Find Peak Element Using Binary Search in DSA C++

Mental Model

A peak element is one that is bigger than its neighbors. We use binary search to quickly find such an element by checking the middle and deciding which half to explore next.

Analogy: Imagine climbing a mountain range where you want to find a peak. Instead of walking the whole range, you look at the middle point. If the slope goes up to the right, you move right; if it goes up to the left, you move left. This way, you quickly find a peak without checking every step.

Index: 0   1   2   3   4   5   6
Value: 1 -> 3 -> 2 -> 5 -> 4 -> 6 -> 3
          ↑
        mid pointer

Dry Run Walkthrough

Input: array: [1, 3, 2, 5, 4, 6, 3]

Goal: Find any peak element index where element is greater than neighbors

Step 1: Calculate mid index as 3, value is 5

Array: 1 -> 3 -> 2 -> [5] -> 4 -> 6 -> 3
mid=3

Why: Start by checking middle element to decide which half to search next

Step 2: Compare mid value 5 with next element 4; since 5 > 4, move search to left half including mid

Search range: indices 0 to 3
Array: 1 -> 3 -> 2 -> [5]

Why: If mid is greater than next, peak must be on left side or at mid

Step 3: Calculate new mid index as 1, value is 3

Array: 1 -> [3] -> 2 -> 5
mid=1

Why: Check middle of left half to narrow down peak location

Step 4: Compare mid value 3 with next element 2; since 3 > 2, move search to left half including mid

Search range: indices 0 to 1
Array: 1 -> [3]

Why: Peak is on left side or at mid because mid is greater than next

Step 5: Calculate new mid index as 0, value is 1

Array: [1] -> 3
mid=0

Why: Check middle of new range to find peak

Step 6: Compare mid value 1 with next element 3; since 1 < 3, move search to right half excluding mid

Search range: indices 1 to 1
Array: [3]

Why: If mid is less than next, peak must be on right side

Step 7: Only one element left at index 1, which is 3; this is a peak

Peak found at index 1 with value 3

Why: Single element left is peak by definition

Result:

Peak element index: 1, value: 3

Annotated Code

DSA C++

#include <iostream>
#include <vector>
using namespace std;

class Solution {
public:
    int findPeakElement(const vector<int>& nums) {
        int left = 0;
        int right = nums.size() - 1;
        while (left < right) {
            int mid = left + (right - left) / 2;
            // Compare mid element with next element
            if (nums[mid] > nums[mid + 1]) {
                // Peak is in left half including mid
                right = mid;
            } else {
                // Peak is in right half excluding mid
                left = mid + 1;
            }
        }
        // left == right is peak index
        return left;
    }
};

int main() {
    Solution sol;
    vector<int> nums = {1, 3, 2, 5, 4, 6, 3};
    int peakIndex = sol.findPeakElement(nums);
    cout << "Peak element index: " << peakIndex << ", value: " << nums[peakIndex] << endl;
    return 0;
}

while (left < right) {

loop until search range narrows to one element

int mid = left + (right - left) / 2;

calculate middle index to check

if (nums[mid] > nums[mid + 1]) {

if mid element is greater than next, peak is on left side including mid

right = mid;

move right boundary to mid to narrow search to left half

else { left = mid + 1; }

otherwise, move left boundary to mid+1 to search right half

return left;

return the index where left meets right, which is peak

Output

Peak element index: 1, value: 3

Complexity Analysis

Time: O(log n) because binary search halves the search space each step

Space: O(1) because only a few variables are used regardless of input size

vs Alternative: Compared to linear scan O(n), binary search is much faster for large arrays

Edge Cases

Single element array

Returns index 0 as peak since only one element exists

DSA C++

while (left < right) {

Strictly increasing array

Peak is last element because each next is bigger; binary search moves right

DSA C++

else { left = mid + 1; }

Strictly decreasing array

Peak is first element because each next is smaller; binary search moves left

DSA C++

if (nums[mid] > nums[mid + 1]) { right = mid; }

When to Use This Pattern

When you need to find a local maximum or peak in an array efficiently, use binary search by comparing mid element with its neighbor to decide search direction.

Common Mistakes

Mistake: Comparing mid element with previous instead of next element
Fix: Always compare mid with next element to decide which half contains a peak

Mistake: Using while (left <= right) causing infinite loop
Fix: Use while (left < right) to ensure loop ends when left meets right

Summary

Finds an index of a peak element where the element is greater than its neighbors.

Use when you want to quickly find a local maximum without checking every element.

The key insight is comparing mid element with its next neighbor to decide which half to search next.