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

Time Complexity: Find Minimum in Rotated Sorted Array

O(log n)

Understanding Time Complexity

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

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

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function findMin(nums) {
  let left = 0;
  let 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];
}
    

This code finds the smallest number in a rotated sorted array by using a smart search that cuts the search space in half each time.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop that compares middle and right elements to narrow down the search.
• How many times: The loop runs until the search space is reduced to one element, roughly halving each time.

How Execution Grows With Input

Each step cuts the list roughly in half, so the number of steps grows slowly even if the list grows a lot.

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

Pattern observation: The steps grow very slowly compared to input size because the search space halves each time.

Final Time Complexity

Time Complexity: O(log n)

This means the time to find the minimum grows slowly as the list gets bigger, making it very efficient.

Common Mistake

[X] Wrong: "The search must check every number to find the minimum, so it takes O(n) time."

[OK] Correct: The code uses a smart method that cuts the search area in half each time, so it doesn't check every number.

Interview Connect

Understanding this search method shows you can use smart ways to find answers quickly, a skill that helps in many coding problems.

Self-Check

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

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