Added Backtracking pattern

✅ Pattern 17: BackTracking.md

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+
+
+Backtracking used for to get all possible solutions.
+
+```
+void backtrack(arguments) {
+	if (condition == true) { // Condition when we should stop our exploration.
+		result.push_back(current);
+		return;
+	}
+	for (int i = num; i <= last; i++) {
+		current.push_back(i); // Choose candidate.
+		backtrack(arguments); // Explore more
+		current.pop_back();   // Remove candidate.
+	}
+}
+```
+**Popular Questions**
+
+ 1. [All Possible Combinations](https://www.interviewbit.com/problems/all-possible-combinations/)
+ 2. [Subset](https://www.interviewbit.com/problems/subset/)
+ 3. [Combination Sum](https://www.interviewbit.com/problems/combination-sum/)
+ 4. [Combination Sum 2](https://www.interviewbit.com/problems/combination-sum-ii/)
+ 5. [Combinations](https://www.interviewbit.com/problems/combinations/)
+ 6. [Subset2](https://www.interviewbit.com/problems/subsets-ii/)
+
+**All Possible Combinations**
+
+Input 1:
+
+```
+A = ['ab', 'cd']
+```
+**Example Output**  
+
+Output 1:
+
+```
+['ac', 'ad', 'bc', 'bd']
+```
+CODE -- ACCORDING TO OUR PATTERN
+![image](https://github.com/user-attachments/assets/f00d62aa-1a01-4080-9dfb-c872e5e6b663)
+
+
+**SUBSET**
+**Example Input**  
+
+A = [1, 2, 3]
+
+
+
+**Example Output**  
+
+[  
+ [],  
+ [1],  
+ [1, 2],  
+ [1, 2, 3],  
+ [1, 3],  
+ [2],  
+ [2, 3],  
+ [3],  
+]
+
+![image](https://github.com/user-attachments/assets/96978d2d-5be0-4b1b-b243-507484208465)
+
+**Combination Sum**
+
+**Example Input**  
+
+Input 1:
+
+A = [2, 3]
+B = 2
+
+Input 2:
+
+A = [2, 3, 6, 7]
+B = 7
+
+**Example Output**  
+
+Output 1:
+
+[ [2] ]
+
+Output 2:
+
+[ [2, 2, 3], [7] ]
+
+![image](https://github.com/user-attachments/assets/8d5d7fdd-b505-4cde-aa50-12084f495ae8)
+
+**Combination Sum2**
+
+Only diff in combination sum and this problem is duplicate not allowed so we move to next index asap
+
+Input and output
+
+![image](https://github.com/user-attachments/assets/ca4e43aa-ee52-4837-b970-4bea3e0718cf)
+
+![image](https://github.com/user-attachments/assets/efe1c936-ba27-4522-925b-eec376c6fd59)
+
+**SubSet2**
+
+The only difference between this and the subset problem is, here we store results in set to avoid duplicates while in the set question we used ArrayList to get all subsets including duplicates
+
+Input
+![image](https://github.com/user-attachments/assets/6eb69ee9-7807-44e7-a5f9-a6f5e82ca8db)
+
+![image](https://github.com/user-attachments/assets/1c500ee1-1200-4d9a-9d44-2aca157c7701)
+
+
+**Letter Phone**
+
+Input
+
+![image](https://github.com/user-attachments/assets/8568e54f-6068-4f5f-823d-c1e8a30b3678)
+
+Output
+
+Used All Possible Sum Approach
+
+![image](https://github.com/user-attachments/assets/e2ee6556-1c43-4797-8f61-dad21457162b)
+
+**Permutation**
+
+Input
+
+![image](https://github.com/user-attachments/assets/44adba86-10e0-4adc-9af9-9c1220b3ab85)
+
+Output
+
+![image](https://github.com/user-attachments/assets/2a8d4705-41ad-40c5-8904-991549d2e877)
+
+**N QUEEN**
+
+![image](https://github.com/user-attachments/assets/0fc7b275-9e0b-4a57-84c1-63ae5a1efb62)
+
+Output
+
+![image](https://github.com/user-attachments/assets/d408d73c-520f-40ad-a029-ffbafed4364a)
+
+
+
+
+
+
+
+

✅ Pattern 18:Graph.md

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+https://leetcode.com/discuss/study-guide/3903992/graph-its-implementation-and-some-popular-bfsdfs-problems
+
+**BFS**
+
+```
+import java.util.*;
+
+class Solution {
+    public List<Integer> bfsOfGraph(int n, List<List<Integer>> adj) {
+        // Create a visited array to keep track of visited nodes
+        boolean[] vis = new boolean[n];
+
+        // 1. Pick a starting node, push it into the queue, and mark it as visited.
+        Queue<Integer> q = new LinkedList<>();
+        vis[0] = true;
+        q.add(0);
+        List<Integer> ans = new ArrayList<>();
+
+        // 4. Repeat steps 2 and 3.
+        while (!q.isEmpty()) {
+
+            // 2. In every iteration, pop out the 'x' node and put it in the solution list.
+            int node = q.poll();
+            ans.add(node);
+
+            // 3. All the unvisited adjacent nodes from 'x' are pushed into the queue.
+            for (int a : adj.get(node)) {
+                if (!vis[a]) {
+                    vis[a] = true;
+                    q.add(a);
+                }
+            }
+        }
+        return ans;
+    }
+
+
+```
+
+**DFS**
+
+```
+class Solution {
+  public:
+
+    void dfs(int node, vector<int>adj[], vector<bool>&vis, vector<int>& ans) {
+        vis[node] = 1;
+        ans.push_back(node);
+
+        for(auto a:adj[node]) {
+            if(!vis[a]) {
+                dfs(a,adj,vis,ans);
+            }
+        }
+    }
+
+    // Function to return a list containing the DFS traversal of the graph.
+    vector<int> dfsOfGraph(int n, vector<int> adj[]) {
+        vector<bool>vis(n,0);
+        int s = 0;
+        vector<int>ans;
+        dfs(s, adj, vis, ans);
+        return ans;
+    }
+};
+```