Find Peak Element Using Binary Search in DSA Go - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Find Peak Element Output

What is the output index of the peak element for the given array using the binary search approach?

DSA Go

package main

import "fmt"

func findPeakElement(nums []int) int {
    left, right := 0, len(nums)-1
    for left < right {
        mid := (left + right) / 2
        if nums[mid] > nums[mid+1] {
            right = mid
        } else {
            left = mid + 1
        }
    }
    return left
}

func main() {
    nums := []int{1, 3, 20, 4, 1, 0}
    peakIndex := findPeakElement(nums)
    fmt.Println(peakIndex)
}

A 2

B 1

C 3

D 5

💡 Hint

Check where the array value is greater than its next neighbor to decide which side to search.

The binary search compares mid element with mid+1 element. Since nums[2] = 20 is greater than nums[3] = 4, the peak lies at or before index 2. The loop narrows down to index 2 as the peak.

❓ Predict Output

intermediate

Find Peak Element Output with Multiple Peaks

What is the output index of the peak element for the array with multiple peaks using the binary search method?

DSA Go

package main

import "fmt"

func findPeakElement(nums []int) int {
    left, right := 0, len(nums)-1
    for left < right {
        mid := (left + right) / 2
        if nums[mid] > nums[mid+1] {
            right = mid
        } else {
            left = mid + 1
        }
    }
    return left
}

func main() {
    nums := []int{1, 2, 1, 3, 5, 6, 4}
    peakIndex := findPeakElement(nums)
    fmt.Println(peakIndex)
}

A 5

B 1

C 4

D 3

💡 Hint

The function returns any peak, not necessarily the first peak.

The binary search finds index 5 where value 6 is a peak (greater than neighbors 5 and 4).

🧠 Conceptual

advanced

Why Binary Search Works for Finding a Peak Element

Why does the binary search approach guarantee finding a peak element in an unsorted array?

A Because the binary search checks all elements one by one to find the peak.

B Because the array is sorted, so the middle element is always a peak.

C Because if nums[mid] > nums[mid+1], a peak must exist on the left side including mid.

D Because the peak is always at the start or end of the array.

💡 Hint

Think about how comparing mid and mid+1 helps decide which side to search.

If nums[mid] > nums[mid+1], then a peak exists on the left side including mid because the slope is descending after mid. Otherwise, the peak is on the right side.

🔧 Debug

advanced

Identify the Bug in Peak Element Binary Search

What error does the following code produce when run, and why?

DSA Go

package main

import "fmt"

func findPeakElement(nums []int) int {
    left, right := 0, len(nums)-1
    for left <= right {
        mid := (left + right) / 2
        if nums[mid] > nums[mid+1] {
            right = mid - 1
        } else {
            left = mid + 1
        }
    }
    return left
}

func main() {
    nums := []int{1, 2, 3, 1}
    peakIndex := findPeakElement(nums)
    fmt.Println(peakIndex)
}

A Compiles but prints -1

B Infinite loop

C Returns incorrect peak index

D Index out of range runtime error

💡 Hint

Check the condition nums[mid+1] when mid is at the last index.

The loop condition allows mid to be the last index, so nums[mid+1] causes index out of range error.

🚀 Application

expert

Find Peak Element in Circular Array

Given a circular array where the end connects to the start, which approach correctly finds a peak element index?

A Use linear scan to check each element against neighbors ignoring circularity.

B Modify binary search to compare nums[mid] with nums[(mid+1) % n] and nums[(mid-1+n) % n] to handle circular neighbors.

C Apply standard binary search without changes since circularity doesn't affect peak definition.

D Sort the array first then find the maximum element index.

💡 Hint

Think about how to access neighbors in a circular array safely.

In a circular array, neighbors wrap around. Using modulo ensures correct neighbor indices for comparison in binary search.

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. An element that is smaller than its neighbors

C. The first element of the array

D. The last element of the array

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 only one side if at the boundary).

2. Step 2: Apply this to binary search context

   Binary search finds such an element by comparing middle elements 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 smaller [OK]

Common Mistakes:

• Thinking peak is smallest element
• Confusing peak with first or last element always
• Ignoring boundary conditions

2. Which Go code snippet correctly initializes the binary search boundaries to find a peak element in array nums?

easy

A. left, right := 1, len(nums)

B. left, right := 0, len(nums) - 1

C. left, right := 0, len(nums)

D. left, right := 1, len(nums) - 1

Solution

1. Step 1: Identify valid index range for array in Go

   Array indices in Go start at 0 and end at len(nums)-1, so boundaries must be set accordingly.

2. Step 2: Check each option for correct boundary initialization

   left, right := 0, len(nums) - 1 correctly sets left=0 and right=len(nums)-1, covering full array indices.

3. Final Answer:

   left, right := 0, len(nums) - 1 -> Option B

4. Quick Check:

   Array indices 0 to len-1 [OK]

Hint: Array indices go from 0 to length-1 in Go [OK]

Common Mistakes:

• Setting right to len(nums) (out of range)
• Starting left at 1 and missing first element
• Using negative indices

3. What will be the output index of the peak element for the array nums := []int{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 candidates in the array

   Array: [1, 3, 20, 4, 1, 0]. 20 at index 2 is greater than neighbors 3 and 4, so it's a peak.

2. Step 2: Confirm binary search finds index 2

   Binary search compares mid elements and moves left or right until it finds peak at index 2.

3. Final Answer:

   2 -> Option C

4. Quick Check:

   Peak at index 2 with value 20 [OK]

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

Common Mistakes:

• Choosing first or last element incorrectly
• Confusing index with value
• Not checking both neighbors

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

func findPeak(nums []int) int {
    left, right := 0, len(nums)-1
    for left < right {
        mid := (left + right) / 2
        if nums[mid] < nums[mid+1] {
            left = mid + 1
        } else {
            right = mid - 1
        }
    }
    return left
}

medium

A. The right boundary update should be right = mid, not mid - 1

B. The left boundary update should be left = mid, not mid + 1

C. The loop condition should be left <= right

D. The function should return right instead of left

Solution

1. Step 1: Analyze the binary search boundary updates

   When nums[mid] >= nums[mid+1], peak is at mid or to the left, so right should be set to mid, not mid - 1 to avoid skipping mid.

2. Step 2: Identify impact of wrong right update

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

3. Final Answer:

   The right boundary update should be right = mid, not mid - 1 -> Option A

4. Quick Check:

   Right boundary must include mid [OK]

Hint: Use right = mid to keep mid in search space [OK]

Common Mistakes:

• Setting right = mid - 1 and skipping mid
• Changing left incorrectly
• Wrong loop condition causing infinite loop

5. You have an array nums := []int{10, 20, 15, 2, 23, 90, 67}. Using binary search to find a peak element, which index will be returned and why?

hard

A. 1, because 20 is a peak greater than neighbors 10 and 15

B. 0, because first element is always a peak

C. 6, because last element is always a peak

D. 5, because 90 is a peak greater than neighbors 23 and 67

Solution

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

   Array: [10, 20, 15, 2, 23, 90, 67]. Peaks are 20 at index 1 and 90 at index 5.

2. Step 2: Understand binary search behavior

   Binary search compares mid and mid+1; it will move towards the higher peak 90 at index 5.

3. Step 3: Confirm returned index

   Binary search will return index 5 where 90 is a peak greater than neighbors 23 and 67.

4. Final Answer:

   5, because 90 is a peak greater than neighbors 23 and 67 -> Option D

5. Quick Check:

   Binary search finds highest local peak at index 5 [OK]

Hint: Binary search finds any peak, often the higher one found first [OK]

Common Mistakes:

• Assuming first or last element always peak
• Choosing smaller peak incorrectly
• Not understanding binary search direction