Peak Index in a Mountain Array — Binary Search on Slope

What does it solve?

A mountain array strictly rises then strictly falls. Find the index of the peak. A linear scan is O(N), but the slope lets binary search reach it in O(log N).

Key insight: compare a[mid] with its right neighbour a[mid+1]. If a[mid] < a[mid+1] you're on the ascending side, the peak is to the right, so lo = mid+1. Otherwise you're on the descending side (or at the peak), so hi = mid. When lo == hi, that's the summit → O(log N).

The core idea

lo=0, hi=n-1
while lo < hi:
  mid=(lo+hi)/2
  if a[mid] < a[mid+1]:
    lo = mid+1 # uphill → go right
  else:
    hi = mid # downhill → go left
return lo

Complexity

Time complexity

O(log N)

Each comparison of neighbours halves the search range. Same shape as the "find peak element" pattern.

Space complexity

O(1)

Two pointers. No recursion needed.

Examples & edge cases

Live animation — follow the slope to the summit

lo mid mid+1 neighbour hi peak

Mountain

Current step detail

lo = — mid = — hi = — peak = —

Step — of —

a[mid]

—

midpoint height

a[mid+1]

—

right neighbour

slope

—

up / down

peak index

—

summit

Select an example and press Play to start.

✓ Done

SpeedMedium

Step-by-step table

Step	lo	mid	hi	a[mid]	a[mid+1]	slope	action

C++ implementation

peak_index_mountain.cpp

#include <bits/stdc++.h>
using namespace std;

// Index of the peak in a strictly increasing-then-decreasing array
int peakIndexInMountainArray(vector<int>& a) {
    int lo = 0, hi = (int)a.size() - 1;
    while (lo < hi) {
        int mid = lo + (hi - lo) / 2;
        if (a[mid] < a[mid + 1])
            lo = mid + 1;   // ascending slope → peak is to the right
        else
            hi = mid;       // descending (or at peak) → peak is here or left
    }
    return lo;            // lo == hi == peak
}

int main() {
    vector<int> a = {0,2,4,3,1};
    cout << peakIndexInMountainArray(a) << "\n"; // 2
    return 0;
}

My Notes

📝Notes✅

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