From 9f3c1e7ac130d903fa580d81e5b5a615b652be4b Mon Sep 17 00:00:00 2001
From: Samiksha Manjunath <ssamma@amazon.com>
Date: Tue, 12 May 2026 10:18:05 -0700
Subject: [PATCH] Done Backtracking-1

---
 combination_sum.java         | 41 ++++++++++++++++++
 expression_add_operator.java | 80 ++++++++++++++++++++++++++++++++++++
 2 files changed, 121 insertions(+)
 create mode 100644 combination_sum.java
 create mode 100644 expression_add_operator.java

diff --git a/combination_sum.java b/combination_sum.java
new file mode 100644
index 00000000..0138d4d0
--- /dev/null
+++ b/combination_sum.java
@@ -0,0 +1,41 @@
+// Time Complexity : O(2^(m + n)) where m is candidates length and n is target
+// 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 in three sentences only
+// We use backtracking to explore two choices at every step: either include the current candidate or skip it and move to the next one.
+// If the target becomes 0, we add a deep copy of the current path to the result since it forms a valid combination.
+// Backtracking removes the last chosen element after recursion so other possible combinations can be explored correctly.
+
+class Solution {
+    List<List<Integer>> result;
+    public List<List<Integer>> combinationSum(int[] candidates, int target) {
+        this.result = new ArrayList<>();
+        helper(candidates, target, 0, new ArrayList<>());
+        return result;    
+    }
+
+    private void helper(int[] candidates, int target, int index, List<Integer> path){
+        // Base case
+        if (target < 0 || index == candidates.length) return;
+
+        // Always return a deep copy
+        if (target == 0) {
+            result.add(new ArrayList<>(path));
+            return;
+        }
+
+        // don't choose
+        helper(candidates, target, index+1, path);
+
+        // choose
+        path.add(candidates[index]);
+        helper(candidates, target-candidates[index], index, path);
+
+        //backtrack
+        path.remove(path.size()-1);
+            
+    }
+}
\ No newline at end of file
diff --git a/expression_add_operator.java b/expression_add_operator.java
new file mode 100644
index 00000000..fbd9bfe7
--- /dev/null
+++ b/expression_add_operator.java
@@ -0,0 +1,80 @@
+// Time Complexity : O(4^n) cos 4 operations - noop, +, -, *
+// 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 in three sentences only
+// We use backtracking to try every possible number split and insert '+', '-', or '*' between numbers.
+// The tail variable stores the previous operand so multiplication precedence can be handled by removing tail and adding tail * curr.
+// We skip numbers with leading zeros and backtrack the StringBuilder after each recursive call to explore other expressions.
+
+class Solution {
+    List<String> result;
+    public List<String> addOperators(String num, int target) {
+        this.result = new ArrayList<>();
+        helper(num, target, 0, 0l, 0l, new StringBuilder());
+        return result;
+        
+    }
+
+    private void helper(String num, int target, int pivot, long calculated_value, long tail, StringBuilder path) {
+        // Base case
+        if (pivot == num.length()) {
+            if (calculated_value == target) {
+                result.add(path.toString());
+            }
+            return;
+        }
+
+        // Logic
+        for (int i = pivot; i < num.length(); i++) {
+            Long curr = Long.parseLong(num.substring(pivot, i+1));
+
+            // Preceding zero
+            if(num.charAt(pivot) == '0' && pivot!=i) {
+                continue;
+            }
+
+            int le = path.length();
+            // Top level
+            if (pivot == 0) {
+                // action
+                path.append(curr);
+                // recurse
+                helper(num, target, i+1, curr, curr, path);
+                //backtrack
+                path.setLength(le);
+            } else {
+                // +
+                // action
+                path.append("+");
+                path.append(curr);
+                // recurse
+                helper(num, target, i+1, calculated_value+curr, curr, path);
+                //backtrack
+                path.setLength(le);
+
+                // -
+                // action
+                path.append("-");
+                path.append(curr);
+                // recurse
+                helper(num, target, i+1, calculated_value-curr, -curr, path);
+                //backtrack
+                path.setLength(le);
+
+                // *
+                // action
+                path.append("*");
+                path.append(curr);
+                // recurse
+                helper(num, target, i+1, calculated_value-tail + curr*tail, curr*tail, path);
+                //backtrack
+                path.setLength(le);
+            }
+        }
+        
+
+    }
+}
\ No newline at end of file