Hard Coding Question

Heading ; The Puzzle Masters of Puzzleville

Problem Statement : A young prodigy Zoey stumbled upon a mysterious challenge left by puzzle master. Arrange the pieces in unique combination that would sum up to a magic number provided by master. No piece can be used more than once. Find the number of unique combos that are possible.

Question

Hard Coding Question

Heading ; The Puzzle Masters of Puzzleville

Problem Statement : A young prodigy Zoey stumbled upon a  mysterious challenge left by puzzle master. Arrange the pieces in unique combination that would sum up to a magic number provided by master. No piece can be used more than once. Find the number of unique combos that are possible.

Puzzle Master · Accepted Answer

def a(b, c, d, e, f, g):
    if c == e:
        g.append(tuple(sorted(f)))
        return
    if c > e or b >= len(d):
        return

    a(b + 1, c + d[b], d, e, f + [d[b]], g)
    a(b + 1, c, d, e, f, g)

def h(i, j, k):
    l = []
    a(0, 0, k, j, [], l)
    
    m = set(l)
    return len(m)

# Example usage
i = 7
j = 8
k = [10, 1, 2, 7, 6, 1, 5]

n = h(i, j, k)
print(n)  # Expected Output: 4

Sumit · Answer

In c++

Vivek Kumar(vivekkumar002216@gmail.com) · Answer

1. The Puzzle Masters of Puzzleville by using the concept of dp we can easly solve ,basically the problem is asking about to find how many unique combinations of pieces sum up to a magic number K. #include #include using namespace std; int count_combinations(int N, int K, vector& pieces) { vector dp(K + 1, 0); dp[0] = 1; for (int piece : pieces) { for (int i = K; i >= piece; --i) { dp[i] += dp[i - piece]; } } return dp[K]; } int main() { int N, K; cin >> N >> K; vector pieces(N); for (int i = 0; i > pieces[i]; } cout << count_combinations(N, K, pieces) << endl; return 0; }

Deltax

Pergunta de entrevista da empresa Deltax

Respostas da entrevista

Empresas seguidas

Buscas de vagas