Skip to content

Combination Sum⚓︎

Link

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.

Note: The frequency of an element x is the number of times it occurs in the array.

The test cases are generated such 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
  • 2 <= candidates[i] <= 40
  • All elements of candidates are distinct.
  • 1 <= target <= 40

Solution⚓︎

Way 1⚓︎

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
class Solution {
private:
    vector<vector<int>> res;
    vector<int> path;
    void backtracking(vector<int>& candidates, int sum, int startIndex) {
        if (sum < 0) return;
        if (sum == 0) {
            res.push_back(path);
            return;
        }
        for (int i = startIndex; i < candidates.size(); i++) {
            path.push_back(candidates[i]);
            // No need to use i+1, meaning that the current number can be read repeatedly
            backtracking(candidates, sum - candidates[i], i);
            path.pop_back();
        }
    }
public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        backtracking(candidates, target, 0);
        return res;
    }
};
  • Time complexity: \(O(n\times 2^n)\);
  • Space complexity: \(O(\mathrm{target})\).

Optimized code with pruning:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
class Solution {
private:
    vector<vector<int>> res;
    vector<int> path;
    void backtracking(vector<int>& candidates, int sum, int startIndex) {
        if (sum < 0) return ;
        if (sum == 0) {
            res.push_back(path);
            return;
        }
        for (int i = startIndex; i < candidates.size(); i++) {
            if (sum - candidates[i] < 0) break;
            path.push_back(candidates[i]);
            backtracking(candidates, sum - candidates[i], i);
            path.pop_back();
        }
    }
public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        sort(candidates.begin(), candidates.end());
        backtracking(candidates, target, 0);
        return res;
    }
};

After sorting the total set, if the sum of the next level (which is the sum + candidates[i] of this level) is already greater than target, we can end this round of traversal of the for loop.

Way 2⚓︎

See reference.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
class Solution {
private:
    vector<vector<int>> res;
    vector<int> path;
    void backtracking(vector<int>& candidates, int sum, int startIndex) {
        if (sum < 0) return;
        if (sum == 0) {
            res.push_back(path);
            return;
        }
        // if index crosses the last index, we will return saying that no more element is left to choose
        if (startIndex == candidates.size()) return;

        // we are not taking the startIndex-th element,
        // so without decreasing sum we will move to next index because it will not contribute in making our sum
        backtracking(candidates, sum, startIndex + 1);

        // we are taking the startIndex-th element and not moving onto the next element 
        // because it may be possible that this element again contribute in making our sum.
        // but we decrease our target sum as we are considering that this will help us in making our target sum
        path.push_back(candidates[startIndex]);
        backtracking(candidates, sum - candidates[startIndex], startIndex);
        path.pop_back();
    }
public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        backtracking(candidates, target, 0);
        return res;
    }
};