| @@ -0,0 +1,15 @@ | |||
| class Solution: | |||
| def peakIndexInMountainArray(self, A): | |||
| # return the peak of the mountain... i.e Where the array starts going down in value | |||
| last = -1 | |||
| for ind, n in enumerate(A): | |||
| if n < last: | |||
| return ind - 1 | |||
| last = n | |||
| # This should never happen | |||
| return -1 | |||
| s = Solution() | |||
| print("Expected: 1") | |||
| print("Got:", s.peakIndexInMountainArray([0, 2, 1, 0])) | |||
| @@ -0,0 +1,23 @@ | |||
| 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. | |||
| @@ -0,0 +1,3 @@ | |||
| #!/bin/bash | |||
| python3 main.py | |||