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852. Peak Index in a Mountain Array

Description

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

Return the index of the peak element.

Your task is to solve it in O(log(n)) 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.

Solutions

Solution 1

Thinking

The array rises strictly then falls strictly; we need the peak index. A linear scan works, but \(n\le 10^5\) and the peak test is monotone: left of the peak \(arr[i]<arr[i+1]\), right of it the opposite.

Binary search \([1,n-2]\): if \(arr[mid]>arr[mid+1]\) the peak is in the left half (including \(mid\)), otherwise in the right. The left endpoint is the peak.

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class Solution:
    def peakIndexInMountainArray(self, arr: List[int]) -> int:
        left, right = 1, len(arr) - 2
        while left < right:
            mid = (left + right) >> 1
            if arr[mid] > arr[mid + 1]:
                right = mid
            else:
                left = mid + 1
        return left

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class Solution {
    public int peakIndexInMountainArray(int[] arr) {
        int left = 1, right = arr.length - 2;
        while (left < right) {
            int mid = (left + right) >> 1;
            if (arr[mid] > arr[mid + 1]) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
}

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class Solution {
public:
    int peakIndexInMountainArray(vector<int>& arr) {
        int left = 1, right = arr.size() - 2;
        while (left < right) {
            int mid = (left + right) >> 1;
            if (arr[mid] > arr[mid + 1])
                right = mid;
            else
                left = mid + 1;
        }
        return left;
    }
};

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func peakIndexInMountainArray(arr []int) int {
    left, right := 1, len(arr)-2
    for left < right {
        mid := (left + right) >> 1
        if arr[mid] > arr[mid+1] {
            right = mid
        } else {
            left = mid + 1
        }
    }
    return left
}

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function peakIndexInMountainArray(arr: number[]): number {
    let left = 1,
        right = arr.length - 2;
    while (left < right) {
        const mid = (left + right) >> 1;
        if (arr[mid] > arr[mid + 1]) {
            right = mid;
        } else {
            left = mid + 1;
        }
    }
    return left;
}

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impl Solution {
    pub fn peak_index_in_mountain_array(arr: Vec<i32>) -> i32 {
        let mut left = 1;
        let mut right = arr.len() - 2;
        while left < right {
            let mid = left + (right - left) / 2;
            if arr[mid] > arr[mid + 1] {
                right = mid;
            } else {
                left = left + 1;
            }
        }
        left as i32
    }
}

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/**
 * @param {number[]} arr
 * @return {number}
 */
var peakIndexInMountainArray = function (arr) {
    let left = 1;
    let right = arr.length - 2;
    while (left < right) {
        const mid = (left + right) >> 1;
        if (arr[mid] < arr[mid + 1]) {
            left = mid + 1;
        } else {
            right = mid;
        }
    }
    return left;
};

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