Peak Index in a Mountain Array

You are given an integer mountain array arr of length n where values increase to a peak and then decrease.

Return the index of the peak element in O(log n) time.

Practice Link

Intuition#

At any index mid in the mountain, the slope tells you exactly where the peak is:

• arr[mid - 1] < arr[mid] — you're on the ascending side; peak is at mid or to the right.
• arr[mid] < arr[mid - 1] — you're on the descending side; peak is to the left.

Binary search on slope direction, always moving toward the peak.

Why low = 1 and high = n-2?
The first and last elements can never be the peak — the array is guaranteed to strictly increase then decrease, so the peak is always interior. Skipping the boundary indices avoids an out-of-bounds access when checking arr[mid - 1].

Solution — Binary Search#

class Solution {
public:
    int peakIndexInMountainArray(vector<int>& arr) {
        int n = arr.size();
        int low = 1, high = n - 2;
        int peak = -1;

        while (low <= high) {
            int mid = low + (high - low) / 2;

            if (arr[mid - 1] <= arr[mid]) {
                // ascending or flat: peak is here or to the right
                peak = mid;
                low = mid + 1;
            } else {
                // descending: peak is to the left
                high = mid - 1;
            }
        }
        return peak;
    }
};

Time Complexity: O(log n)
Space Complexity: O(1)

Trace — arr = [0, 10, 5, 2]#

Return peak = 1 ✓

Relation to Find Peak Element#

Both problems binary-search on slope direction, but they differ in guarantees: