# Sort List by Reversing Once - Amazon Top Interview Questions

### Problem Statement :

```You are given a list of integers nums. Given that you can first reverse one sublist in nums, return whether you can make the resulting list be arranged in ascending order.

Constraints

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

Example 1

Input

nums = [1, 3, 3, 7, 6, 9]

Output

True

Explanation

If we reverse the sublist [7, 6], then we can sort the list in ascending order: [1, 3, 3, 6, 7, 9].

Example 2

Input

nums = [1, 3, 9, 8, 2]

Output

False

Explanation

There's no way to reverse any sublist to sort nums in ascending order.

Example 3

Input

nums = [1, 2, 3, 4]

Output

True

Explanation

This list is already sorted in ascending order so we can reverse any sublist of length 1. For example, reverse [2] to get the same [1, 2, 3, 4].```

### Solution :

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

bool solve(vector<int>& nums) {
vector<int> v;
v = nums;
sort(v.begin(), v.end());
int l = -1, r = -1;
for (int i = 0; i < nums.size(); i += 1) {
if (v[i] != nums[i]) {
if (l == -1) l = i;
r = i;
}
}
if (l == -1) return true;
while (l < r) {
swap(nums[l], nums[r]);
l += 1, r -= 1;
}
return nums == v;
}```
```

```                        ```Solution in Python :

class Solution:
def solve(self, nums):
issorted = lambda x: all(nums[i] <= nums[i + 1] for i in range(len(nums) - 1))
if issorted(nums):
return True
lhs = 0
rhs = 1
while nums[lhs] <= nums[rhs]:
lhs += 1
rhs += 1
while lhs > 0 and nums[lhs - 1] >= nums[lhs]:
lhs -= 1
while rhs + 1 < len(nums) and nums[rhs] >= nums[rhs + 1]:
rhs += 1
while lhs <= rhs:
temp = nums[lhs]
nums[lhs] = nums[rhs]
nums[rhs] = temp
lhs += 1
rhs -= 1
return issorted(nums)```
```

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