Codility 'GenomicRangeQuery' Solution

Martin Kysel · August 5, 2014

Short Problem Definition:

Find the minimal nucleotide from a range of sequence DNA.



expected worst-case time complexity is O(N+M)

expected worst-case space complexity is O(N)


Remember the last position on which was the genome (A, C, G, T) was seen. If the distance between Q and P is lower than the distance to the last seen genome, we have found the right candidate.


def writeCharToList(S, last_seen, c, idx):
    if S[idx] == c:
        last_seen[idx] = idx
    elif idx > 0:
        last_seen[idx] = last_seen[idx -1]

def solution(S, P, Q):
    if len(P) != len(Q):
        raise Exception("Invalid input")
    last_seen_A = [-1] * len(S)
    last_seen_C = [-1] * len(S)
    last_seen_G = [-1] * len(S)
    last_seen_T = [-1] * len(S)
    for idx in xrange(len(S)):
        writeCharToList(S, last_seen_A, 'A', idx)
        writeCharToList(S, last_seen_C, 'C', idx)
        writeCharToList(S, last_seen_G, 'G', idx)
        writeCharToList(S, last_seen_T, 'T', idx)
    solution = [0] * len(Q)
    for idx in xrange(len(Q)):
        if last_seen_A[Q[idx]] >= P[idx]:
            solution[idx] = 1
        elif last_seen_C[Q[idx]] >= P[idx]:
            solution[idx] = 2
        elif last_seen_G[Q[idx]] >= P[idx]:
            solution[idx] = 3
        elif last_seen_T[Q[idx]] >= P[idx]:
            solution[idx] = 4
            raise Exception("Should never happen")
    return solution

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