Problem #46: Permutations

Intuition & Strategy Breakdown

Use recursive backtracking. Maintain a state of used elements or swap elements in-place to construct all permutations of length N.

Python Solution

class Solution:
    def permute(self, nums: list[int]) -> list[list[int]]:
        res = []
        
        def backtrack(start):
            if start == len(nums):
                res.append(nums[:])
                return
            
            for i in range(start, len(nums)):
                # Swap
                nums[start], nums[i] = nums[i], nums[start]
                # Recurse
                backtrack(start + 1)
                # Backtrack
                nums[start], nums[i] = nums[i], nums[start]
        
        backtrack(0)
        return res

Java Solution

class Solution {
    public List> permute(int[] nums) {
        List> res = new ArrayList<>();
        backtrack(0, nums, res);
        return res;
    }
    
    private void backtrack(int start, int[] nums, List> res) {
        if (start == nums.length) {
            List list = new ArrayList<>();
            for (int num : nums) list.add(num);
            res.add(list);
            return;
        }
        
        for (int i = start; i < nums.length; i++) {
            swap(nums, start, i);
            backtrack(start + 1, nums, res);
            swap(nums, start, i);
        }
    }
    
    private void swap(int[] nums, int i, int j) {
        int temp = nums[i];
        nums[i] = nums[j];
        nums[j] = temp;
    }
}