Find Minimum in Rotated Sorted Array in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the left pointer to the start of the array.

DSA C

int left = [1];

Drag options to blanks, or click blank then click option'

A0

B1

C-1

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Starting left at 1 skips the first element.

Using -1 causes invalid indexing.

Setting left to n goes beyond array bounds.

The left pointer should start at index 0 to cover the whole array.

2fill in blank

medium

Complete the code to calculate the middle index between left and right.

DSA C

int mid = left + ([1] - left) / 2;

Drag options to blanks, or click blank then click option'

Aright

Bmid

Cleft

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Using mid inside its own calculation causes errors.

Using n is unrelated to midpoint calculation.

Using left twice does not find midpoint.

We calculate mid as left plus half the distance to right to avoid overflow.

3fill in blank

hard

Fix the error in the condition to check if the middle element is greater than the right element.

DSA C

if (nums[mid] [1] nums[right]) {

Drag options to blanks, or click blank then click option'

A<

B<=

C==

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using <= reverses the logic.

Using == does not help find minimum.

Using < leads to wrong half selection.

If nums[mid] is greater than nums[right], the minimum is in the right half.

4fill in blank

hard

Fill both blanks to update pointers correctly when nums[mid] is greater than nums[right].

DSA C

if (nums[mid] > nums[right]) {
    left = [1];
} else {
    right = [2];
}

Drag options to blanks, or click blank then click option'

Amid + 1

Bmid - 1

Cmid

Dright - 1

Attempts:

3 left

💡 Hint

Common Mistakes

Setting left = mid skips the minimum if it is mid + 1.

Setting right = mid - 1 may skip the minimum.

Using right - 1 is incorrect here.

When nums[mid] > nums[right], minimum is right of mid, so left = mid + 1. Else, minimum is at mid or left, so right = mid.

5fill in blank

hard

Fill all three blanks to complete the function that finds the minimum in a rotated sorted array.

DSA C

int findMin(int* nums, int numsSize) {
    int left = [1], right = numsSize - 1;
    while (left < right) {
        int mid = left + (right - left) / 2;
        if (nums[mid] > nums[right]) {
            left = [2];
        } else {
            right = [3];
        }
    }
    return nums[left];
}

Drag options to blanks, or click blank then click option'

A0

Bmid + 1

Cmid

DnumsSize

Attempts:

3 left

💡 Hint

Common Mistakes

Starting left at numsSize causes out-of-bounds.

Setting left = mid skips minimum if it is mid + 1.

Setting right = mid - 1 may miss the minimum.

Initialize left to 0. If nums[mid] > nums[right], move left to mid + 1. Else, move right to mid. Finally, return nums[left] as minimum.

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