Find Minimum in Rotated Sorted Array in DSA Javascript - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Find the minimum element output

What is the output of this JavaScript code that finds the minimum in a rotated sorted array?

DSA Javascript

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]));

A 4

B 7

C 1

D 0

💡 Hint

Think about how the binary search narrows down the smallest element in the rotated array.

The code uses binary search to find the smallest element. Since 0 is the smallest in the rotated array [4,5,6,7,0,1,2], it returns 0.

❓ Predict Output

intermediate

Minimum element in a non-rotated sorted array

What will the function output for a sorted array that is not rotated?

DSA Javascript

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([1,2,3,4,5,6,7]));

A 7

B 1

C 4

D 3

💡 Hint

If the array is not rotated, the smallest element is the first element.

Since the array is sorted and not rotated, the smallest element is the first element, which is 1.

🔧 Debug

advanced

Identify the error in this minimum finder

What error will this code produce when trying to find the minimum in a rotated sorted array?

DSA Javascript

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 - 1;
    }
  }
  return nums[left];
}

console.log(findMin([3,1,2]));

A Returns wrong minimum value 3

B IndexError

C TypeError

D Returns correct minimum value 1

💡 Hint

Check the loop condition and how right pointer is updated.

The loop condition and right pointer update cause the function to miss the correct minimum and return 3 instead of 1.

❓ Predict Output

advanced

Output of minimum finder with duplicates

What is the output of this code when the rotated sorted array contains duplicates?

DSA Javascript

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 if (nums[mid] < nums[right]) {
      right = mid;
    } else {
      right--;
    }
  }
  return nums[left];
}

console.log(findMin([2,2,2,0,1,2]));

A 0

B 1

C 2

D Error

💡 Hint

Duplicates require careful handling to avoid missing the minimum.

The code handles duplicates by decreasing right pointer when equal, eventually finding the minimum 0.

🧠 Conceptual

expert

Why binary search works for rotated sorted arrays

Why does binary search still work to find the minimum element in a rotated sorted array?

A Because linear search is too slow and binary search guesses the minimum randomly

B Because the array is fully sorted and binary search always works on sorted arrays

C Because the array is partially sorted and we can compare middle and right elements to decide search direction

D Because the minimum element is always at the middle index after rotation

💡 Hint

Think about how the rotated array keeps some sorted order properties.

Binary search works because the rotated array is partially sorted, allowing comparison between middle and right elements to decide which half contains the minimum.

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