Peak Index In Mountain Array

find the peak or pivot element in the mountain array

class Solution {
public:

    // Function to find the peak or pivot element in the mountain array
    int find_pivot(vector<int> v) {
        int s = 0, e = v.size() - 1; // Initialize start (s) and end (e) indices
        int mid = (s + e) / 2;       // Calculate the middle index

        // Binary search loop
        while (s < e) {
            if (v[mid] < v[mid + 1]) { // If mid element is smaller than the next one
                s = mid + 1;           // Move the start to the right (look for a larger element)
            } else {                   // If mid element is larger than the next one
                e = mid;               // Move the end to the left (since mid could be the peak)
            }

            mid = (s + e) / 2;         // Recalculate the middle index
        }
        return s; // The start and end will converge to the peak index
    }

    // Main function that calls find_pivot to get the peak index
    int peakIndexInMountainArray(vector<int>& arr) {
        return find_pivot(arr); // Find and return the peak index
    }
};

Example

For arr = {0, 2, 4, 6, 3, 1}:

1. Initial state:
   • s = 0, e = 5, mid = (0 + 5) / 2 = 2.
   • v[mid] = 4, and v[mid + 1] = 6.
   • Since v[mid] < v[mid + 1], move s = mid + 1 = 3.
2. Next iteration:
   • s = 3, e = 5, mid = (3 + 5) / 2 = 4.
   • v[mid] = 3, and v[mid + 1] = 1.
   • Since v[mid] > v[mid + 1], move e = mid = 4.
3. Next iteration:
   • s = 3, e = 4, mid = (3 + 4) / 2 = 3.
   • v[mid] = 6, and v[mid + 1] = 3.
   • Since v[mid] > v[mid + 1], move e = mid = 3.
4. Final state:
   • s = e = 3.
   • The loop exits, and the peak index 3 is returned.