Find Minimum in Rotated Sorted Array in DSA C - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is a rotated sorted array?

A rotated sorted array is an array that was originally sorted in ascending order but then rotated (shifted) at some pivot point. For example, [4,5,6,7,0,1,2] is a rotated version of [0,1,2,4,5,6,7].

beginner

Why can't we just scan the entire rotated sorted array to find the minimum?

Scanning the entire array takes O(n) time. Using the special property of rotation, we can find the minimum in O(log n) time using binary search, which is faster for large arrays.

intermediate

In the binary search approach to find the minimum in a rotated sorted array, what does it mean if the middle element is greater than the rightmost element?

It means the minimum is in the right half of the array because the rotation point is to the right of mid.

beginner

What is the stopping condition for the binary search when finding the minimum in a rotated sorted array?

When the search space is reduced to one element, that element is the minimum.

intermediate

Show the C code snippet for the binary search to find the minimum in a rotated sorted array.

int findMin(int* nums, int numsSize) { int left = 0, right = numsSize - 1; while (left < right) { int mid = left + (right - left) / 2; if (nums[mid] > nums[right]) { left = mid + 1; } else { right = mid; } } return nums[left]; }

What is the time complexity of finding the minimum in a rotated sorted array using binary search?

A O(1)

B O(n)

C O(n log n)

D O(log n)

Binary search divides the search space in half each time, leading to O(log n) time complexity.

If nums[mid] is less than nums[right], where is the minimum located?

A At right + 1

B In the right half excluding mid

C In the left half including mid

D At mid + 1

If nums[mid] < nums[right], the minimum is at mid or to the left of mid.

What does it mean if nums[mid] > nums[right] in the rotated sorted array?

A Minimum is in the left half

B Minimum is in the right half

C Array is not rotated

D Minimum is at mid

If nums[mid] > nums[right], the rotation point and minimum are in the right half.

What is the initial search range for the binary search in the rotated sorted array?

A From 0 to numsSize - 1

B From 1 to numsSize

C From numsSize/2 to numsSize - 1

D From 0 to numsSize/2

We start searching from the first to the last index of the array.

What is the output of findMin for the array [3,4,5,1,2]?

A 1

B 2

C 3

D 5

1 is the smallest element and the minimum in the rotated sorted array.

Explain how binary search helps find the minimum element in a rotated sorted array.

Think about which half contains the smallest value based on mid and right values.

Write the steps to find the minimum in a rotated sorted array without scanning all elements.

Use binary search logic focusing on the rotation property.

Practice

1. What is the main goal when finding the minimum in a rotated sorted array like [4, 5, 6, 7, 0, 1, 2]?

easy

A. To find the smallest element which is the rotation point

B. To find the largest element in the array

C. To count how many times the array is rotated

D. To sort the array again

Solution

1. Step 1: Understand the rotated sorted array

   A rotated sorted array is originally sorted but then shifted. The smallest element is where the rotation happened.

2. Step 2: Identify the minimum element

   The minimum element is the smallest number and marks the rotation point in the array.

3. Final Answer:

   To find the smallest element which is the rotation point -> Option A

4. Quick Check:

   Minimum element = rotation point [OK]

Hint: Minimum is the rotation point in the array [OK]

Common Mistakes:

• Looking for the largest element instead
• Trying to count rotations instead of finding minimum
• Sorting the array again unnecessarily

2. Which of the following C code snippets correctly initializes variables to find the minimum in a rotated sorted array nums of size n?

easy

A. int left = 0, right = n; int mid;

B. int left = 1, right = n; int mid;

C. int left = 0, right = n - 1; int mid;

D. int left = -1, right = n - 1; int mid;

Solution

1. Step 1: Set pointers to array bounds

   In C, array indices start at 0, so left should be 0 and right should be n-1.

2. Step 2: Declare mid variable

   Mid is used to find the middle index between left and right.

3. Final Answer:

   int left = 0, right = n - 1; int mid; -> Option C

4. Quick Check:

   Array indices start at 0, so left=0 and right=n-1 [OK]

Hint: Array indices start at 0, so right = n - 1 [OK]

Common Mistakes:

• Setting left or right outside valid index range
• Using right = n which is out of bounds
• Starting left at 1 instead of 0

3. What will be the output of the following C code snippet that finds the minimum in a rotated sorted array?

int nums[] = {4, 5, 6, 7, 0, 1, 2};
int left = 0, right = 6, mid;
while (left < right) {
    mid = left + (right - left) / 2;
    if (nums[mid] > nums[right]) {
        left = mid + 1;
    } else {
        right = mid;
    }
}
printf("%d", nums[left]);

medium

A. 2

B. 0

C. 7

D. 4

Solution

1. Step 1: Understand the binary search logic

   The code compares middle element with right element to decide which half contains the minimum.

2. Step 2: Trace the loop to find minimum

   Loop narrows down to index 4 where value is 0, the smallest element.

3. Final Answer:

   0 -> Option B

4. Quick Check:

   Minimum element found = 0 [OK]

Hint: Compare mid with right to find minimum side [OK]

Common Mistakes:

• Printing nums[mid] instead of nums[left]
• Confusing greater than and less than in condition
• Not updating left or right correctly

4. Identify the error in the following C code that attempts to find the minimum in a rotated sorted array nums of size n:

int left = 0, right = n - 1, mid;
while (left <= right) {
    mid = left + (right - left) / 2;
    if (nums[mid] > nums[right]) {
        left = mid + 1;
    } else {
        right = mid - 1;
    }
}
return nums[left];

medium

A. The mid calculation is incorrect

B. The if condition should compare nums[mid] < nums[right]

C. The return statement should be nums[right]

D. The loop condition should be left < right, not left <= right

Solution

1. Step 1: Analyze loop condition

   Using left <= right can cause infinite loop or out-of-bound access because left can surpass right.

2. Step 2: Correct loop condition

   Changing to left < right ensures the loop stops when left meets right, correctly locating minimum.

3. Final Answer:

   The loop condition should be left < right, not left <= right -> Option D

4. Quick Check:

   Loop condition left < right avoids infinite loops [OK]

Hint: Use left < right to avoid infinite loops [OK]

Common Mistakes:

• Using <= in loop condition causing infinite loops
• Incorrectly adjusting right pointer by mid - 1
• Returning nums[left] without proper loop exit

5. Given a rotated sorted array nums that may contain duplicates, which modification to the binary search approach helps correctly find the minimum element? A) Always move left pointer when nums[mid] == nums[right] B) Increase left pointer by one when nums[mid] == nums[right] C) Decrease right pointer by one when nums[mid] == nums[right] D) Return nums[mid] immediately when nums[mid] == nums[right]

hard

A. Decrease right pointer by one when nums[mid] == nums[right]

B. Increase left pointer by one when nums[mid] == nums[right]

C. Always move left pointer when nums[mid] == nums[right]

D. Return nums[mid] immediately when nums[mid] == nums[right]

Solution

1. Step 1: Understand duplicates in rotated array

   Duplicates can cause ambiguity when nums[mid] equals nums[right], so normal binary search fails.

2. Step 2: Adjust right pointer carefully

   When nums[mid] == nums[right], decreasing right by one helps skip duplicate safely without missing minimum.

3. Final Answer:

   Decrease right pointer by one when nums[mid] == nums[right] -> Option A

4. Quick Check:

   Handle duplicates by reducing right pointer [OK]

Hint: When equal, reduce right pointer to skip duplicates [OK]

Common Mistakes:

• Moving left pointer instead causing missed minimum
• Returning mid value prematurely
• Not handling duplicates causing wrong answer