Feb 28, 2022·

# Problem statement

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: Backtracking

If a candidate `c` belongs to a satisfying combination that sums to `target`, then you just need to solve the problem with a smaller `target = target - c`.

This leads to a backtracking technique to find all possible solutions for this problem.

For unique combinations, you might put the candidates in ascending order.

## Example 1

For `candidates = [2,3,6,7]` (already sorted) and `target = 7`.

1. Initialize an empty solution: `solution = []`.
2. Starting with candidate `c = 2`. It might belong to some solution since it is smaller than current `target = 7`. Put it into current building solution, `solution = [2]`. Repeat the finding with `target = target - c = 5`.
3. `c = 2` is still a valid candidate with `target = 5`. Put it to the current building solution `solution = [2, 2]`. Repeat the finding with `target = 5 - 2 = 3`.
4. `c = 2` might still be a valid candidate with `target = 3`. Put it to the current building solution `solution = [2, 2, 2]`. Repeat the finding with `target = 3 - 2 = 1`.
5. There is no candidate satisfied for `target = 1` at this point. The last candidate `2` of the current building solution `[2, 2, 2]` is not a good one. Remove it from the `solution`, i.e. `solution = [2, 2]` and `target = 3`. Repeat the finding with another candidate (This is called backtracking).
6. `c = 3` is a perfect candidate with `target = 3`. This make `solution = [2, 2, 3]` a satisfied one for the problem.

## Code

``````#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

void solve(vector<int>& candidates, int target, vector<vector<int>>& result, vector<int>& solution) {
if (target == 0) {
result.push_back(solution);
return;
}
for (auto c : candidates) {
if (c <= target) {
if (!solution.empty() && c < solution.back()) {
continue;
}
solution.push_back(c);
target -= c;
solve(candidates, target, result, solution);
if (!solution.empty()) {
target += solution.back();
solution.pop_back();
}
}
}
}
vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
sort(candidates.begin(), candidates.end());
vector<vector<int>> result;
vector<int> solution;
solve(candidates, target, result, solution);
return result;
}
void print(vector<vector<int>>& result) {
for (auto& v : result) {
cout << "[";
for (int a: v) {
cout << a << ",";
}
cout << "]";
}
cout << endl;
}
int main() {
vector<int> candidates{2,3,6,7};
auto result = combinationSum(candidates, 7);
print(result);
candidates = {2,3,5};
result = combinationSum(candidates, 8);
print(result);
}
``````
``````Output:
[2,2,3,][7,]
[2,2,2,2,][2,3,3,][3,5,]
``````

## Complexity

• Runtime: `O(2^N)`, where `N = candidates.length`.
• Extra space: `O(1)` (if not count the space to store the result, which is `O(N^2)`).