Question

852. Peak Index in a Mountain Array

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Description

An array arr is 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.

Solutions

Solution 1

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

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;
  }
}

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;
  }
};

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
}

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;
}

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
  }
}

/**
 * @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;
};