Last updated: Nov 27, 2022
Difficulty | Runtime | Faster_than | Memory | Lesser_than |
---|---|---|---|---|
medium | 88 ms | 84.93 % | 51.1 mb | 87.55 % |
Difficulty : medium
Runtime : 88 ms Faster than 84.93 %
Memory : 51.1 mb Lesser than 87.55 %
852. Peak Index in a Mountain Array
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 a mountain array arr, return the index i such that arr[0] < arr[1] < ... < arr[i - 1] < arr[i] > arr[i + 1] > ... > arr[arr.length - 1].
You must solve it in O(log(arr.length)) time complexity.
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 <= 105
- 0 <= arr[i] <= 106
- arr is guaranteed to be a mountain array.
Solution:
/**
* @param {number[]} arr
* @return {number}
*/
var peakIndexInMountainArray = function(arr) {
var l = 0
var r = arr.length - 1
var mid = Math.floor((r - l) / 2)
while(l < r) {
if(arr[mid] > arr[mid - 1] && arr[mid] > arr[mid + 1]) {
return mid
} else if (arr[mid] < arr[mid -1]) mid--
else mid++
l++
}
return mid
};