Find Peak Element Using Binary Search in DSA Javascript - Time & Space Complexity

Time Complexity: Find Peak Element Using Binary Search

O(log n)

Understanding Time Complexity

We want to know how fast the binary search method finds a peak element in an array.

How does the number of steps change as the array gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

This code finds a peak element by repeatedly narrowing the search range using binary search.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop that halves the search range each time.
• How many times: About log base 2 of the input size times (because the range halves each step).

How Execution Grows With Input

Each step cuts the search area in half, so the number of steps grows slowly as the array gets bigger.

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

Pattern observation: Doubling the input size adds only one extra step.

Final Time Complexity

Time Complexity: O(log n)

This means the steps needed grow slowly, making the search very efficient even for large arrays.

Common Mistake

[X] Wrong: "The loop runs once for each element, so it is O(n)."

[OK] Correct: The search range halves each time, so the loop runs much fewer times than the number of elements.

Interview Connect

Understanding how binary search reduces work quickly is a key skill that shows you can write fast and smart code.

Self-Check

"What if we changed the array to be sorted and wanted to find a specific value instead of a peak? How would the time complexity change?"

Practice

1. What is a peak element in an array when using binary search to find it?

easy

A. An element that is greater than its neighbors

B. The smallest element in the array

C. The first element of the array

D. An element equal to its neighbors

Solution

1. Step 1: Understand the definition of a peak element

   A peak element is one that is strictly greater than its neighbors on both sides (or one side if at the boundary).

2. Step 2: Compare options with the definition

   Only An element that is greater than its neighbors matches this definition exactly.

3. Final Answer:

   An element that is greater than its neighbors -> Option A

4. Quick Check:

   Peak element = greater than neighbors [OK]

Hint: Peak element is bigger than neighbors, not smallest or equal [OK]

Common Mistakes:

• Thinking peak means smallest element
• Confusing peak with first or last element
• Assuming peak equals neighbors

2. Which of the following is the correct way to calculate the middle index mid in binary search in JavaScript?

easy

A. const mid = (left + right) / 2;

B. const mid = Math.ceil((left + right) / 2);

C. const mid = Math.floor((left + right) / 2);

D. const mid = left + right / 2;

Solution

1. Step 1: Understand integer division in JavaScript

   JavaScript division returns a float, so we use Math.floor to get an integer index.

2. Step 2: Check each option

   const mid = Math.floor((left + right) / 2); uses Math.floor on the average of left and right, which is correct for binary search.

3. Final Answer:

   const mid = Math.floor((left + right) / 2); -> Option C

4. Quick Check:

   Use Math.floor for integer mid index [OK]

Hint: Use Math.floor((left + right) / 2) for mid index [OK]

Common Mistakes:

• Not using Math.floor causes float index
• Using only division without parentheses
• Incorrect operator precedence in calculation

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

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

console.log(findPeak([1, 3, 2, 4, 1]));

medium

A. 3

B. 1

C. 2

D. 0

Solution

1. Step 1: Trace the binary search steps on array [1,3,2,4,1]

   Initial left=0, right=4. mid=2 (value=2). Compare nums[2]=2 and nums[3]=4. Since 2 < 4, left=mid+1=3.

2. Step 2: Next iteration with left=3, right=4

   mid=3 (value=4). Compare nums[3]=4 and nums[4]=1. Since 4 > 1, right=mid=3.

3. Step 3: Loop ends as left=right=3

   Return left=3, which is index of peak element 4.

4. Final Answer:

   3 -> Option A

5. Quick Check:

   Peak index = 3 [OK]

Hint: Compare mid and mid+1 to decide left or right [OK]

Common Mistakes:

• Returning value instead of index
• Not updating left or right correctly
• Confusing mid calculation

4. Identify the bug in this binary search code to find a peak element:

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

medium

A. The comparison should be nums[mid] > nums[mid + 1] instead of <.

B. The mid calculation is incorrect.

C. The return statement should return right instead of left.

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

Solution

1. Step 1: Analyze the loop condition

   The loop uses left <= right, which can cause mid+1 to go out of bounds when mid = right.

2. Step 2: Correct loop condition for safe mid+1 access

   Changing to left < right ensures mid+1 is always valid inside the loop.

3. Final Answer:

   The loop condition should be left < right instead of <=. -> Option D

4. Quick Check:

   Loop condition left < right avoids out-of-bounds [OK]

Hint: Use left < right to avoid mid+1 overflow [OK]

Common Mistakes:

• Using <= causes index out of range error
• Changing mid calculation unnecessarily
• Returning wrong index

5. Given a strictly increasing then strictly decreasing array nums, how can you modify the binary search to find the peak element efficiently?
Choose the best approach.

hard

A. Use binary search comparing nums[mid] and nums[mid-1], move left or right accordingly.

B. Use binary search comparing nums[mid] and nums[mid+1], move left or right accordingly.

C. Scan the array linearly from start to find the maximum element.

D. Sort the array first, then pick the middle element.

Solution

1. Step 1: Understand the array shape

   The array increases then decreases, so the peak is the maximum element where nums[mid] > nums[mid+1].

2. Step 2: Use binary search comparing nums[mid] and nums[mid+1]

   If nums[mid] > nums[mid+1], move right pointer to mid; else move left pointer to mid+1.

3. Step 3: Avoid linear scan or sorting

   Linear scan is O(n), sorting changes order and is unnecessary.

4. Final Answer:

   Use binary search comparing nums[mid] and nums[mid+1], move left or right accordingly. -> Option B

5. Quick Check:

   Compare mid and mid+1 for peak in bitonic array [OK]

Hint: Compare mid and mid+1 to find peak in bitonic array [OK]

Common Mistakes:

• Using linear scan instead of binary search
• Sorting array destroys peak property
• Comparing mid and mid-1 incorrectly