Find Minimum in Rotated Sorted Array in DSA Javascript - 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 by moving some numbers from the front to the back. Your job is to find the smallest number in this rotated list.Imagine a circular queue of numbers where the smallest number is the point where the rotation happened.

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

📋 What You'll Learn

Create an array called rotatedArray with the exact values [4, 5, 6, 7, 0, 1, 2].

Create a variable called minValue and set it to the first element of rotatedArray.

Use a for loop with variable i to go through each element in rotatedArray.

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, like circular buffers or time-based logs. Finding the minimum quickly helps in searching and sorting tasks.

💼 Career

Understanding how to find minimum values in rotated arrays is useful for software engineers working on search algorithms, data processing, and optimization problems.

1

Create the rotated sorted array

Create an array called rotatedArray with these exact values: [4, 5, 6, 7, 0, 1, 2].

DSA Javascript

// Your code here

Hint

Use const rotatedArray = [...] to create the array with the exact numbers.

2

Set up the minimum value variable

Create a variable called minValue and set it to the first element of rotatedArray.

DSA Javascript

const rotatedArray = [4, 5, 6, 7, 0, 1, 2];
// Your code here

Hint

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

3

Find the minimum value using a loop

Use a for loop with variable i to go through each element in rotatedArray. Inside the loop, update minValue if the current element is smaller than minValue.

DSA Javascript

const rotatedArray = [4, 5, 6, 7, 0, 1, 2];
let minValue = rotatedArray[0];
// Your code here

Hint

Use a for loop to check each number. If a number is smaller than minValue, update minValue.

4

Print the minimum value

Write console.log(minValue) to print the smallest number found in rotatedArray.

DSA Javascript

const rotatedArray = [4, 5, 6, 7, 0, 1, 2];
let minValue = rotatedArray[0];
for (let i = 0; i < rotatedArray.length; i++) {
    if (rotatedArray[i] < minValue) {
        minValue = rotatedArray[i];
    }
}
// Your code here

Hint

Use console.log(minValue); to show the smallest number.

Practice

1. What is the main goal when finding the minimum in a rotated sorted array?

easy

A. To find the largest element in the array

B. To count the number of rotations

C. To find the smallest element after the rotation point

D. To sort the array again

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 goal

   The goal is to find the smallest element, which is the minimum after rotation.

3. Final Answer:

   To find the smallest element after the rotation point -> Option C

4. Quick Check:

   Minimum element = smallest after rotation [OK]

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

Common Mistakes:

• Confusing minimum with maximum
• Trying to count rotations instead of finding minimum
• Assuming array is fully sorted without rotation

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

easy

A. while(left < right) { ... }

B. for(let left = 0, right = nums.length - 1; left <= right; )

C. for(let i = 0; i < nums.length; i++)

D. if(nums.length === 0) return -1

Solution

1. Step 1: Identify the loop type for binary search

   Binary search typically uses a while loop with condition left < right to narrow down the search space.

2. Step 2: Check options

   while(left < right) { ... } uses while(left < right), which is correct for binary search in this problem.

3. Final Answer:

   while(left < right) { ... } -> Option A

4. Quick Check:

   Binary search loop uses while(left < right) [OK]

Hint: Use while loop with left < right for binary search [OK]

Common Mistakes:

• Using for loop incorrectly for binary search
• Using <= instead of < causing infinite loops
• Not initializing left and right properly

3. What will be the output of the following code?

function findMin(nums) {
  let left = 0, right = nums.length - 1;
  while (left < right) {
    let mid = Math.floor((left + right) / 2);
    if (nums[mid] > nums[right]) {
      left = mid + 1;
    } else {
      right = mid;
    }
  }
  return nums[left];
}

console.log(findMin([4,5,6,7,0,1,2]));

medium

A. 1

B. 4

C. 7

D. 0

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 code with input [4,5,6,7,0,1,2]

   Initially left=0, right=6. Mid=3 (value 7). Since 7 > 2, left = mid+1 = 4. Now left=4, right=6. Mid=5 (value 1). 1 <= 2, so right=mid=5. Next left=4, right=5. Mid=4 (value 0). 0 <= 1, right=mid=4. Now left=4, right=4, loop ends. Return nums[4] = 0.

3. Final Answer:

   0 -> Option D

4. Quick Check:

   Minimum element found = 0 [OK]

Hint: Compare mid with right to decide search half [OK]

Common Mistakes:

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

4. Identify the bug in this code snippet for finding minimum in rotated sorted array:

function findMin(nums) {
  let left = 0, right = nums.length - 1;
  while (left <= right) {
    let mid = Math.floor((left + right) / 2);
    if (nums[mid] > nums[right]) {
      left = mid + 1;
    } else {
      right = mid;
    }
  }
  return nums[left];
}

medium

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

B. The mid calculation is incorrect

C. The return statement should be nums[right]

D. The comparison should be nums[mid] < nums[right]

Solution

1. Step 1: Analyze loop condition

   The loop uses left <= right, which can cause infinite loop or skip the correct minimum index.

2. Step 2: Correct loop condition

   Binary search for minimum in rotated array requires left < right to narrow search correctly without missing elements.

3. Final Answer:

   The loop condition should be left < right, not left <= right -> Option A

4. Quick Check:

   Use while(left < right) for correct binary search [OK]

Hint: Use strict less than in binary search loops [OK]

Common Mistakes:

• Using <= causing infinite loops
• Changing mid calculation unnecessarily
• Returning wrong index after loop

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?

hard

A. Return nums[mid] immediately when nums[mid] == nums[right]

B. When nums[mid] == nums[right], reduce right by 1 to skip duplicate

C. Always move left to mid + 1 when nums[mid] >= nums[right]

D. Sort the array first before applying binary search

Solution

1. Step 1: Understand duplicates effect

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

2. Step 2: Modify binary search to handle duplicates

   When nums[mid] == nums[right], safely reduce right by 1 to skip duplicate and continue search.

3. Final Answer:

   When nums[mid] == nums[right], reduce right by 1 to skip duplicate -> Option B

4. Quick Check:

   Handle duplicates by shrinking right pointer [OK]

Hint: Skip duplicates by decrementing right when equal [OK]

Common Mistakes:

• Ignoring duplicates causing wrong minimum
• Moving left pointer incorrectly on equality
• Sorting array unnecessarily