LeetCode: Peak Index in a Mountain Array

Posted on June 17, 2018July 26, 2020 by braindenny

Peak Index in a Mountain Array

Similar Problems:

Let’s call an array A a mountain if the following properties hold:

A.length >= 3
There exists some 0 < i < A.length – 1 such that A[0] < A[1] < … A[i-1] < A[i] > A[i+1] > … > A[A.length – 1]

Given an array that is definitely a mountain, return any i such that A[0] < A[1] < … A[i-1] < A[i] > A[i+1] > … > A[A.length – 1].

Example 1:

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

Example 2:

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

Note:

3 <= A.length <= 10000
0 <= A[i] <= 10^6
A is a mountain, as defined above.

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

Solution: binary search

// https://code.dennyzhang.com/peak-index-in-a-mountain-array
// Basic Ideas: binary search
//  monotonic function: A[i]<A[i+1]
//  T, T, T, F, F, F
// Complexity: Time O(log(n)), Space O(1)
func peakIndexInMountainArray(A []int) int {
    low, high := 0, len(A)-1
    // always exists
    for low<high {
        mid := (high-low)/2 + low
        if A[mid] < A[mid+1] {
            low = mid+1
        } else {
            high = mid
        }
    }
    return low
}

Solution: linear search

// https://code.dennyzhang.com/peak-index-in-a-mountain-array
// Basic Ideas: Too straightforward
// Complexity: Time O(n), Space O(1)
func peakIndexInMountainArray(A []int) int {
    for i:=1; i<len(A)-1; i++ {
        if A[i]>A[i-1] && A[i]>A[i+1] {
            return i
        }
    }
    return -1
}

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