**Next Smaller Permutation - Facebook Top Interview Questions**

### Problem Statement :

You are given a list of integers nums. Given that you must first swap any two numbers in nums, return the lexicographically largest list which is smaller than nums. If there is no solution, then return the same list. Constraints n ≤ 100,000 where n is the length of nums Example 1 Input nums = [2, 0, 1] Output [1, 0, 2] Explanation The largest permutation we can make that's less than nums is [1, 0, 2]. We swap 2 and 1. Note that [1, 2, 0] is not possible since getting this permutation requires 2 swaps. Example 2 Input nums = [1, 2, 3] Output [1, 2, 3] Explanation There's no way to get a smaller permutation so we return the same list.

### Solution :

` ````
Solution in C++ :
vector<int> solve(vector<int>& nums) {
int i, j;
for (i = nums.size() - 2; i >= 0; i--) {
if (nums[i] > nums[i + 1]) break;
}
if (i < 0) return nums;
for (j = nums.size() - 1; j >= 0; j--) {
if (j > 0 && nums[j] == nums[j - 1]) continue;
if (nums[j] < nums[i]) break;
}
swap(nums[i], nums[j]);
return nums;
}
```

` ````
Solution in Python :
class Solution:
def solve(self, nums):
nums_len = len(nums)
for i in range(nums_len - 2, -1, -1):
if nums[i] <= nums[i + 1]:
continue
pos = i + 1
swap_pos = None
while pos < nums_len and nums[pos] < nums[i]:
if not swap_pos or nums[pos] > nums[pos - 1]:
swap_pos = pos
pos += 1
nums[i], nums[swap_pos] = nums[swap_pos], nums[i]
return nums
return nums
```

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