Why Find Peak Element Using Binary Search in DSA Typescript?

The Big Idea

What if you could find the highest hill without climbing them all?

The Scenario

Imagine you have a long row of hills, and you want to find the highest hill without climbing every single one.

Checking each hill one by one takes a lot of time and effort.

The Problem

Going hill by hill is slow and tiring.

You might miss the highest hill if you get tired or lose focus.

It's like searching for a needle in a haystack by looking at every straw.

The Solution

Using binary search, you can quickly jump to the middle hill and decide which side to explore next.

This way, you skip many hills and find a peak much faster.

Before vs After

✗ Before

function findPeakElement(nums: number[]): number {
  for (let i = 0; i < nums.length; i++) {
    if ((i === 0 || nums[i] > nums[i - 1]) && (i === nums.length - 1 || nums[i] > nums[i + 1])) {
      return i;
    }
  }
  return -1;
}

✓ After

function findPeakElement(nums: number[]): number {
  let left = 0;
  let 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;
}

What It Enables

This method lets you find a peak quickly even in very large lists, saving time and effort.

Real Life Example

Finding the highest point in a mountain range using fewer steps, or locating a local maximum in stock prices efficiently.

Key Takeaways

Manual search checks every element, which is slow.

Binary search jumps smartly to find a peak faster.

This approach works well for large data sets.

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 just one side if at the boundary).

2. Step 2: Apply this to binary search context

   Binary search finds such an element by comparing mid element with neighbors to decide which half to search next.

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 [OK]

Common Mistakes:

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

2. Which of the following is the correct TypeScript syntax to start a binary search loop for finding a peak element in array nums?

easy

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

B. if(nums[mid] > nums[mid+1]) { ... }

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

D. switch(nums[mid]) { ... }

Solution

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

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

2. Step 2: Check options for correct binary search syntax

   while(left < right) { ... } shows the correct while loop structure for binary search.

3. Final Answer:

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

4. Quick Check:

   Binary search uses while loop [OK]

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

Common Mistakes:

• Using for loop instead of while
• Using if instead of loop for search
• Using switch incorrectly for binary search

3. What will be the output index of a peak element for the array nums = [1, 3, 20, 4, 1, 0] using binary search peak finding?

medium

A. 3

B. 1

C. 2

D. 0

Solution

1. Step 1: Identify peak elements in the array

   Check neighbors: 20 at index 2 is greater than 3 and 4, so it's a peak.

2. Step 2: Binary search will find index 2 as peak

   Binary search compares mid with neighbors and narrows to index 2.

3. Final Answer:

   2 -> Option C

4. Quick Check:

   Peak at index 2 = 20 [OK]

Hint: Peak is local max, here 20 at index 2 [OK]

Common Mistakes:

• Choosing first or last element incorrectly
• Confusing peak with global max always
• Ignoring neighbors in peak definition

4. Given this TypeScript code snippet to find a peak element, what is the bug?

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

medium

A. Return should be right instead of left

B. The loop condition should be while(left <= right)

C. The mid calculation is incorrect

D. The else branch should set right = mid, not mid - 1

Solution

1. Step 1: Analyze the else branch update of right pointer

   Setting right = mid - 1 skips the mid element, possibly missing peak.

2. Step 2: Correct approach is right = mid to keep mid in range

   To ensure peak is not skipped, right should be set to mid, not mid - 1.

3. Final Answer:

   The else branch should set right = mid, not mid - 1 -> Option D

4. Quick Check:

   Keep mid in range by right = mid [OK]

Hint: Use right = mid to avoid skipping mid element [OK]

Common Mistakes:

• Setting right = mid - 1 causing missed peaks
• Changing loop condition unnecessarily
• Miscomputing mid index

5. You have an array nums = [10, 20, 15, 2, 23, 90, 67]. Using binary search to find a peak element, which index could be returned as a valid peak?

hard

A. 2

B. 5

C. 6

D. 3

Solution

1. Step 1: Identify all peak elements in the array

   Peaks are elements greater than neighbors: 20 at index 1 and 90 at index 5.

2. Step 2: Binary search can return any peak; index 5 (value 90) is a valid peak

   90 is greater than 23 and 67, so index 5 is a valid peak.

3. Final Answer:

   5 -> Option B

4. Quick Check:

   90 at index 5 is peak [OK]

Hint: Binary search may return any peak, here index 5 is valid [OK]

Common Mistakes:

• Assuming only one peak exists
• Choosing non-peak elements
• Ignoring neighbors when checking peak