feat: alternative AC solution and TLE solution implemented, also wrote the tutorial for the problem and improved its description

flowers/src/alternative_ac.cpp

@@ -0,0 +1,45 @@
+#include <bits/stdc++.h>
+typedef long long ll;
+using namespace std;
+
+int N, M;
+
+const int MOD = 1e9 + 7;
+
+ll memo[10001][1001][2];
+
+ll flowers(int last, int count = 1, int size = 1) {
+    if (size == N) return 1;
+    /*
+        if sequences ends with X we have to cases:
+            if count(X) < M we can continue the sequence by appending another X in its end
+            we can also start and entirely new sequence starting with Y.
+    */
+   if (memo[size][count][last] == 0) {
+       if (last == 0) {
+           if (count < M)
+               memo[size][count][0] += flowers(0, count + 1, size + 1);
+           memo[size][count][0] += flowers(1, 1, size + 1);
+       } else {
+           if (count < M)
+               memo[size][count][1] += flowers(1, count + 1, size + 1);
+           memo[size][count][1] += flowers(0, 1, size + 1);     
+       }
+   }
+
+    return memo[size][count][last] %= MOD;
+}
+
+int main(){
+   cin >> N >> M;
+
+    for (int i = 0; i <= 10000; i++) {
+        for (int j = 0; j <= 1000; j++) {
+            memo[i][j][0] = memo[i][j][1] = 0;
+        }
+    }
+
+    cout << (flowers(0) + flowers(1)) % MOD << endl;
+
+   return 0;
+}