3_combinationSumI.md

Combination Sum

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.

Backtracking

• for a number we have 2 choices either:
  • we choose it
  • we don't choose it
• if current element is less than target we can choose it.
• else we can't choose it.
• Base case arises when target is 0, we add the vector in answer.

Code-1

class Solution {
private:
    void combinationSumBack(vector<int>& candidates, int target, int i, vector<vector<int>>& ans, vector<int>& temp)
    {
        if (target == 0) {
            ans.push_back(temp);
            return;
        }
        while (i < candidates.size() && candidates[i] <= target) {
            temp.push_back(candidates[i]);
            combinationSumBack(candidates, target - candidates[i], i, ans, temp);
            temp.pop_back();
            i++;
        }
    }

public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target)
    {
        int n = candidates.size();
        sort(candidates.begin(),candidates.end());
        vector<vector<int>> ans;
        vector<int> temp = {};
        combinationSumBack(candidates, target, 0, ans, temp);
        return ans;
    }
};

Code-2 (100%-AC)

class Solution {
private:
    void combinationSumBack(vector<int>& candidates, int target, int i, vector<vector<int>>& ans, vector<int>& temp)
    {
        if (target == 0) {
            ans.push_back(temp);
            return;
        }
        if (i >= candidates.size())
            return;
        if (candidates[i] <= target) {
            temp.push_back(candidates[i]);
            combinationSumBack(candidates, target - candidates[i], i, ans, temp);
            temp.pop_back();
        }
        combinationSumBack(candidates, target, i + 1, ans, temp);
    }

public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target)
    {
        int n = candidates.size();
        vector<vector<int>> ans;
        vector<int> temp = {};
        combinationSumBack(candidates, target, 0, ans, temp);
        return ans;
    }
};

Code - 3

class Solution {
private:
    void backtrack(vector<int>& cand, int i, int target, vector<int>& temp, vector<vector<int>>& ans)
    {
        if (target == 0) {
            ans.push_back(temp);
            return;
        }
        if (i == cand.size() || target < 0)
            return;

        temp.push_back(cand[i]);
        backtrack(cand, i, target - cand[i], temp, ans);
        temp.pop_back();
        backtrack(cand, i + 1, target, temp, ans);
    }

public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target)
    {
        int n = candidates.size();
        vector<vector<int>> ans;
        vector<int> temp = {};
        sort(candidates.begin(), candidates.end());
        backtrack(candidates, 0, target, temp, ans);
        return ans;
    }
};

MUST READ:

• A general approach to backtracking questions in Java (Subsets, Permutations, Combination Sum, Palindrome Partitioning)
  • (1) Why use sort() & (2) Why pass start arugument & (3) what is tempList.remove(tempList.size() - 1) used for?