852. Peak Index in a Mountain Array

Let's call an array `arr` 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 an integer array `arr` that is guaranteed to be a mountain, return any `i` such that `arr[0] < arr[1] < ... arr[i - 1] < arr[i] > arr[i + 1] > ... > arr[arr.length - 1]`.

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
```

Example 4:

```Input: arr = [3,4,5,1]
Output: 2
```

Example 5:

```Input: arr = [24,69,100,99,79,78,67,36,26,19]
Output: 2
```

Constraints:

• `3 <= arr.length <= 104`
• `0 <= arr[i] <= 106`
• `arr` is guaranteed to be a mountain array.

Follow up: Finding the `O(n)` is straightforward, could you find an `O(log(n))` solution?

852. Peak Index in a Mountain Array
``````struct Solution;

impl Solution {
fn peak_index_in_mountain_array(a: Vec<i32>) -> i32 {
let mut l: usize = 0;
let mut r: usize = a.len() - 1;
while l < r {
let m = (l + r) / 2;
if a[m] < a[m + 1] {
l = m + 1;
} else {
r = m;
}
}
l as i32
}
}

#[test]
fn test() {
let a = vec![0, 1, 0];
assert_eq!(Solution::peak_index_in_mountain_array(a), 1);
let a = vec![0, 2, 1, 0];
assert_eq!(Solution::peak_index_in_mountain_array(a), 1);
}
``````