diff --git a/05- May/13- Count Ways To Build Good Strings/13- Count Ways To Build Good Strings (Ahmed Gamal).cpp b/05- May/13- Count Ways To Build Good Strings/13- Count Ways To Build Good Strings (Ahmed Gamal).cpp
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+// Author: Ahmed Gamal
+
+// for this problem we will use dp (memoization)
+// the idea will be the same as the iterative solution, you can see the iterative solution for more details about the idea
+// we will use dp[i] to be the number of good strings we can build from i to n (n is any number in the range [low, high])
+
+class Solution {
+public:
+    int countGoodStrings(int low, int high, int zero, int one) {
+        // MOD is the modulo we will use to avoid overflow
+        // memo is the dp array we will use to store the number of good strings we can build from i to n
+        const int MOD = 1e9 + 7;
+        vector<int> memo(high + 5, -1);
+        
+        // add is a lambda function to add two numbers and take the modulo
+        auto add = [&](int& a, int b) -> void {
+            a += b;
+            if(a >= MOD) {
+                a -= MOD;
+            }
+        };
+        
+        // dp is a lambda function to calculate the number of good strings we can build from i to n
+        auto dp = [&](auto dp, int len) -> int {
+            // if we reach the end of the string, we will return 1 if the length is in the range [low, high] and 0 otherwise
+            if(len >= high) {
+                return len == high;
+            }
+
+            // if we already calculated the number of good strings we can build from i to n, we will return it
+            int& ret = memo[len];
+            if(~ret) {
+                return ret;
+            }
+
+            // ret will be the number of good strings we can build from i to n
+            // before we add the number of good strings we can build from i to n, we will check if the length is in the range [low, high]
+            // if the length is in the range, then we have a good string, so we will add 1 to the number of good strings we can build and continue building the rest of it
+            ret = len >= low;
+
+            // we will try to add a zero and a one to the string and continue building the rest of it
+            add(ret, dp(dp, len + zero));
+            add(ret, dp(dp, len + one));
+
+            return ret;
+        };
+
+        // we will start building the string from the first character
+        return dp(dp, 0);
+    }
+};
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