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0852 - Peak Index in a Mountain Array (Easy)

https://leetcode.com/problems/peak-index-in-a-mountain-array/

Problem Statement​

Let's call an array arr a mountain if the following properties hold:

  • arr.length >= 3
  • There exists some i with 0 < i < arr.length - 1 such that:
    • arr[0] < arr[1] < ... arr[i-1] < arr[i]
    • arr[i] > arr[i+1] > ... > arr[arr.length - 1]

Given an integer array arr that is guaranteed to be a mountain, return any i such that arr[0] < arr[1] < ... arr[i - 1] < arr[i] > arr[i + 1] > ... > arr[arr.length - 1].

Example 1:

Input: arr = [0,1,0]
Output: 1

Example 2:

Input: arr = [0,2,1,0]
Output: 1

Example 3:

Input: arr = [0,10,5,2]
Output: 1

Constraints:

  • 3 <= arr.length <= 10^4
  • 0 <= arr[i] <= 10^6
  • arr is guaranteed to be a mountain array.

Follow up: Finding the O(n) is straightforward, could you find an O(log(n)) solution?\

Prerequisite
Written by @wingkwong
class Solution {
public:
    int peakIndexInMountainArray(vector<int>& arr) {
        // arr[i] < arr[i + 1]
        // [T, T, T, .., T, F, F, F, .., F]
        int l = 0, r = arr.size() - 1;
        while (l < r) {
            int m = l + (r - l) / 2;
            if (arr[m] < arr[m + 1]) l = m + 1;
            else r = m;
        }
        return l;
    }
};