Find Minimum in Rotated Sorted Array in DSA C - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Find the minimum element output

What is the output of the following C code that finds the minimum element in a rotated sorted array?

DSA C

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];
}

int main() {
    int arr[] = {4,5,6,7,0,1,2};
    int size = sizeof(arr) / sizeof(arr[0]);
    int min = findMin(arr, size);
    printf("%d\n", min);
    return 0;
}

A 0

B 1

C 2

D 4

💡 Hint

Think about where the smallest element lies in a rotated sorted array.

The code uses binary search to find the minimum element. Since the array is rotated, the minimum is the only element smaller than its previous. The algorithm narrows down the search space until left meets right at the minimum element, which is 0 here.

❓ Predict Output

intermediate

Output for single element array

What will be the output of the code when the input array has only one element?

DSA C

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];
}

int main() {
    int arr[] = {10};
    int size = sizeof(arr) / sizeof(arr[0]);
    int min = findMin(arr, size);
    printf("%d\n", min);
    return 0;
}

A 1

B 0

C 10

D Undefined behavior

💡 Hint

Consider the array size and how the loop behaves.

With only one element, the loop condition left < right is false initially, so the function returns the only element, 10.

🔧 Debug

advanced

Identify the error in this modified code

What error will the following code produce when run?

DSA C

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 - 1;
        }
    }
    return nums[left];
}

int main() {
    int arr[] = {};
    int size = sizeof(arr) / sizeof(arr[0]);
    int min = findMin(arr, size);
    printf("%d\n", min);
    return 0;
}

A Infinite loop

B Runtime error: accessing out of bounds

C Prints 1 correctly

D Compilation error

💡 Hint

Check how the pointers move and the loop condition.

The loop condition and pointer updates cause left to go beyond array bounds, leading to out-of-bounds access when returning nums[left].

🧠 Conceptual

advanced

Why binary search works on rotated sorted arrays

Why does the binary search approach work to find the minimum element in a rotated sorted array?

A Because the minimum element is always at the middle index

B Because the array elements are all unique

C Because the array is fully sorted and binary search always works on sorted arrays

D Because the array is partially sorted and we can discard half the array each time by comparing middle and right elements

💡 Hint

Think about how the rotation affects the sorted order.

The rotated array is partially sorted. By comparing the middle element with the rightmost element, we can decide which half contains the minimum and discard the other half, reducing the search space.

❓ Predict Output

expert

Output for array with duplicates

What is the output of the following code when the input array contains duplicates?

DSA C

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 if (nums[mid] < nums[right]) {
            right = mid;
        } else {
            right--;
        }
    }
    return nums[left];
}

int main() {
    int arr[] = {2,2,2,0,1,2};
    int size = sizeof(arr) / sizeof(arr[0]);
    int min = findMin(arr, size);
    printf("%d\n", min);
    return 0;
}

A 0

B 1

C 2

D Runtime error

💡 Hint

Duplicates require a slight modification in the binary search approach.

The code handles duplicates by decrementing right when nums[mid] equals nums[right], ensuring the minimum is found correctly. The minimum element is 0.

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