Q852 Peak Index in a Mountain Array

Peak Index in a Mountain Array

Question

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.

Approach 1: Linear Scan

Intuition and Algorithm

The mountain increases until it doesn't. The point at which it stops increasing is the peak.

class Solution(object):
    def peakIndexInMountainArray(self, A):
        for i in range(0,len(A)):
            if A[i]>A[i+1]: break
        return i

def main():
    A=[0,1,2,0]
    solution=Solution().peakIndexInMountainArray(A)
    print(solution)


if __name__ == "__main__":
    main()

Complexity Analysis

• Time Complexity: O(N), where N is the length of A.
• Space Complexity: O(1).
• Runtime: 56 ms, faster than 95.41% of Python online submissions for Peak Index in a Mountain Array.
• Memory Usage: 12.5 MB, less than 99.37% of Python online submissions for Peak Index in a Mountain Array.

Approach 2: Binary Search

Intuition and Algorithm

The comparison A[i] < A[i+1] in a mountain array looks like [True, True, True, ..., True, False, False, ..., False]: 1 or more boolean Trues, followed by 1 or more boolean False. For example, in the mountain array [1, 2, 3, 4, 1], the comparisons A[i] < A[i+1] would be True, True, True, False.

We can binary search over this array of comparisons, to find the largest index i such that A[i] < A[i+1]. For more on binary search, see the LeetCode explore topic here.

class Solution(object):
    def peakIndexInMountainArray(self, A):
        lo, hi = 0, len(A) - 1
        while lo < hi:
            mi = (lo + hi) / 2
            if A[mi] < A[mi + 1]:
                lo = mi + 1
            else:
                hi = mi
        return lo

def main():
    A=[0,1,2,0]
    solution=Solution().peakIndexInMountainArray(A)
    print(solution)


if __name__ == "__main__":
    main()

Complexity Analysis

• Time Complexity: O(N), where N is the length of A.
• Space Complexity: O(1).
• Runtime: 56 ms, faster than 95.41% of Python online submissions for Peak Index in a Mountain Array.
• Memory Usage: 12.5 MB, less than 99.37% of Python online submissions for Peak Index in a Mountain Array.