39

39. Combination Sum (Medium)

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

Companies:
Facebook, Amazon, Airbnb, Microsoft, Apple, Snapchat, Adobe, Goldman Sachs, Salesforce

Related Topics:
Array, Backtracking

Similar Questions:

• Letter Combinations of a Phone Number (Medium)
• Combination Sum II (Medium)
• Combinations (Medium)
• Combination Sum III (Medium)
• Factor Combinations (Medium)
• Combination Sum IV (Medium)

Solution 1. Backtracking

// OJ: https://leetcode.com/problems/combination-sum
// Author: github.com/lzl124631x
// Time: O(N^(T/M+1)) where N is the length of A, T is target and M is the minimum value of A[i]
// Space: O(T/M)
class Solution {
public:
    vector<vector<int>> combinationSum(vector<int>& A, int target) {
        vector<vector<int>> ans;
        vector<int> tmp;
        function<void(int, int)> dfs = [&](int i, int target) {
            if (target == 0) {
                ans.push_back(tmp);
                return;
            }
            if (i == A.size()) return;
            int cnt = 0;
            do {
                dfs(i + 1, target);
                tmp.push_back(A[i]);
                target -= A[i];
                ++cnt;
            } while (target >= 0);
            while (cnt-- > 0) tmp.pop_back();
        };
        dfs(0, target);
        return ans;
    }
};

Solution 2. Backtracking

// OJ: https://leetcode.com/problems/combination-sum
// Author: github.com/lzl124631x
// Time: O(N^(T/M+1)) where N is the length of A, T is target and M is the minimum value of A[i]
// Space: O(T/M)
class Solution {
public:
    vector<vector<int>> combinationSum(vector<int>& A, int target) {
        vector<vector<int>> ans;
        vector<int> tmp;
        function<void(int, int)> dfs = [&](int start, int target) {
            if (target == 0) {
                ans.push_back(tmp);
            } else if (target < 0) return;
            for (int i = start; i < A.size(); ++i) {
                tmp.push_back(A[i]);
                dfs(i, target - A[i]);
                tmp.pop_back();
            }
        };
        dfs(0, target);
        return ans;
    }
};

Or

// OJ: https://leetcode.com/problems/combination-sum/
// Author: github.com/lzl124631x
// Time: O(N^(T/M+1)) where N is the length of A, T is target and M is the minimum value of A[i]
// Space: O(T/M)
// Ref: https://discuss.leetcode.com/topic/14654/accepted-16ms-c-solution-use-backtracking-easy-understand
class Solution {
public:
    vector<vector<int>> combinationSum(vector<int>& A, int target) {
        vector<vector<int>> ans;
        vector<int> tmp;
        sort(begin(A), end(A)); // sorting is a must for this algorithm
        function<void(int, int)> dfs = [&](int start, int target) {
            if (target == 0) {
                ans.push_back(tmp);
            }
            for (int i = start; i < A.size() && target - A[i] >= 0; ++i) {
                tmp.push_back(A[i]);
                dfs(i, target - A[i]);
                tmp.pop_back();
            }
        };
        dfs(0, target);
        return ans;
    }
};