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, it was rotated at some point, so the smallest number is not at the start anymore.For example, the list [4, 5, 6, 7, 0, 1, 2] was rotated. 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 an integer array called nums with the exact values [4, 5, 6, 7, 0, 1, 2].

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

Use a for loop with an integer variable i to go through each element in nums.

Inside the loop, update minValue if the current element is smaller.

Print the value of minValue.

💡 Why This Matters

🌍 Real World

Rotated sorted arrays appear in systems where data is shifted or rotated, such as circular buffers or rotated logs.

💼 Career

Understanding how to find minimum values in rotated arrays helps in optimizing search algorithms and solving real-world problems in software development.

1

Create the rotated sorted array

Create an integer array called nums with these exact values: 4, 5, 6, 7, 0, 1, 2.

DSA C#

// Create the array nums with the values 4, 5, 6, 7, 0, 1, 2
// Your code here

Hint

Use int[] nums = { ... }; to create the array with the exact numbers.

2

Set the initial minimum value

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

DSA C#

int[] nums = {4, 5, 6, 7, 0, 1, 2};
// Create minValue and set it to nums[0]
// Your code here

Hint

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

3

Find the minimum value using a loop

Use a for loop with an integer variable i from 0 to nums.Length. Inside the loop, if nums[i] is less than minValue, update minValue to nums[i].

DSA C#

int[] nums = {4, 5, 6, 7, 0, 1, 2};
int minValue = nums[0];
// Use a for loop to check each element and update minValue
// Your code here

Hint

Use a for loop and inside it use an if statement to compare and update minValue.

4

Print the minimum value

Write Console.WriteLine(minValue); to print the minimum value found.

DSA C#

int[] nums = {4, 5, 6, 7, 0, 1, 2};
int minValue = nums[0];
for (int i = 0; i < nums.Length; i++)
{
    if (nums[i] < minValue)
    {
        minValue = nums[i];
    }
}
// Print the minValue
// Your code here

Hint

Use Console.WriteLine(minValue); to show the smallest number.

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 sort the array again

D. To count how many times the array is rotated

Solution

1. Step 1: Understand the rotated sorted array

   A rotated sorted array is a sorted array shifted at some pivot. The smallest element is the rotation point.

2. Step 2: Identify the minimum element

   The minimum element is the smallest number where the rotation happened, which is the goal to find.

3. Final Answer:

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

4. Quick Check:

   Minimum element = rotation point [OK]

Hint: Look for the smallest number where order breaks [OK]

Common Mistakes:

• Looking for the largest element instead
• Trying to sort the array again
• Counting rotations instead of finding minimum

2. Which of the following C# code snippets correctly initializes an array named nums with values [3, 4, 5, 1, 2]?

easy

A. int nums[] = new int[5] {3, 4, 5, 1, 2};

B. int nums = {3, 4, 5, 1, 2};

C. int[] nums = new int[4] {3, 4, 5, 1, 2};

D. int[] nums = new int[] {3, 4, 5, 1, 2};

Solution

1. Step 1: Check array declaration syntax in C#

   Correct syntax requires specifying type and using new int[] with initializer.

2. Step 2: Validate each option

   int[] nums = new int[] {3, 4, 5, 1, 2}; uses correct syntax. int nums = {3, 4, 5, 1, 2}; misses array type. int[] nums = new int[4] {3, 4, 5, 1, 2}; has size mismatch (4 vs 5 values). int nums[] = new int[5] {3, 4, 5, 1, 2}; mixes syntax incorrectly.

3. Final Answer:

   int[] nums = new int[] {3, 4, 5, 1, 2}; -> Option D

4. Quick Check:

   Correct array initialization syntax [OK]

Hint: Use 'new int[]' when initializing arrays with values [OK]

Common Mistakes:

• Omitting 'new int[]' keyword
• Declaring array without type
• Using wrong brackets or missing semicolon

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

int[] nums = {4, 5, 6, 7, 0, 1, 2};
int left = 0, right = nums.Length - 1;
while (left < right) {
  int mid = left + (right - left) / 2;
  if (nums[mid] > nums[right]) {
    left = mid + 1;
  } else {
    right = mid;
  }
}
Console.WriteLine(nums[left]);

medium

A. 4

B. 7

C. 0

D. 2

Solution

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

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

2. Step 2: Trace the loop to find minimum index

   Loop narrows down to index of smallest element, which is 0 at index 4.

3. Final Answer:

   0 -> Option C

4. Quick Check:

   Minimum element found = 0 [OK]

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

Common Mistakes:

• Returning nums[mid] instead of nums[left]
• Confusing comparison direction
• Not updating pointers correctly

4. Identify the error in this C# code snippet intended to find the minimum in a rotated sorted array:

int[] nums = {3, 4, 5, 1, 2};
int left = 0, right = nums.Length - 1;
while (left <= right) {
  int mid = left + (right - left) / 2;
  if (nums[mid] > nums[right]) {
    left = mid + 1;
  } else {
    right = mid - 1;
  }
}
Console.WriteLine(nums[left]);

medium

A. The mid calculation is incorrect

B. The loop condition should be 'left < right' instead of 'left <= right'

C. The array initialization is wrong

D. The print statement should use 'right' instead of 'left'

Solution

1. Step 1: Analyze loop condition impact

   Using 'left <= right' can cause infinite loop or index out of range because pointers cross.

2. Step 2: Correct loop condition for binary search

   Changing to 'left < right' ensures loop stops when pointers meet at minimum.

3. Final Answer:

   The loop condition should be 'left < right' instead of 'left <= right' -> Option B

4. Quick Check:

   Proper loop condition avoids infinite loops [OK]

Hint: Use 'left < right' for binary search loops [OK]

Common Mistakes:

• Using <= causes pointer crossing errors
• Changing mid calculation unnecessarily
• Printing wrong index after loop

5. Given a rotated sorted array nums = [10, 12, 15, 1, 3, 6, 8], which code snippet correctly finds the minimum element using binary search?

hard

A. int left = 0, right = nums.Length - 1; while (left < right) { int mid = left + (right - left) / 2; if (nums[mid] > nums[right]) left = mid + 1; else right = mid; } Console.WriteLine(nums[left]);

B. int left = 0, right = nums.Length - 1; while (left <= right) { int mid = (left + right) / 2; if (nums[mid] < nums[right]) right = mid - 1; else left = mid + 1; } Console.WriteLine(nums[left]);

C. int min = nums[0]; for (int i = 1; i < nums.Length; i++) { if (nums[i] < min) min = nums[i]; } Console.WriteLine(min); // Binary search used here

D. int left = 0, right = nums.Length - 1; while (left < right) { int mid = left + (right - left) / 2; if (nums[mid] < nums[left]) right = mid; else left = mid + 1; } Console.WriteLine(nums[right]);

Solution

1. Step 1: Review binary search logic for minimum in rotated array

   The correct approach compares mid with right and adjusts pointers accordingly.

2. Step 2: Evaluate each option's correctness

   int left = 0, right = nums.Length - 1; while (left < right) { int mid = left + (right - left) / 2; if (nums[mid] > nums[right]) left = mid + 1; else right = mid; } Console.WriteLine(nums[left]); correctly implements binary search. int left = 0, right = nums.Length - 1; while (left <= right) { int mid = (left + right) / 2; if (nums[mid] < nums[right]) right = mid - 1; else left = mid + 1; } Console.WriteLine(nums[left]); uses wrong loop condition and pointer updates. int min = nums[0]; for (int i = 1; i < nums.Length; i++) { if (nums[i] < min) min = nums[i]; } Console.WriteLine(min); // Binary search used here uses linear search, not binary. int left = 0, right = nums.Length - 1; while (left < right) { int mid = left + (right - left) / 2; if (nums[mid] < nums[left]) right = mid; else left = mid + 1; } Console.WriteLine(nums[right]); compares mid with left incorrectly.

3. Final Answer:

   Option A code snippet -> Option A

4. Quick Check:

   Binary search with mid vs right comparison [OK]

Hint: Compare mid with right, move pointers to find minimum [OK]

Common Mistakes:

• Using linear search instead of binary
• Wrong loop conditions causing infinite loops
• Comparing mid with left instead of right