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by Sumit Chourasia | Oct 03, 2020 | Category :coding | Tags : algorithm binary-search data-structure easy leetcode

Peak Index in a Mountain Array - Array - Easy - LeetCode

Peak Index in a Mountain Array - Array - Easy - LeetCode

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
Example 4:

Input: arr = [3,4,5,1]
Output: 2
Example 5:

Input: arr = [24,69,100,99,79,78,67,36,26,19]
Output: 2
 

Constraints:

3 <= arr.length <= 104
0 <= arr[i] <= 106
arr is guaranteed to be a mountain array.

public class Solution {
    public int PeakIndexInMountainArray(int[] arr) {
        int start = 1;
        int end = arr.Length-2;
        while(start<=end){
            int mid = start + (end-start)/2;
            if(arr[mid-1]<arr[mid] && arr[mid]<arr[mid+1]){
                start = mid + 1;
            }
            else if(arr[mid-1]>arr[mid] && arr[mid]>arr[mid+1]){
                end = mid - 1;
            }
            else{
                return mid;
            }
        }
        
        return -1;
    }
}

Time Complexity: O(logn)

Space Complexity: O(1)

Contributed By: Sumit Chourasia
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