# Strictly Alternating List - Facebook Top Interview Questions

### Problem Statement :

```You are given a list of integers nums.

Return whether the list alternates from first strictly increasing to strictly decreasing and then strictly increasing etc.

If a list is only strictly increasing, return true.

Constraints

2 ≤ n ≤ 100,000 where n is the length of nums

Example 1

Input

nums = [1, 2, 5, 7, 4, 1, 6, 8, 3, 2]

Output

True

Explanation

This list strictly increases, strictly decreases, strictly increases, then strictly decreases.

Example 2

Input

nums = [1, 1, 2, 3, 2, 1]

Output

False

Explanation

This list increases and then decreases, but is not strictly increasing at first.

Example 3

Input

nums = [1, 3, 5, 7]

Output

True

Explanation

This list is strictly increasing.

Example 4

Input

nums = [5, 3, 1, 5, 7]

Output

False

Explanation

This list is not strictly increasing at first.```

### Solution :

```                        ```Solution in C++ :

bool solve(vector<int>& nums) {
int n = nums.size();
for (int i = 1; i < n; i++) {
if (nums[i] == nums[i - 1]) return false;
if (i == 1 and nums[i] < nums[i - 1]) return false;
}
return true;
}```
```

```                        ```Solution in Java :

import java.util.*;

class Solution {
public boolean solve(int[] nums) {
if (nums.length <= 1)
return true;
if (nums[1] <= nums[0])
return false;
for (int i = 1; i < nums.length; i++) {
if (nums[i] == nums[i - 1]) {
return false;
}
}
return true;
}
}```
```

```                        ```Solution in Python :

class Solution:
def solve(self, a):
return a[0] < a[1] and all(a[i - 1] != a[i] for i in range(1, len(a)))```
```

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