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  3. Probability
  4. Sherlock and Probability
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Sherlock and Probability

  • daniel_v_oconnor
     + 0 comments

    The trickiest part of this problem is that one must use the "rolling sum" trick to obtain an O(N) solution that meets the runtime requirements. Here's a relatively simple Python solution:

    def solve(N, K, S):   
    S = [int(b) for b in S]
    rolling_sum = sum(S[0:K])
    numerator = 0
    for i in range(N):
        if i > K:
            rolling_sum = rolling_sum - S[i-K-1]
        if i < N - K:
            rolling_sum = rolling_sum + S[i+K]
        if S[i]:
            numerator = numerator + rolling_sum
            
    denominator = N**2
    d = math.gcd(numerator, denominator)    
    return f'{numerator//d}/{denominator//d}'