Python 3.14 (Coding): Permutations Test

The solution approach of the problem is to use the backtracking algorithm. Lets implement a function which checks if n is less than or equal to 1. If so, it means there is only one element (or no elements) in nums, so the function directly returns a vector containing nums as the only permutation. If n is greater than 1, the function initializes an empty 2D vector ret to store the permutations. It enters a for loop that iterates over each index i from 0 to n - 1. This loop represents choosing the current element to be placed at the first position of the permutation. Inside the loop, it initializes the variable cur with the value at index i in nums. This represents the current element chosen for the first position. It swaps the current element cur with the element at index i in nums. This ensures that the current element is moved to the first position temporarily for generating permutations. The code then removes the last element from nums using pop_back() since the first element is already fixed. It recursively calls the permute function with the modified nums (without the first element) and stores the returned permutations in the sub vector. It enters a for loop that iterates over each permutation s in sub. Inside the loop, it appends the current element cur to the permutation s, representing the fixed first element. The updated permutation s with the fixed first element is appended to the ret vector. After the loop ends, the first element cur is added back to nums using push_back() and is swapped back to its original position using swap(). Finally, the function returns the resulting 2D vector ret containing all possible permutations of nums.