← Back to DSA Animator
Permutations LC #46 Medium Backtracking · Used Array
Problem

Given an array nums of distinct integers, return all the possible permutations. You can return the answer in any order.

Example 1
Input: nums = [1,2,3]
Output: [[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]
Explanation: 3! = 6 permutations.
Example 2
Input: nums = [0,1]
Output: [[0,1],[1,0]]
Constraints: 1 ≤ nums.length ≤ 6  |  −10 ≤ nums[i] ≤ 10  |  All integers are unique.
Try Examples
Custom:
Approach
Backtracking with used[] Array
Track which elements are in the current path with a boolean[] used. At each level, try ALL indices (not just start) but skip used[i]=true. Base: curr.size()==n → found permutation.
Input Array
Current Permutation
curr = [ ] (empty)
Results 0 found
Variables
curr
[ ]
used[]
—
total found
0
Step Logic
Press ▶ Play or Next Step to begin.
✅
Ready
0 / 0
Select an example above and press Play.
Algorithm
1
Use boolean[] used to track which elements are in current permutation
2
At each level, try ALL indices 0..n-1; skip if used[i]=true
3
Base case: curr.size() == n → permutation complete, add to results
4
n! total permutations for n distinct elements
Time
O(n·n!)
Space
O(n)
Why Backtracking Works

Unlike subsets, permutations use every element exactly once, so we iterate from index 0 each time (not a start index) but skip used[i] elements. The used array prevents including the same element twice in one permutation. Backtracking resets used[i]=false to try the next choice.