diff --git a/05- May/13- Count Ways To Build Good Strings/13- Count Ways To Build Good Strings (Mahmoud Aboelsoud).cpp b/05- May/13- Count Ways To Build Good Strings/13- Count Ways To Build Good Strings (Mahmoud Aboelsoud).cpp
new file mode 100644
index 000000000..4dcd1b7bf
--- /dev/null
+++ b/05- May/13- Count Ways To Build Good Strings/13- Count Ways To Build Good Strings (Mahmoud Aboelsoud).cpp	
@@ -0,0 +1,49 @@
+// Author: Mahmoud Aboelsoud
+
+class Solution {
+public:
+    // we need to count the number of ways to build a string of length between low and high
+    // we will use dp to count the number of ways to build a string of length n
+    // we will use a recursive function to count the number of ways to build a string of length n
+    // the recursive function will return 0 if n > high
+
+
+    // dp: dp[i] -> number of ways to build a string of length i
+    // low: the minimum length of the string
+    // high: the maximum length of the string
+    // zero: the number of zeros you can append to the string
+    // one: the number of ones you can append to the string
+    int dp[100005], low, high, zero, one, mod = 1e9 + 7;
+
+    // recursive function to count the number of ways to build a string of length n
+    int cnt_ways(int n){
+        // if n > high return 0
+        if(n > high) return 0;
+
+        // if n is calculated before return the value
+        if(dp[n] != -1) return dp[n];
+
+        // if n is in the range [low, high] then add 1 to the answer
+        int ans = (n >= low);
+
+        // add the number of ways to build a string of length n + zero
+        ans += cnt_ways(n + zero);
+        ans %= mod;
+        // add the number of ways to build a string of length n + one
+        ans += cnt_ways(n + one);
+        ans %= mod;
+
+        // return the answer
+        return dp[n] = ans;
+    }
+
+    int countGoodStrings(int low, int high, int zero, int one){
+        // initialize dp with -1
+        memset(dp, -1, sizeof(dp));
+        // initialize the global variables    
+        this -> low = low, this -> high = high, this -> zero = zero, this -> one = one;
+
+        // return the number of ways to build a string of length between low and high starting from length 0
+        return cnt_ways(0);
+    }
+};
diff --git a/05- May/14- Maximize Score After N Operations/14- Maximize Score After N Operations (Ahmed Hossam).cpp b/05- May/14- Maximize Score After N Operations/14- Maximize Score After N Operations (Ahmed Hossam).cpp
new file mode 100644
index 000000000..5ce3a0d5c
--- /dev/null
+++ b/05- May/14- Maximize Score After N Operations/14- Maximize Score After N Operations (Ahmed Hossam).cpp	
@@ -0,0 +1,49 @@
+// Author: Ahmed Hossam
+
+class Solution {
+public:
+
+    // declare variables
+    int n, full_mask;
+    // vectors to store numbers and dynamic programming results
+    vector<int> nums, dp;
+
+    // function to check if a bit is empty in a binary mask
+    bool is_empty_bit(int bit, int mask){
+        return !(mask & (1 << bit));
+    }
+
+    // recursive function to calculate the maximum score
+    int max_score(int mask){
+        // if all bits are filled, return 0
+        if(mask == full_mask) return 0;
+        // use memoization to avoid redundant computations
+        int& ret = dp[mask];
+        // calculate the current index based on the number of filled bits
+        int idx = __builtin_popcount(mask) / 2 + 1;
+        // if the result has already been calculated, return it
+        if(~ret) return ret;
+        ret = 0;
+        // iterate over all pairs of numbers
+        for(int i = 0; i < n; i++)
+            for(int j = i + 1; j < n; j++)
+                // if both numbers are empty, calculate the score
+                if(is_empty_bit(i, mask) && is_empty_bit(j, mask))
+                    ret = max(ret, (idx * __gcd(nums[i], nums[j])) + max_score(mask | (1 << i) | (1 << j)));
+        // return the maximum score
+        return ret;
+    }
+
+    int maxScore(vector<int>& nums) {
+        // set the number of elements
+        n = nums.size();
+        // set the vector of numbers
+        this -> nums = nums;
+        // set the full binary mask
+        full_mask = (1 << n) - 1;
+        // initialize the dynamic programming vector with -1
+        dp = vector<int>(1 << n, -1);
+        // return the maximum score starting with an empty mask
+        return max_score(0);
+    }
+};
diff --git a/05- May/14- Maximize Score After N Operations/14- Maximize Score After N Operations (Lama Salah).cpp b/05- May/14- Maximize Score After N Operations/14- Maximize Score After N Operations (Lama Salah).cpp
new file mode 100644
index 000000000..51b2c0d31
--- /dev/null
+++ b/05- May/14- Maximize Score After N Operations/14- Maximize Score After N Operations (Lama Salah).cpp	
@@ -0,0 +1,56 @@
+// Author: Lama Salah
+
+class Solution {
+public:
+    // Declare the size of the list and memoization array.
+    int n, memo[8][100000]; 
+
+    int dp(int i, int mask, vector <int>& nums){
+        // If we've performed all operations, return 0 - (the base-case).
+        if (i > n) return 0; 
+
+        // Get a reference to the memoized result for this i and mask and check if the result is already memoized.
+        int& ans = memo[i][mask]; 
+        if (~ans) return ans; 
+
+        // Loop over all possible unremoved pairs.
+        for (int k = 0; k < 2*n; k++){
+            // Check if the k-th bit in the mask is 0 (the k-th integer has not been removed).
+            if (!((mask >> k) & 1)){ 
+                // Set the k-th bit in the mask to 1 (remove the k-th integer).
+                mask |= (1 << k); 
+             
+                // Loop over all possible unremoved integers.
+                for (int l = 0; l < 2*n; l++){
+                    // check If the l-th bit in the mask is 0 (the l-th integer has not been removed).
+                    if (!((mask >> l) & 1)){ 
+
+                        // Set the l-th bit in the mask to 1 (remove the l-th integer).
+                        mask |= (1 << l); 
+
+                        // Compute the score for using the k-th and l-th integers.
+                        ans = max(ans, (i* gcd(nums[k], nums[l])) + dp(i+1, mask, nums));
+                        
+                        // Unset the l-th bit in the mask.
+                        mask ^= (1 << l);
+                    }
+                }
+
+                // Unset the k-th bit in the mask.
+                mask ^= (1 << k);
+            }
+        }
+
+        return ans; 
+    }
+
+    int maxScore(vector<int>& nums) {
+        this -> n = nums.size()/2;
+
+        // Initialize the memo array with -1 values using the memset() function.
+        memset(memo, -1, sizeof memo);
+        
+        // Compute the maximum score after N operations.
+        return dp(1, 0, nums); 
+    }
+};
diff --git a/05- May/14- Maximize Score After N Operations/14- Maximize Score After N Operations (Mohamed Emara).cpp b/05- May/14- Maximize Score After N Operations/14- Maximize Score After N Operations (Mohamed Emara).cpp
new file mode 100644
index 000000000..879868b30
--- /dev/null
+++ b/05- May/14- Maximize Score After N Operations/14- Maximize Score After N Operations (Mohamed Emara).cpp	
@@ -0,0 +1,64 @@
+
+// Author: MohamedEmara
+
+class Solution {
+public:
+   
+    int n;
+    int dp[20][20000];
+    int arr[20];
+
+    int rec(int idx, int mask)
+    {
+        // the base case WHEN all bits are set to one
+        if(mask == (1<<(2*n))-1)
+            return 0;
+
+        int &ret = dp[idx][mask];
+        if(~ret)
+            return ret;
+        
+
+        ret = 0;
+
+        for(int i=0; i<2*n; i++)
+        {
+            for(int j=i+1; j<2*n; j++)
+            {
+                // if the two bits i , j not taken yet in the mask
+                // try to take them and maximize the value.
+                if(!(mask & (1 << j)) && !(mask & (1 << i)))
+                {
+                    // set two bits of one ie: add the value of these two bits
+                    int tmp = mask + (1 << j) + (1 << i);
+                    ret = fmax(ret, idx * __gcd(arr[i], arr[j]) + rec(idx+1, tmp));
+                }
+            }
+        }
+
+        return ret;
+    }
+
+
+    int maxScore(vector<int>& nums) {
+         ios_base::sync_with_stdio(false);
+cin.tie(NULL);
+
+
+        n = nums.size() / 2;
+        
+        // At every idx(ith operation)
+        // We will try to choose two distinct numbers
+        // An save the taken elements in a mask that is intially all zeros
+        // the two taken element --> set their bits to one and continue trying other elements
+        // till the mask becoms all ones of lenght 2*N  -->  (1 << (2*n) - 1)
+
+        memset(dp, -1, sizeof(dp));
+        for(int i=0; i<2*n; i++)
+            arr[i] = nums[i];
+
+        return rec(1, 0);
+    }
+};
+
+
diff --git a/05- May/14- Maximize Score After N Operations/14- Maximize Score After N Operations (Omar Sanad).cpp b/05- May/14- Maximize Score After N Operations/14- Maximize Score After N Operations (Omar Sanad).cpp
new file mode 100644
index 000000000..bcae90d85
--- /dev/null
+++ b/05- May/14- Maximize Score After N Operations/14- Maximize Score After N Operations (Omar Sanad).cpp	
@@ -0,0 +1,55 @@
+// author : Omar Sanad
+
+class Solution {
+public:
+    // declare the array nums in the class itslef, to use it anywhere in the class
+    vector < int > nums;
+
+    // declare a 2D array for the memoization
+    // declare the number of steps (n)
+    // declare the size of the array which equals 2 * n
+    int dp[15][1 << 14], n, the_sz;
+    int rec(int idx, int mask) {
+        
+        // if we've already done the n operations, then we return
+        if (idx > n)
+            return 0;
+
+        // assign the dp[idx][mask] to ret by reference to use it easily
+        int &ret = dp[idx][mask];
+
+        // if this state was already been calculated, then we return its answer
+        if (~ret)   return ret;
+
+        // otherwise we initialize the answer with 0
+        ret = 0;
+
+        // we iterate over the elements of the array, and for every two unvisited elements we try taking them for this step
+        for (int i = 0; i < the_sz; i++)
+            if ((mask & (1 << i)) == 0)
+                for (int j = i + 1; j < the_sz; j++)
+                    if ((mask & (1 << j)) == 0)
+                        // if we try to take nums[i], nums[j]
+                        // then we mark them as visited in the mask --> mask | (1 << j) | (1 << i)
+                        ret = max(ret, idx * gcd(nums[i], nums[j]) + rec(idx + 1, mask | (1 << j) | (1 << i)));
+
+
+        // return the answer for this state
+        return ret;
+    }
+    int maxScore(vector<int>& nums) {
+
+        // assign the passed array nums to the array nums which is declared inside the class itself
+        this->nums = nums;
+
+        this->the_sz = nums.size(); // the size of the array
+        this->n = the_sz / 2;       // the number of operations to be done
+        
+        // initialize the array dp with -1, in other words mark as not calculated
+        memset(dp, -1, sizeof(dp));
+
+
+        // return the answer to the problem
+        return rec(1, 0);
+    }
+};
diff --git a/05- May/README.md b/05- May/README.md
index c314c9784..0c09bf734 100644
--- a/05- May/README.md	
+++ b/05- May/README.md	
@@ -34,6 +34,7 @@
 1. **[Uncrossed Lines](#11--uncrossed-lines)**
 1. **[Solving Questions With Brainpower](#12--solving-questions-with-brainpower)**
 1. **[Count Ways To Build Good Strings](#13--count-ways-to-build-good-strings)**
+1. **[Maximize Score After N Operations](#14--maximize-score-after-n-operations)**
 
 <hr>
 <br><br>
@@ -727,4 +728,70 @@ public:
     }
 };
 ```
+    
+<hr>
+<br><br>
+
+## 14)  [Maximize Score After N Operations](https://leetcode.com/problems/maximize-score-after-n-operations/)
+
+### Difficulty
+
+![](https://img.shields.io/badge/Hard-red?style=for-the-badge)
+
+### Related Topic
+
+`Array` `Math` `Dynamic Programming` `Backtracking` `Bit Manipulation` `Number Theory` `Bitmask`
+
+### Code
+
+
+```cpp
+class Solution {
+public:
+
+    // declare variables
+    int n, full_mask;
+    // vectors to store numbers and dynamic programming results
+    vector<int> nums, dp;
+
+    // function to check if a bit is empty in a binary mask
+    bool is_empty_bit(int bit, int mask){
+        return !(mask & (1 << bit));
+    }
+
+    // recursive function to calculate the maximum score
+    int max_score(int mask){
+        // if all bits are filled, return 0
+        if(mask == full_mask) return 0;
+        // use memoization to avoid redundant computations
+        int& ret = dp[mask];
+        // calculate the current index based on the number of filled bits
+        int idx = __builtin_popcount(mask) / 2 + 1;
+        // if the result has already been calculated, return it
+        if(~ret) return ret;
+        ret = 0;
+        // iterate over all pairs of numbers
+        for(int i = 0; i < n; i++)
+            for(int j = i + 1; j < n; j++)
+                // if both numbers are empty, calculate the score
+                if(is_empty_bit(i, mask) && is_empty_bit(j, mask))
+                    ret = max(ret, (idx * __gcd(nums[i], nums[j])) + max_score(mask | (1 << i) | (1 << j)));
+        // return the maximum score
+        return ret;
+    }
+
+    int maxScore(vector<int>& nums) {
+        // set the number of elements
+        n = nums.size();
+        // set the vector of numbers
+        this -> nums = nums;
+        // set the full binary mask
+        full_mask = (1 << n) - 1;
+        // initialize the dynamic programming vector with -1
+        dp = vector<int>(1 << n, -1);
+        // return the maximum score starting with an empty mask
+        return max_score(0);
+    }
+};
+```
     
\ No newline at end of file