Find Minimum in Rotated Sorted Array in DSA C++ - Build from Scratch

Find Minimum in Rotated Sorted Array

📖 Scenario: You have a list of numbers that was originally sorted from smallest to largest. Then, someone rotated it at some point, moving some numbers from the front to the back. Your task is to find the smallest number in this rotated list.

🎯 Goal: Build a program that finds the minimum number in a rotated sorted array using a simple loop.

📋 What You'll Learn

Create a vector called nums with the exact values {4, 5, 6, 7, 0, 1, 2}

Create an integer variable called min_val and set it to the first element of nums

Use a for loop with variable num to go through each element in nums

Inside the loop, update min_val if num is smaller

Print the value of min_val

💡 Why This Matters

🌍 Real World

Finding the minimum in a rotated sorted array is useful in systems where data is rotated or shifted, such as circular buffers or rotated logs.

💼 Career

Understanding this problem helps in technical interviews and shows your ability to work with arrays and loops efficiently.

1

Create the rotated sorted array

Create a vector called nums with these exact values: {4, 5, 6, 7, 0, 1, 2}

DSA C++

#include 
#include 

int main() {

    // Create vector nums with values {4, 5, 6, 7, 0, 1, 2}
    // Your code here

    return 0;
}

Hint

Use std::vector<int> and initialize it with the given numbers.

2

Initialize the minimum value

Create an integer variable called min_val and set it to the first element of nums using nums[0]

DSA C++

#include 
#include 

int main() {
    std::vector nums = {4, 5, 6, 7, 0, 1, 2};

    // Create int min_val and set it to nums[0]
    // Your code here

    return 0;
}

Hint

Use int min_val = nums[0]; to start with the first number as minimum.

3

Find the minimum value using a loop

Use a for loop with variable num to go through each element in nums. Inside the loop, if num is smaller than min_val, update min_val to num

DSA C++

#include 
#include 

int main() {
    std::vector nums = {4, 5, 6, 7, 0, 1, 2};
    int min_val = nums[0];

    // Use for loop with num to check each element in nums
    // Update min_val if num is smaller
    // Your code here

    return 0;
}

Hint

Use a range-based for loop and an if statement to update min_val.

4

Print the minimum value

Print the value of min_val using std::cout

DSA C++

#include 
#include 

int main() {
    std::vector nums = {4, 5, 6, 7, 0, 1, 2};
    int min_val = nums[0];

    for (int num : nums) {
        if (num < min_val) {
            min_val = num;
        }
    }

    // Print min_val
    // Your code here

    return 0;
}

Hint

Use std::cout << min_val << std::endl; to print the minimum value.

Practice

1. What is the main idea behind finding the minimum element in a rotated sorted array?

easy

A. Use a hash map to store elements and find the minimum

B. Sort the array again and pick the first element

C. Traverse the array linearly and pick the smallest element

D. Use binary search to find the point where the order breaks

Solution

1. Step 1: Understand the rotated sorted array property

   A rotated sorted array is sorted but rotated at some pivot point, causing a break in order.

2. Step 2: Use binary search to find the break point

   Binary search helps find the pivot where the order breaks, which is the minimum element.

3. Final Answer:

   Use binary search to find the point where the order breaks -> Option D

4. Quick Check:

   Binary search finds minimum in rotated array [OK]

Hint: Look for the pivot where order breaks using binary search [OK]

Common Mistakes:

• Trying to sort the array again wastes time
• Linear search is slower than binary search
• Using extra space unnecessarily

2. Which of the following is the correct syntax to start a binary search loop in C++ for finding minimum in a rotated sorted array named nums?

easy

A. while (left <= right) { int mid = left + (right - left) / 2; }

B. while (left < right) { int mid = (left + right) / 2; }

C. for (int left = 0, right = nums.size() - 1; left <= right; )

D. for (int i = 0; i < nums.size(); i++)

Solution

1. Step 1: Understand binary search loop conditions

   For finding minimum in a rotated sorted array, the binary search loop typically uses while (left < right) to avoid infinite loops and correctly narrow down the search space.

2. Step 2: Check correct mid calculation and loop condition

   while (left < right) { int mid = (left + right) / 2; } is the common and correct approach for this problem.

3. Final Answer:

   while (left < right) { int mid = (left + right) / 2; } -> Option B

4. Quick Check:

   Correct binary search loop syntax for this problem [OK]

Hint: Use while with left < right and mid calculation [OK]

Common Mistakes:

• Using for loop instead of while for binary search
• Calculating mid as left + (right - left) / 2 with wrong loop condition
• Using wrong loop condition like left <= right causing infinite loop

3. What will be the output of the following C++ code snippet?

std::vector<int> nums = {4,5,6,7,0,1,2};
int left = 0, right = nums.size() - 1;
while (left < right) {
  int mid = left + (right - left) / 2;
  if (nums[mid] > nums[right]) left = mid + 1;
  else right = mid;
}
std::cout << nums[left] << std::endl;

medium

A. 0

B. 4

C. 7

D. 2

Solution

1. Step 1: Understand the binary search logic on rotated array

   The code compares nums[mid] with nums[right] to decide which half to keep.

2. Step 2: Trace the loop to find minimum

   Since nums[mid] > nums[right], left moves to mid+1 until left == right at index of minimum (0).

3. Final Answer:

   0 -> Option A

4. Quick Check:

   Minimum element found is 0 [OK]

Hint: Compare mid and right to narrow search for minimum [OK]

Common Mistakes:

• Confusing mid and right comparison direction
• Printing nums[right] instead of nums[left]
• Not updating left or right correctly

4. Identify the error in the following code snippet that tries to find the minimum in a rotated sorted array:

int findMin(std::vector<int>& nums) {
  int left = 0, right = nums.size() - 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];
}

medium

A. The else block should set right = mid, not right = mid - 1

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

C. The mid calculation causes overflow

D. The return statement should return nums[right], not nums[left]

Solution

1. Step 1: Analyze the binary search update rules

   When nums[mid] <= nums[right], the minimum is at mid or left side, so right should be set to mid.

2. Step 2: Identify incorrect update causing skipping minimum

   Setting right = mid - 1 skips mid, possibly missing the minimum element.

3. Final Answer:

   The else block should set right = mid, not right = mid - 1 -> Option A

4. Quick Check:

   Correct right update avoids missing minimum [OK]

Hint: Set right = mid to include mid in search [OK]

Common Mistakes:

• Using left <= right causes infinite loop
• Returning nums[left] is correct
• Mid calculation is safe here

5. Given a rotated sorted array nums that may contain duplicates, which modification is needed in the binary search approach to correctly find the minimum element?

hard

A. If nums[mid] == nums[right], return nums[mid] immediately

B. If nums[mid] == nums[right], increase left by one to skip duplicate

C. If nums[mid] == nums[right], reduce right by one to skip duplicate

D. Duplicates do not affect the binary search logic

Solution

1. Step 1: Understand impact of duplicates on binary search

   Duplicates can cause ambiguity when nums[mid] == nums[right], making it unclear which side to choose.

2. Step 2: Modify search by reducing right to skip duplicate

   Reducing right by one helps shrink the search space safely without missing the minimum.

3. Final Answer:

   If nums[mid] == nums[right], reduce right by one to skip duplicate -> Option C

4. Quick Check:

   Handle duplicates by shrinking right pointer [OK]

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

Common Mistakes:

• Returning immediately on equality misses minimum
• Increasing left instead of reducing right causes errors
• Ignoring duplicates leads to wrong answers