Backtracking-1 Done

CombinationSum.java

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+/*
+Time Complexity : O(2^T), T is target value
+Space Complexity : O(T)
+Did this code successfully run on Leetcode : yes
+Any problem you faced while coding this : no
+Approach :
+We need to use recursion with backtracking to find all possible combinations that sums to target. In 0/1 recursion,
+we add a number to the path and reduce the target, if the target become equal to 0, it is valid combination and is added to result.
+If not, remove the number from the path after recursion is done and recurse without choose this number.
+*/
+import java.util.ArrayList;
+import java.util.List;
+
+public class CombinationSum {
+    public List<List<Integer>> combinationSum(int[] candidates, int target) {
+        List<List<Integer>> result = new ArrayList<>();
+        combinationRecursion(candidates, target, 0, new ArrayList<>(), result);
+      //  combination01Recursion(candidates, target, 0, new ArrayList<>(), result);
+        return result;
+    }
+
+    private void combination01Recursion(int[] candidates, int target,int index, ArrayList<Integer> path, List<List<Integer>> result) {
+        if(index == candidates.length) return;
+        if(target < 0) return;
+        if(target == 0){
+            result.add(new ArrayList<>(path));
+            return;
+        }
+        // not choose
+        combination01Recursion(candidates, target, index+1, path, result);
+
+        // choose
+        path.add(candidates[index]);
+        combination01Recursion(candidates, target-candidates[index], index, path, result);
+        path.remove(path.size()-1);
+    }
+
+    // Time Complexity : O(N^D), where N is size of candidates array and D is depth of tree, could be T in worst case O(N^T)
+    // Space Complexity : O(T),
+    private void combinationRecursion(int[] candidates, int target, int pivot, ArrayList<Integer> path, List<List<Integer>> result) {
+        if(pivot == candidates.length) return;
+        if(target < 0) return;
+        if(target == 0){
+            result.add(new ArrayList<>(path));
+            return;
+        }
+
+        for (int i = pivot; i < candidates.length; i++){
+            combinationRecursion(candidates, target, i+1, path, result);
+
+            path.add(candidates[i]);
+            combinationRecursion(candidates, target-candidates[i], i, path, result);
+            path.remove(path.size()-1);
+        }
+    }
+}

ExpressionAndOperators.java

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+/*
+Time Complexity : O(4^N), 4 choices between numbers no operator, +,- and *
+Space Complexity : O(N)
+Did this code successfully run on Leetcode : yes
+Any problem you faced while coding this : no
+Approach :
+Main objective is to first create digits and numbers out of the number string. We will need to use recursion for that.
+We can sub-string the number recursively and create valid combination.
+
+After the digits are created, we need to evaluate them by putting operators between them +,- and *.
+We calculate the value in each recursion call and keep pass it to the next child. If the entire number is evaluated and
+if the calculated is equal to the target, we add the expression to the result.
+
+When evaluating a path, we will need to back track the last value and the operator that got added, so we will need to
+maintain tail variable to identify the last operator/expression appended.
+operator    tail        calculated
++           +curr       calculated+curr
+-           -curr       calculated-curr
+*           tail*curr   calculated-tail+tail*curr
+/           tail/curr   calculated-tail+tail/curr
+
+CAVEATS:
+1. At index = 0, we cant use an operator(+4, -4, *4) of the expression, so, we just call recursion on the next index.
+2. Preceding 0 problem, "05" is not valid, "10" is valid.
+*/
+
+import java.util.ArrayList;
+import java.util.List;
+
+public class ExpressionAndOperators {
+    List<String> result = new ArrayList<>();
+
+    public List<String> addOperators(String num, int target) {
+        StringBuilder sb = new StringBuilder();
+        findExpression(0, num, target, 0, 0, sb);
+        return result;
+    }
+
+    void findExpression(int index, String num, int target, long calculated, long tail, StringBuilder sb) {
+        int length = num.length();
+        if (length == index) {
+            if (target == calculated) {
+                result.add(sb.toString());
+                return;
+            }
+        }
+
+        int len = sb.length();
+        for (int i = index; i < length; i++) {
+            // To solve the preceding zero problem
+            if(num.charAt(index) == '0' && i > index)  break;
+            long curr = Long.parseLong(num.substring(index, i + 1));
+
+            if (index == 0) {
+                sb.append(curr);
+                // At this point curr is just number, without any operator
+                findExpression(i+1, num, target, curr, curr, sb);
+                sb.setLength(len);
+            } else {
+                sb.append("+"+curr);
+                findExpression(i + 1, num, target, calculated + curr, curr, sb);
+                sb.setLength(len);
+
+                sb.append("-"+curr);
+                findExpression(i + 1, num, target, calculated - curr, -curr, sb);
+                sb.setLength(len);
+
+                sb.append("*"+curr);
+                findExpression(i + 1, num, target, calculated - tail + (tail * curr), tail * curr, sb);
+                sb.setLength(len);
+            }
+        }
+    }
+}