39.cpp added

39.cpp

@@ -0,0 +1,55 @@
+//https://leetcode.com/problems/combination-sum/
+//Difficulty Level: Medium
+//Tags: Array, Backtracking
+//since we have to print all, and not just count or max or min, we won't use DP but backtracking
+
+class Solution {
+public:
+    vector<vector<int>> combinationSum(vector<int>& candidates, int target) 
+    {
+        sort(candidates.begin(), candidates.end());
+
+        //Note that we don't have duplicates in the array, else we had to delete the duplicates here bcoz 
+        //we can already take each element multiple times, so duplicate elements don't make a difference
+
+        int index = candidates.size()-1;
+        while(index >=0 && candidates[index] > target)
+        {
+            index--;
+        }
+
+        vector<int> v;
+        vector<vector<int>> res;  //stores result
+        backtrack(candidates, target, 0, index, v, res);
+        return res;
+    }
+
+    void backtrack(vector<int> candidates, int target, int curr_sum, int index, vector<int> v, vector<vector<int>>& res)
+    {
+        if(curr_sum == target)     //if the sum of elements of v add up to target, push v to result vector
+        {
+            res.push_back(v);
+            return;
+        }
+
+        //check all the elements <= target - curr_sum
+        for(int i=index; i>=0; i--)
+        {
+            curr_sum += candidates[i];
+
+            if(curr_sum > target)    //don't include the element if sum is exceeding the target
+            {
+                curr_sum -= candidates[i];
+                continue;
+            }
+            v.push_back(candidates[i]);
+
+            //backtrack to find rest of the elements of v
+            //note that we have passed 'i' and not 'i+1' since we could include the same element any no. of times
+            backtrack(candidates, target, curr_sum, i, v, res);   
+
+            curr_sum -= candidates[i];
+            v.pop_back();
+        }
+    }
+};