LeetCode - Permutations
Problem statement
Given an array nums of distinct integers, return all the possible permutations. You can return the answer in any order.
Problem statement taken from: https://leetcode.com/problems/permutations
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]]
Example 2:
Input: nums = [0, 1]
Output: [[0, 1], [1, 0]]
Example 3:
Input: nums = [1]
Output: [[1]]
Constraints:
- 1 <= nums.length <= 6
- -10 <= nums[i] <= 10
- All the integers of nums are unique.
Explanation
Backtracking
When we are required to generate a permutation or a sequence, recursion is the best approach to use. The recursion for this problem will be a bit different compared to the standard recursion approach.
One approach to solve this problem is to keep track of the element we have visited and generate permutations for the rest of the array elements. But, we can solve this by swapping the array elements.
Let's jump to the algorithm to understand it better.
- set result = [[]]
- call _getPermutations(result, nums, 0, nums.length - 1)
- return result
// _getPermutations(result, nums, l, r)
- if l == r
- push the current nums permutation in the result
- result.push(nums)
- else
- loop for i = l; i <= r; i++
- swap(nums[l], nums[i])
- _getPermutations(result, nums, l + 1, r)
- swap(nums[l], nums[i])
- end if
Let's check our algorithm in C++, Golang, and Javascript.
C++ solution
class Solution {
public:
vector<vector<int>> permute(vector<int>& nums) {
vector<vector<int>> result;
_getPermutations(result, nums, 0, nums.size() - 1);
return result;
}
void _getPermutations(vector<vector<int>>& result, vector<int> nums, int l, int r){
if(l == r){
result.push_back(nums);
return;
} else {
for(int i = l; i <= r; i++){
swap(nums[l], nums[i]);
_getPermutations(result, nums, l + 1, r);
swap(nums[l], nums[i]);
}
}
}
};Golang solution
func permute(nums []int) [][]int {
result := [][]int{}
_getPermutations(&result, nums, 0, len(nums) - 1)
return result
}
func _getPermutations(result *[][]int, nums []int, l, r int) {
if l == r {
cp := make([]int, len(nums))
copy(cp, nums)
*result = append(*result, cp)
} else {
for i := l; i <= r; i++ {
nums[l], nums[i] = nums[i], nums[l]
_getPermutations(result, nums, l + 1, r)
nums[l], nums[i] = nums[i], nums[l]
}
}
}Javascript solution
var permute = function(nums) {
const result = [];
_getPermutations(result, nums, 0, nums.length - 1);
return result;
};
function _getPermutations(result, nums, l, r) {
if(l === r) {
result.push(nums.slice(0));
return;
} else {
for(let i = l; i <= r; i++) {
[nums[l], nums[i]] = [nums[i], nums[l]];
_getPermutations(result, nums, l + 1, r);
[nums[l], nums[i]] = [nums[i], nums[l]];
}
}
}Dry Run
Let's dry-run our algorithm for Example 1.
Input: nums = [1, 2, 3]
// in permute function
Step 1: vector<vector<int>> result
Step 2: _getPermutations(result, nums, 0, nums.size() - 1)
_getPermutations(result, nums, 0, 2)
// in _getPermutations function
Step 3: if l == r
0 == 2
false
else
loop for i = l; i <= r
i = 0
0 <= 2
true
swap(nums[l], nums[i])
swap(nums[0], nums[0])
nums = [1, 2, 3]
_getPermutations(result, nums, l + 1, r)
_getPermutations(result, nums, 0 + 1, 2)
_getPermutations(result, nums, 1, 2)
Step 4: if l == r
1 == 2
false
else
loop for i = l; i <= r
i = 1
1 <= 2
true
swap(nums[l], nums[i])
swap(nums[1], nums[1])
nums = [1, 2, 3]
_getPermutations(result, nums, l + 1, r)
_getPermutations(result, nums, 1 + 1, 2)
_getPermutations(result, nums, 2, 2)
Step 5: if l == r
2 == 2
true
result.push_back(nums)
result = [[1, 2, 3]]
return
// We return to step 4
Step 6: swap(nums[l], nums[i])
swap(nums[1], nums[1])
nums = [1, 2, 3]
i++
i = 2
loop for i <= r
i = 2
2 <= 2
true
swap(nums[l], nums[i])
swap(nums[1], nums[2])
nums = [1, 3, 2]
_getPermutations(result, nums, l + 1, r)
_getPermutations(result, nums, 1 + 1, 2)
_getPermutations(result, nums, 2, 2)
Step 7: if l == r
2 == 2
true
result.push_back(nums)
result = [[1, 2, 3], [1, 3, 2]]
return
// We return to step 6
Step 8: swap(nums[l], nums[i])
swap(nums[1], nums[2])
nums = [1, 2, 3]
i++
i = 3
loop for i <= r
i = 3
3 <= 2
false
// we backtrack to step 3
Step 9: swap(nums[l], nums[i])
swap(nums[0], nums[0])
nums = [1, 2, 3]
i++
i = 1
loop for i <= r
i = 1
1 <= 2
true
swap(nums[l], nums[i])
swap(nums[0], nums[1])
nums = [2, 1, 3]
_getPermutations(result, nums, l + 1, r)
_getPermutations(result, nums, 0 + 1, 2)
_getPermutations(result, nums, 1, 2)
Step 10: if l == r
1 == 2
false
else
for i = l; i <= r
i = 1
1 <= 2
true
swap(nums[l], nums[i])
swap(nums[1], nums[1])
nums = [2, 1, 3]
_getPermutations(result, nums, l + 1, r)
_getPermutations(result, nums, 1 + 1, 2)
_getPermutations(result, nums, 2, 2)
Step 11: if l == r
2 == 2
true
result.push_back(nums)
result = [[1, 2, 3], [1, 3, 2], [2, 1, 3]]
return
// We return to step 10
We similarly backtrack to generate the rest of the solution
We return the solution as
[[1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1]]