feat: new problem formatted

edit-distance/src/ac.cpp

@@ -0,0 +1,48 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+
+int memo[510][510];
+
+int solve(string &w1, string &w2, int i = 0, int j = 0)
+{
+   // all letters of w1 processed, we must add the remaining letters in w2
+   if (i >= w1.size())
+   {
+      return w2.size() - j;
+   }
+   // all letters of w2 processed, we must remove the left over letter in w1
+   if (j >= w2.size())
+   {
+      return w1.size() - i;
+   }
+
+   if (memo[i][j] != -1)
+   {
+      return memo[i][j];
+   }
+
+   // if letters are the same we just need to find the min number of operations
+   // for the remaining letters in w1 and w2
+   if (w1[i] == w2[j])
+   {
+      return memo[i][j] = solve(w1, w2, i + 1, j + 1);
+   }
+
+   // if letters are not the same we consider three cases
+   // 1. Deleting the current letter in w1
+   // 2. Replacing the current letter in w1 to be the same as the letter in w2
+   // 3. Inserting the current letter of w2 in w1 and continue with the same letters in w1
+   return memo[i][j] = min({solve(w1, w2, i + 1, j), solve(w1, w2, i + 1, j + 1), solve(w1, w2, i, j + 1)}) + 1;
+}
+
+int main()
+{
+   int n, m;
+   cin >> n >> m;
+   string w1, w2;
+   cin >> w1 >> w2;
+   memset(memo, -1, sizeof(memo));
+   cout << solve(w1, w2) << endl;
+   return 0;
+}