here are some key insights gained from analyzing the constraints:

  • The small size limit allows brute force approaches without requiring complex optimizations.

  • Integer values simplify handling and comparisons without precision issues.

  • Only outputting the minimum value keeps storage and logic simple.

  • Immutability of the input array frees us to copy or rearrange as needed.

  • Undefined duplicate positioning provides flexibility in handling edge cases.

  • No specified efficiency constraints gives leeway to optimize for time or space as needed.

  • Maintained sorted order within subarrays enables exploiting binary search ideas.

  • Self-contained problem without external dependencies simplifies assumptions.

In summary, the constraints enable simple brute force solutions while also allowing optimized binary search approaches by permitting mutability and providing sorted substructures. The flexibility and lack of complex requirements allows room for creativity.

Case Analysis

Here are some additional test cases covering different scenarios:

Basic Case

Input: [4, 5, 6, 7, 0, 1, 2]

Output: 0

Reasoning: Odd number of elements, single rotation.

Duplicates

Input: [3,3,1,3,3]

Output: 1

Reasoning: Handles duplicate values obscuring rotation.

Nearly Sorted

Input: [1, 2, 3, 4, 0]

Output: 0

Reasoning: Minimum close to start, edge case pivot points.

Reversed

Input: [5, 4, 3, 2, 1]

Output: 1

Reasoning: Test descending sorted order.

Boundary Cases:

  • 1 element array
  • 2 element array
  • Entirely duplicates

Edge Cases:

  • Minimum at start/end
  • Multiple duplicates
  • Many rotations

Testing these cases validates correctness, identifies assumptions, and stresses edge conditions.

Here are some ideas for visualizing test cases for this problem of finding minimum in rotated sorted array:

Basic Case

  • Array shown in circular rotated fashion with minimum highlighted

Duplicates

  • Repeated elements stacked to illustrate duplication

Nearly Sorted

  • Array almost in order but slightly rotated

Reversed

  • Descending array values visualized

1 Element

  • Single element array

All Duplicates

  • Array of all identical values

Multiple Rotations

  • Array rotated multiple times

Min at End

  • Minimum highlighted at end of array

In general:

  • Ascending/descending slope lines to illustrate sorted subsequences
  • Arrows and animation to show rotation
  • Highlight found minimum

Visualizations help interpret test cases and edge conditions correctly. Animations can clarify rotations.

Here are some key insights gained from analyzing these test cases:

  • The basic case validates the core algorithm and rotation handling logic.

  • Duplicates reveal complications in identifying rotation divides.

  • Nearly sorted cases stress test minimum finding near start.

  • Reversed array ensures descending order is handled.

  • Boundary cases like 1 element simplify edge logic.

  • Edge cases stress test issues like duplicate heavy or multi-rotated arrays.

  • Visualizing cases provides more intuitive understanding of rotations.

  • Animations can clarify how ordering changes upon rotation.

  • A wide variety of tests is needed to thoroughly validate correctness.

Overall, these insights show that comprehensive test coverage is crucial for robust code - errors, implicit assumptions, and edge conditions are revealed through careful examination of a diverse input space.

Identification of Applicable Theoretical Concepts

Here are some mathematical and algorithmic concepts that could help optimize finding the minimum in a rotated sorted array:

  • Binary search - Can adapt classic binary search to handle unknown rotation.

  • Divide and conquer - Break into subproblems of searching each sorted half.

  • Recursion - Recursive binary search provides divide and conquer.

  • Amortized analysis - Helps analyze binary search cost with duplicates.

  • Randomization - Quickly find approximate minimum as pivot using random sampling.

  • Range minimum queries - Use specialized data structures for efficient range min lookups.

  • Tree data structures - Balanced trees can efficiently support searches and range queries.

  • Aggregate analysis - Apply probability theory to expect number of comparisons needed.

  • Comparator abstraction - Allow flexible comparisons to handle duplicates.

Applying algorithmic concepts like binary search and recursion as well as data structures like trees and analysis with probability can provide insights into efficiently solving this problem.

Simple Explanation

Here’s how I would explain this problem of finding the minimum in a rotated sorted array in simple terms:

Think of a giant Ferris wheel with seats numbered sequentially from 1 to 100 (metaphor). The seats start numbered 1 to 100 from the bottom, going clockwise.

Now the Ferris wheel rotated forward 10 seats. So seat 1 is no longer at the bottom, but somewhere in the middle.

But all the seats are still in order - just this orderly section starting at 1 has been rotated forward some unknown number of seats.

Some seats also have duplicated numbers. So there may be more than 1 seat numbered 50 for example.

How could you efficiently find the lowest numbered seat on the Ferris wheel, given this unknown rotation of the sorted seat numbers?

This is like an array where the lowest number could be anywhere based on an unknown rotation pivot point. We need to leverage the sorted seats on each half to search quickly.

Hopefully this helps relate the problem to a real-world rotated sorting analogy! Let me know if any part needs more clarification.

Problem Breakdown and Solution Methodology

Here is a detailed step-by-step approach I would take to solve this problem of finding the minimum in a rotated sorted array:

  1. Compare the first and last elements of the array to determine if it is already sorted or rotated.

  2. If sorted, return the first element as minimum. Otherwise, use binary search to handle rotation.

  3. For binary search, start by finding the midpoint index of the array.

  4. Compare the element at midpoint with the first element. This gives insight on which half the minimum lies in.

  5. If midpoint element > first element, minimum is in the first half. Recursively search the first half.

  6. Else, minimum is in the second half. Recursively search the second half.

  7. When the subarray size is 2, compare the two elements to find minimum.

  8. Return the minimum element found after recursive calls end.

Increasing array size expands the search space but binary search limits this to O(log n). Duplicates could affect comparisons.

Let’s walk through array [6, 7, 8, 9, 1, 2, 3]

  • Midpoint is 8, greater than first element 6
  • Thus, minimum is in first half [6, 7]
  • Compare the two, minimum is 6

This leverages binary search, adapting it to pivot unknown sorted halves and hone in on the minimum efficiently.

Inference of Problem-Solving Approach from the Problem Statement

Here are the key terms and how they guide the solution approach:

  • Sorted array - Means we can adapt binary search to leverage ordering.

  • Rotation - Divides array into sorted halves. We can binary search within each half.

  • Unknown pivot - Requires comparing against array ends to determine which half.

  • Duplicates - Adds ambiguity for half determination. Must handle carefully.

  • Minimum - We only care about finding the single minimum value. Simplifies goal.

  • Efficiency - Need fast search time. Leads us toward binary search for O(log n).

The core ideas of a sorted array with unknown rotation pivot point the application of modified binary search techniques. The efficiency goal and simplified output of just the minimum value allow focusing the binary search logic. The complications of duplicates and unknown pivot require special handling.

Here are some ways to visualize the properties and concepts of this problem using diagrams:

Sorted Rotation

  • Array shown in circular fashion with ordering
  • Arrows indicate unknown rotation point

Binary Search

  • Annotate binary search steps on array
  • Circle comparisons to determine which half

Duplicate Handling

  • Repeated elements shown stacked vertically
  • Indicate ambiguity for search

Recursion Tree

  • Show recursive calls bifurcating search ranges
  • Leading to minimum element at leaf

Time Complexity

  • Plot number of operations vs input size
  • Show O(log n) curve for binary search

These types of visual models help reinforce concepts like the division into sorted halves, narrowing search range by recursion, duplicate ambiguity, and logarithmic time complexity.

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

Here is a stepwise refinement for finding the minimum in a rotated sorted array:

  1. High-level approach: Use modified binary search handling unknown rotation

  2. Break into steps:

  • Check if array is already sorted
  • Use binary search approach comparing against array ends
  • Recursively search left or right half based on comparisons
  • Handle duplicates carefully when comparing
  • Return minimum once found
  1. Independent sub-problems:
  • Determining which half to search
  • Handling duplicate comparisons
  • Base case check when two elements left
  1. Repeatable patterns:
  • Recursive binary search on left/right halves
  • Narrowing search range each recursion
  • Comparing against array ends to determine half

The key is adapting binary search to handle the unknown rotation pivot, leveraging the divided sorted halves. Comparisons, duplicates, and base cases can be handled independently.

Solution Approach and Analysis

Here is a detailed step-by-step approach to finding the minimum in a rotated sorted array:

  1. Compare the first and last elements. If last < first, array is rotated.

  2. If not rotated, return the first element as minimum. Else use binary search.

  3. Find the midpoint index and compare midpoint to first element.

  4. If midpoint > first, search the first half recursively, adjusting the end index.

  5. Else, search the second half recursively, adjusting the start index.

  6. Base case is when 2 elements left. Compare and return min.

  7. Return the minimum element from recursive calls.

Increasing array size increases recursion depth and duplicate comparisons become harder. Can optimize with better pivot finding.

Let’s walk through array [6, 7, 8, 9, 1, 2, 3]:

  • Mid is 8 > first element 6, so search first half [6, 7]
  • This gives 6, which is the minimum.

This leverages modified binary search on the rotated sorted halves to efficiently hone in on the minimum element.

Identify Invariant

The key invariant in this problem of finding the minimum in a rotated sorted array is:

At each step in the binary search, the current search range is a sorted contiguous subarray.

This means that after each recursive call splitting the range in half, the working subarray under consideration remains sorted.

We can maintain this invariant because:

  • The overall input array is sorted, even after unknown rotation.

  • The unknown rotation point splits the array into two sorted halves.

  • When we bisect the range, we always bisect it at an element within a sorted half.

  • So the left and right subranges split on each recursive call remain sorted subarrays.

This invariant is crucial to allow the binary search logic to work correctly - it relies on the structure of sorted subarrays to allow comparing against the ends and hone in on the minimum.

Identify Loop Invariant

The loop invariant in the binary search solution for finding the minimum in a rotated sorted array is:

At the start of each loop iteration, the subarray defined by start and end indices is guaranteed to be sorted.

This means that before evaluating the loop condition and making recursive calls, the current range being searched from start to end must be a contiguous sorted subarray.

To maintain this invariant:

  • The initial array is globally sorted, allowing the initial range to be sorted.

  • When bifurcating the range, we select a midpoint inside a sorted half to split on, maintaining sorted subarrays.

  • The start and end points delimit a subarray sorted in ascending order (or descending if reversed).

This invariant is crucial because the binary search logic depends on only searching sorted ranges. By preserving sorted subarrays within the changing start to end range each iteration, we can correctly hone in on the minimum.

Yes, in this problem of finding the minimum in a rotated sorted array, the invariant and loop invariant are essentially the same concept:

The invariant is the key property that must hold true for the algorithm to function correctly.

Specifically, the invariant is that the current search range is always a sorted subarray.

The loop invariant is the invariant property that must be maintained before and after each iteration of the loop in the binary search solution.

In this case, the loop invariant is identical to the overall invariant - that the start to end range delimits a sorted subarray.

This must be true before the loop condition, so the recursive calls work properly. And it is restored after each iteration by the bifurcation logic.

So the invariant and loop invariant are equivalent - both referring to the core property that the search range contains a sorted subarray at every step.

This equivalence occurs because the overarching invariant aligns directly with the behavior of the iterative loop logic itself. They refer to the same key property about the state of the algorithm.

Here is how we can prove correctness using a loop invariant for the binary search solution to finding the minimum in a rotated sorted array:

Initialization: Prior to the first iteration of the loop, we have start set to 0 and end set to the last index n-1. This delimits the entire input array, which is globally sorted due to rotation maintaining ordering. Therefore, the initial loop invariant that the subarray from start to end is sorted holds prior to the first loop iteration.

Maintenance: To see that each iteration maintains the loop invariant, the key lines are:

mid = start + (end - start) // 2

if nums[mid] > nums[start]: start = mid + 1 else: end = mid

By only setting start to mid+1 when the condition holds, and end to mid otherwise, the subarray from start to end remains sorted after updating the indices.

Termination: At termination, start and end converge on the minimum index such that the subarray bounds a single element. This element must be the minimum by the maintenance of the sorted subarray invariant.

Therefore, by initializing a sorted subarray, carefully updating the bounds each iteration to maintain sorted order, and terminating on the minimum, the invariant proves the approach finds the minimum under rotation.

Thought Process

Here is how I would approach solving this problem of finding the minimum in a rotated sorted array:

The cues from the problem:

  • Sorted array is provided
  • Unknown rotation point
  • Need to find just the minimum value
  • Want optimal efficiency

These cues suggest:

  • Leverage the sorted structure through binary search
  • Handle unknown rotation pivot point
  • Recursively divide search space in half
  • Compare against ends to determine which half

Steps:

  1. Check if array is rotated by comparing first and last element
  2. If not rotated, return first element
  3. Find midpoint index
  4. Compare midpoint to first element
  5. Recursively search left or right half accordingly
  6. Repeat until find minimum

Python code:

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def find_min(nums):

  if nums[0] < nums[-1]: # Not rotated
    return nums[0] 
  
  start = 0
  end = len(nums) - 1
  
  while start < end:
    mid = start + (end - start) // 2
    
    if nums[mid] > nums[0]:
      start = mid + 1 
    else:
      end = mid
  
  return nums[start]

This implements a binary search approach, using comparisons against array ends to handle the unknown rotation point and hone in on the minimum efficiently.

PENDING TEST

Establishing Preconditions and Postconditions

Here are the details for the findMin() method:

  1. Parameters:
  • nums: List[int] - The rotated sorted array of integers
  1. Preconditions:
  • nums is not empty
  • nums contains only integers
  • nums is rotated sorted
  1. Functionality:
  • Finds the minimum value in the rotated sorted array
  • Returns the integer minimum element
  1. Postconditions:
  • The minimum value is identified
  • The minimum integer is returned
  1. Error handling:
  • Empty nums raises ValueError
  • Invalid value in nums raises TypeError
  • If not rotated sorted, raises RuntimeError
  • On other errors, raises RuntimeError

So in summary, this takes the rotated sorted array, leverages the structure to efficiently find the minimum element, and returns it. Input validation is performed and edge cases handled through exceptions.

Problem Decomposition

Here’s how I would break down this problem of finding the minimum in a rotated sorted array:

  1. Given a sorted array that has an unknown rotation point, efficiently find the minimum value in the array.

  2. Major parts are:

    • Determining rotation
    • Adapted binary search
    • Duplicates handling
    • Base case check

  3. Breakdown:

    • Check if rotated
    • Compare against ends
    • Recursive search left/right half
    • Careful duplicate value comparisons
    • Terminate when subarray size is 2
    • Return minimum value

  4. Generalized tasks:

    • Rotation identification
    • Binary search modifications
    • Duplicate handling
    • Base case identification

  5. Abstractions look good.

  6. Method names are clear.

  7. Order:

    • Check for rotation
    • Modified binary search
    • Handle duplicates
    • Base case check
    • Return minimum

The key is identifying rotation, adapting binary search accordingly, handling edge cases like duplicates carefully, and terminating correctly when minimum is found. The breakdown structures the approach well.

From Brute Force to Optimal Solution

Here is a brute force approach and optimizations for finding the minimum in a rotated sorted array:

Brute force:

  • Loop through the array, comparing each element to find the minimum

This requires O(n) time complexity.

Optimizations:

  1. Check if already sorted:
  • Avoid full scan if no rotation. O(1) time.
  1. Binary search:
  • Ignore one half based on comparisons against ends. O(log n) time.
  1. Recursion:
  • Refine search space recursively. O(log n) time.
  1. Duplicate handling:
  • Careful comparisons dealing with duplicates. O(log n) time.
  1. Base case:
  • Terminate early when subarray size is 2. O(1) time.

These optimizations improve the overall time complexity from O(n) to O(log n) by leveraging the sorted rotation structure to quickly hone in on the minimum recursively.

No additional space is used, so space remains O(1).

In summary, comparing array ends, recursive binary search, duplicate handling, and base case shortcuts together optimize the brute force approach to find the minimum in logarithmic time.

Code Explanation and Design Decisions

Here is an analysis of the logic behind this minimum finding code:

  1. The parameters start and end represent the current search bounds, which is the problem state.

  2. The while loop iterates to narrow the bounds to hone in on minimum. Each iteration separates the search space.

  3. The if condition checks the midpoint against start to determine which half to search next. This handles the unknown rotation.

  4. Start and end indices are updated to maintain the sorted subarray loop invariant.

  5. The invariant of sorted subarray within bounds enables correct comparisons.

  6. The final start value points to the minimum element, satisfying the problem goal.

Overall, the loop drives the recursive binary search, progressively dividing the space based on comparisons against the start. The intent is to leverage the sorted invariant to optimally narrow down to the minimum element.

Coding Constructs

Here is an analysis of the high-level logic of this minimum finding code:

  1. It uses a binary search approach, optimally narrowing down search range.

  2. I would explain the code splits the data in half each round to quickly find the smallest value.

  3. The core logic includes loops, recursion, conditionals, variables and methods.

  4. The algorithm recursively divides the data in two based on comparisons, eliminating portions until the minimum is found.

  5. The key steps are determining rotation, recursively searching halves, handling duplicates, and terminating correctly. This efficiently hones in on the minimum.

  6. The overall strategy is a modified binary search using recursion and comparisons to optimize searching the rotated sorted data structure.

This analysis extracts the semantics around efficiently leveraging the structure of the data to optimize search using binary divide and conquer approaches. The focus is on the high-level patterns rather than language-specific details.

Language Agnostic Coding Drills

  1. Working with Lists or Arrays: Lists or Arrays are fundamental data structures in nearly all programming languages. Understanding how to access (read or write) elements in an array, finding the length of the array, and understanding zero-based indexing are important skills.

    Drill: Create an array of integers. Write a function to return the length of the array. Write another function to return the nth element of the array.

  2. Binary Search: The code employs a variant of the binary search algorithm, which is a divide-and-conquer strategy used to find elements in a sorted list.

    Drill: Write a function that takes a sorted list of integers and a target integer, and uses binary search to return the index of the target in the list. If the target is not in the list, return -1.

  3. Control Flow: This code makes extensive use of conditional (if/else) statements. Understanding how to write and use these constructs is essential for controlling the flow of execution in a program.

    Drill: Write a function that takes an integer as input and returns a string. If the integer is divisible by 3, it should return “Fizz”. If the integer is divisible by 5, it should return “Buzz”. If it’s divisible by both, it should return “FizzBuzz”. Otherwise, it should return the string representation of the number itself.

  4. Iteration: Understanding how to write and control loops is a fundamental programming skill. This particular code uses a while loop to repeatedly perform an action until a certain condition is met.

    Drill: Write a function that takes a list of integers and a target integer, and uses a while loop to find and return the index of the target in the list. If the target is not in the list, it should return -1.

  5. Integer Division and Modulus Operations: The code uses integer division (//) and modulus (%) operations, which are common in many languages. Integer division rounds the result down to the nearest integer, and modulus gives the remainder when one number is divided by another.

    Drill: Write a function that takes two integers and returns their quotient and remainder as a pair (i.e., a two-element list).

  6. Comparison Operators: The code uses comparison operators (==, >, <), which are used to compare two values and return a Boolean result.

    Drill: Write a function that takes two integers and returns whether the first is greater than, less than, or equal to the second.

By practicing these drills, one can build the fundamental skills necessary to understand and implement the provided solution to the problem of finding the minimum in a rotated sorted array. The solution uses a binary search approach to efficiently find the minimum element, carefully handling the case where the array has been “rotated” such that the elements are not entirely in sorted order.

Targeted Drills in Python

  1. Use of Lists or Arrays: In Python, lists are used to store multiple items in a single variable.

    Drill: Create a list of integers. Write a function to print the list, add a new element at the end of the list and another function to remove an element from the list based on the index provided.

    1
2
3
4
5
6
7
8

    def manipulate_list(nums):
    print(nums)  # print the list
    nums.append(7)  # add 7 at the end
    print(nums)  # print the list after appending
    nums.pop(1)  # remove element at index 1
    print(nums)  # print the list after removing

manipulate_list([1, 2, 3, 4, 5, 6])

  2. Control Flow: The code has if-elif-else statements. The control flow of a Python program is regulated by conditional statements, loops, and function calls.

    Drill: Write a program that accepts an integer and checks whether it’s positive, negative or zero.

     1
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10
11

    def check_number(num):
    if num > 0:
        print("Positive")
    elif num < 0:
        print("Negative")
    else:
        print("Zero")

check_number(5)
check_number(-3)
check_number(0)

  3. Binary Search: Binary Search is a searching algorithm for finding an element’s position in a sorted array. In this approach, the element is always searched in the middle of a portion of the array.

    Drill: Implement a simple binary search. Given a sorted list and a target, return the index of the target in the list. If the target is not present, return -1.

     1
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10
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12
13

    def binary_search(nums, target):
    low, high = 0, len(nums) - 1
    while low <= high:
        mid = (low + high) // 2
        if nums[mid] < target:
            low = mid + 1
        elif nums[mid] > target:
            high = mid - 1
        else:
            return mid
    return -1

print(binary_search([1, 2, 3, 4, 5, 6], 4))

  4. While Loop: The Python while loop runs as long as the condition remains true.

    Drill: Write a function that prints the first n numbers using a while loop.

    1
2
3
4
5
6
7

    def print_numbers(n):
    i = 1
    while i <= n:
        print(i)
        i += 1

print_numbers(5)

  5. Integer Division and Modulo Operations:

    Drill: Write a function that accepts two integers and returns their integer division and modulo results.

    1
2
3
4
5

    def compute_div_mod(a, b):
    print("Integer Division:", a // b)
    print("Modulo:", a % b)

compute_div_mod(10, 3)

These drills will help in understanding the different aspects of Python programming that are used in the given code.

Here are some Python coding drills for each of those key concepts:

  1. Arrays:
1
2
3
4
5

def array_length(arr):
  return len(arr)

def get_element(arr, n):
  return arr[n]
  1. Binary Search:
1
2
3

def binary_search(nums, target):
  # Binary search implementation
  return index
  1. Control Flow:
1
2
3
4
5
6
7
8
9

def fizzbuzz(num):
  if num % 3 == 0 and num % 5 == 0:
    return "FizzBuzz"
  elif num % 3 == 0:  
    return "Fizz"
  elif num % 5 == 0:
    return "Buzz"
  else:
    return str(num)
  1. Iteration:
1
2
3
4
5

def find_num(nums, target):
   for i in range(len(nums)):
     if nums[i] == target:
        return i
   return -1
  1. Integer Operations:
1
2

def quotient_remainder(a, b):
  return [a // b, a % b]
  1. Comparisons:
1
2
3
4
5
6
7

def compare(a, b):
  if a > b:
    return 1
  elif a < b:
    return -1
  else:
    return 0