completed Backtracking-1

problem1.cpp

@@ -0,0 +1,38 @@
+// Time Complexity : O(2^(m+n)), n = target, m = size of candidates array
+// Space Complexity : O(n)
+// Did this code successfully run on Leetcode : yes
+// Any problem you faced while coding this : no 
+
+// Your code here along with comments explaining your approach
+// Go over all combinations recursively
+// AT every step either pick or no pick the number
+// Store all valid combos which result in target sum
+
+class Solution {
+private:
+    void helper(vector<int>& candidates, int target, int idx, vector<int>& path, vector<vector<int>>& result) {
+        // base:
+        if(target < 0 || idx == candidates.size()) {
+            return;
+        } 
+        if(target == 0) {
+            result.push_back(path);
+            return;
+        }
+        // logic:
+        // pick
+        path.push_back(candidates[idx]);
+        helper(candidates, target - candidates[idx], idx, path, result);
+        path.pop_back();
+        // no pick
+        helper(candidates, target, idx + 1, path, result);
+
+    }
+public:
+    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
+        vector<vector<int>> result;
+        vector<int> path;
+        helper(candidates, target, 0, path, result);
+        return result;
+    }
+};

problem2.cpp

@@ -0,0 +1,48 @@
+// Time Complexity : O(4^n)
+// Space Complexity : O(n)
+// Did this code successfully run on Leetcode : yes
+// Any problem you faced while coding this : no 
+
+// Your code here along with comments explaining your approach:
+// Explore all the ways to insert the operators between the digits
+// Maintain current total, last number and add to result when expression = target
+// Special case: * -> since * has higher precedence, backtrack the last operation 
+// For *, we backtrack the last addition using tail and recalculate properly because * has higher precedence.
+
+class Solution {
+private:
+    void helper(string num, int pivot, long calc, long tail, string path, int target, vector<string>& result) {
+        // base:
+        if (pivot == num.length()) {
+            if (calc == target) {
+                result.push_back(path);
+            }
+            return;
+        }
+
+        // logic:
+        for (int i = pivot; i < num.length(); i++) {
+            if (num[pivot] == '0' && i > pivot) break;
+
+            long curr = stol(num.substr(pivot, i - pivot + 1));
+
+            if (pivot == 0) {
+                helper(num, i + 1, curr, curr, to_string(curr), target, result);
+            } else {
+                // +
+                helper(num, i + 1, calc + curr, curr, path + "+" + to_string(curr), target, result);
+                // -
+                helper(num, i + 1, calc - curr, -curr, path + "-" + to_string(curr), target, result);
+                // *
+                helper(num, i + 1, calc - tail + tail * curr, tail * curr, path + "*" + to_string(curr), target, result);
+            }
+        }
+    }
+
+public:
+    vector<string> addOperators(string num, int target) {
+        vector<string> result;
+        helper(num, 0, 0, 0, "", target, result);
+        return result;
+    }
+};