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Red John is Back
Red John is Back
+ 0 comments Here is my solution in java, javascript, python, c ,C++, Csharp HackerRank Red John is Back Problem Solution
+ 0 comments I have the least time execution dealing with prime numbers
def isprime(n): prime = [True for i in range(n + 1)] p = 2 while p * p <= n: if prime[p]: for i in range(p * 2, n + 1, p): prime[i] = False p += 1 prime_numbers = [] for p in range(2, n): if prime[p]: prime_numbers.append(p) return len(prime_numbers) def redJohn(n): m = 0 for i in range(n // 4 + 1): pl = n - (i * 3) m += int(math.factorial(pl) / (math.factorial(i) * math.factorial(pl - i))) return isprime(m+1)
+ 0 comments Here is Red John is Back problem solution - https://programs.programmingoneonone.com/2021/07/hackerrank-red-john-is-back-problem-solution.html
+ 0 comments public static int redJohn(int n) { int countWays = numberOfWays(n); int totalPrimes = sieve(countWays); return totalPrimes; } public static int numberOfWays(int N){ if (N<=3) return 1; if (N==4) return 2; return numberOfWays(N-1) + numberOfWays(N-4); } public static int sieve(int n) { int count = 0; boolean prime[] = new boolean[n + 1]; for (int i = 0; i <= n; i++) prime[i] = true; for (int p = 2; p * p <= n; p++) if (prime[p] == true) for (int i = p * 2; i <= n; i += p) prime[i] = false; for (int i = 2; i <= n; i++) if (prime[i] == true) count++; return count;
+ 0 comments public static int redJohn(int n) { int countWays = numberOfWays(n); int totalPrimes = sieve(countWays);
return totalPrimes; } public static int numberOfWays(int N){ if (N<=3) return 1; if (N==4) return 2; return numberOfWays(N-1) + numberOfWays(N-4); } public static int sieve(int n) { int count = 0; boolean prime[] = new boolean[n + 1]; for (int i = 0; i <= n; i++) prime[i] = true; for (int p = 2; p * p <= n; p++) if (prime[p] == true) for (int i = p * 2; i <= n; i += p) prime[i] = false; for (int i = 2; i <= n; i++) if (prime[i] == true) count++; return count; }
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