Find Minimum in Rotated Sorted Array in DSA C - Time & Space Complexity

Time Complexity: Find Minimum in Rotated Sorted Array

O(log n)

Understanding Time Complexity

We want to understand how fast we can find the smallest number in a rotated sorted array.

How does the time needed change as the array gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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];
}
    

This code finds the minimum element by using a binary search approach on a rotated sorted array.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop that halves the search range each time.
• How many times: The loop runs until the search range is reduced to one element, roughly log base 2 of the array size.

How Execution Grows With Input

Each step cuts the search area in half, so the number of steps grows slowly as the array grows.

Input Size (n)	Approx. Operations
10	About 4 steps
100	About 7 steps
1000	About 10 steps

Pattern observation: Doubling the input size adds only one extra step, showing slow growth.

Final Time Complexity

Time Complexity: O(log n)

This means the time to find the minimum grows slowly, increasing by one step each time the array size doubles.

Common Mistake

[X] Wrong: "The search takes linear time because the array is rotated and not fully sorted."

[OK] Correct: The array is still partially sorted, allowing binary search to work and find the minimum quickly without checking every element.

Interview Connect

Understanding this pattern helps you show how to use binary search in tricky situations, a skill valued in many coding challenges.

Self-Check

"What if the array was not rotated but fully sorted? How would the time complexity change?"

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