# Find all permuations using backtracking

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

```
public List<List<Integer>> permute(int[] nums) {
List<List<Integer>> list = new ArrayList<>();
// Arrays.sort(nums); // not necessary
backtrack(list, new ArrayList<>(), nums);
return list;
}
private void backtrack(List<List<Integer>> list, List<Integer> tempList, int [] nums){
if(tempList.size() == nums.length){
list.add(new ArrayList<>(tempList));
} else{
for(int i = 0; i < nums.length; i++){
if(tempList.contains(nums[i])) continue; // element already exists, skip
tempList.add(nums[i]);
backtrack(list, tempList, nums);
tempList.remove(tempList.size() - 1);
}
}
}
```

## Related Problems

Given an array of **distinct** integers candidates and a target integer target, return *a list of all ***unique combinations*** of *candidates* where the chosen numbers sum to *target*.* You may return the combinations in **any order**.

The **same** number may be chosen from candidates an **unlimited number of times**. Two combinations are unique if the frequency of at least one of the chosen numbers is different.

There are a total of numCourses courses you have to take, labeled from 0 to numCourses - 1. You are given an array prerequisites where prerequisites[i] = [ai, bi] indicates that you **must** take course bi first if you want to take course ai.

For example, the pair [0, 1], indicates that to take course 0 you have to first take course 1.

Return true if you can finish all courses. Otherwise, return false.

Given a sorted array of distinct integers and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.

You must write an algorithm with O(log n) runtime complexity.

Write an efficient algorithm that searches for a value target in an m x n integer matrix matrix. This matrix has the following properties:

Integers in each row are sorted in ascending from left to right.

Integers in each column are sorted in ascending from top to bottom.