# 2022-06-06

## 47. Permutations II

### Description

Given a collection of numbers, `nums`, that might contain duplicates, return all possible unique permutations in any order.

Example 1:

``````Input: nums = [1,1,2]
Output:
[[1,1,2],
[1,2,1],
[2,1,1]]``````

Example 2:

``````Input: nums = [1,2,3]
Output: [[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]``````

Constraints:

• `1 <= nums.length <= 8`

• `-10 <= nums[i] <= 10`

### Solution

#### Approach #0

``````func permuteUnique(nums []int) (ans [][]int) {
sort.Ints(nums)
n := len(nums)
vis := make([]bool, len(nums))
var tmp []int
var backtrack func(int)
backtrack = func(cur int) {
if cur == n {
ans = append(ans, append([]int(nil), tmp...))
return
}
for i, num := range nums {
if vis[i] || i > 0 && !vis[i-1] && nums[i-1] == num {
continue
}
vis[i] = true
tmp = append(tmp, num)
backtrack(cur + 1)
tmp = tmp[:len(tmp)-1]
vis[i] = false
}
}
backtrack(0)
return
}``````

## 39. Combination Sum

### Description

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.

It is guaranteed that the number of unique combinations that sum up to `target` is less than `150` combinations for the given input.

Example 1:

``````Input: candidates = [2,3,6,7], target = 7
Output: [[2,2,3],[7]]
Explanation:
2 and 3 are candidates, and 2 + 2 + 3 = 7. Note that 2 can be used multiple times.
7 is a candidate, and 7 = 7.
These are the only two combinations.``````

Example 2:

``````Input: candidates = [2,3,5], target = 8
Output: [[2,2,2,2],[2,3,3],[3,5]]``````

Example 3:

``````Input: candidates = [2], target = 1
Output: []``````

Constraints:

• `1 <= candidates.length <= 30`

• `1 <= candidates[i] <= 200`

• All elements of `candidates` are distinct.

• `1 <= target <= 500`

### Solution

#### Approach #0

``````func combinationSum(candidates []int, target int) (ans [][]int) {
sort.Ints(candidates)
n := len(candidates)
var t []int
var backtrack func(cur, index int)
backtrack = func(cur, index int) {
if cur == target {
ans = append(ans, append([]int(nil), t...))
return
}
if cur > target {
return
}
for i := index; i < n; i++ {
t = append(t, candidates[i])
backtrack(cur+candidates[i], i)
t = t[:len(t)-1]
}
}
backtrack(0, 0)
return
}``````

## 40. Combination Sum II

### Description

Given a collection of candidate numbers (`candidates`) and a target number (`target`), find all unique combinations in `candidates` where the candidate numbers sum to `target`.

Each number in `candidates` may only be used once in the combination.

Note: The solution set must not contain duplicate combinations.

Example 1:

``````Input: candidates = [10,1,2,7,6,1,5], target = 8
Output:
[
[1,1,6],
[1,2,5],
[1,7],
[2,6]
]``````

Example 2:

``````Input: candidates = [2,5,2,1,2], target = 5
Output:
[
[1,2,2],
[5]
]``````

Constraints:

• `1 <= candidates.length <= 100`

• `1 <= candidates[i] <= 50`

• `1 <= target <= 30`

### Solution

#### Approach #0

``````func combinationSum2(candidates []int, target int) (ans [][]int) {
sort.Ints(candidates)
n := len(candidates)
vis := make([]bool, n)
var t []int
var backtrack func(index, cur int)
backtrack = func(index, cur int) {
if cur == target {
ans = append(ans, append([]int(nil), t...))
return
}
if cur > target {
return
}
for i := index; i < n; i++ {
if i > 0 && !vis[i-1] && candidates[i-1] == candidates[i] {
continue
}
t = append(t, candidates[i])
vis[i] = true
backtrack(i+1, cur+candidates[i])
vis[i] = false
t = t[:len(t)-1]
}
}
backtrack(0, 0)
return
}``````

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