diff --git a/BackTracking/sudoku.cpp b/BackTracking/sudoku.cpp
new file mode 100644
index 0000000..60d9de4
--- /dev/null
+++ b/BackTracking/sudoku.cpp
@@ -0,0 +1,64 @@
+// Problem from GFG
+
+bool isSafe(int i, int j,int num,int grid[9][9]){
+    for(int p=0;p<9;p++){
+        if(grid[i][p]==num){
+            return false;
+        }
+        if(grid[p][j]==num){
+            return false;
+        }
+    }
+  
+    int s = sqrt(9);
+    i = i - (i%s);
+    j = j - (j%s);
+    for(int l =0;l<s;l++){
+        for(int m=0;m<s;m++){
+            if(grid[l+i][m+j]==num){
+                return false;
+            }
+        }
+    }
+    return true;
+}
+bool SolveSudoku(int grid[N][N])  
+{ 
+    bool flag = false;
+    int i,j;
+    for( i=0;i<N;i++){
+        for( j=0;j<N;j++){
+            if(grid[i][j]==0){
+                flag = true;
+                break;
+            }
+        }
+        if(flag){
+            break;
+        }
+    }
+    if(i==N && j==N){
+        return true;
+    }
+    for(int k=1;k<=N;k++){
+        if(isSafe(i,j,k,grid)){
+            grid[i][j] = k;
+            if(SolveSudoku(grid)){
+                 return true;
+            }
+               
+            grid[i][j] = 0;    
+        }
+    }
+ return false;   
+}
+
+void printGrid (int grid[N][N]) 
+{
+    for(int i=0;i<N;i++){
+        for(int j=0;j<N;j++){
+            cout<<grid[i][j]<<" ";
+        }
+    }
+   
+}
diff --git a/Dynamic_Programming/sum_of_substrings_of_a_given_num.cpp b/Dynamic_Programming/sum_of_substrings_of_a_given_num.cpp
new file mode 100644
index 0000000..004dcd0
--- /dev/null
+++ b/Dynamic_Programming/sum_of_substrings_of_a_given_num.cpp
@@ -0,0 +1,18 @@
+// Algorithm to find the sum of all substrings of a given number 
+// Example: Input: S = 1234 Output: 1670 Explanation: Sum = 1 + 2 + 3 + 4 + 12 + …
+
+long long sumSubstrings(string s){
+    const int p = 1000000007;
+    int n = s.length();
+    long long  dp[n];
+    dp[0] = (s[0]-'0');
+    long long sum=dp[0];
+    for(int i=1;i<n;i++){
+        dp[i] = ((i+1)*(s[i]-'0') + 10*dp[i-1])%p;
+        sum+=dp[i];
+        if(sum>p){
+            sum = (sum%p);
+        }
+    }
+    return sum;
+}
diff --git a/Hashing/Group_anagrams_together.cpp b/Hashing/Group_anagrams_together.cpp
new file mode 100644
index 0000000..e9979f2
--- /dev/null
+++ b/Hashing/Group_anagrams_together.cpp
@@ -0,0 +1,37 @@
+// Practice que from gfg
+// Given an array of words, print the count of all anagrams together in sorted order (increasing order of counts).
+// For example, if the given array is {“cat”, “dog”, “tac”, “god”, “act”}, then grouped anagrams are “(dog, god) (cat, tac, act)”. So the output will be 2 3.
+
+
+
+#include <iostream>
+#include <bits/stdc++.h>
+using namespace std;
+
+int main() {
+    int t;
+    cin>>t;
+    while(t--){
+        vector<int> v;
+        int N;
+        cin>>N;
+        string a[N];
+        for(int i=0;i<N;i++){
+            cin>>a[i];
+        }
+        unordered_map<string,int> um;
+        for(int i=0;i<N;i++){
+            sort(a[i].begin(),a[i].end());
+            um[a[i]]++;
+        }
+        for(auto x:um){
+            v.push_back(x.second);
+        }
+        sort(v.begin(),v.end());
+        for(int i=0;i<v.size();i++){
+            cout<<v[i]<<" ";
+        }
+        cout<<endl;
+    }
+	return 0;
+}