recurse(i, j) that takes indices i and j as arguments.j is equal to the length of t, return 1 (base case).i is equal to the length of s, return 0 (base case).recurse(i, j) is already in the memoization dictionary. If yes, return the cached result.s[i] matches t[j], recursively call recurse(i + 1, j + 1) to move both pointers forward and add the result to recurse(i + 1, j) where only the i pointer is moved.s[i] does not match t[j], recursively call recurse(i + 1, j).recurse(0, 0) to start the recursion with both pointers at the beginning of s and t.