PAST4F

from Fourth Algorithm Practical Skills Test PAST4F F - Parsing

• AMBIGUOUS if the number of occurrences is the same compared to above or below. python

def solve(N, K, SS):
    from collections import Counter
    c = Counter(SS).most_common()
    K -= 1
    ck, ckv = c[K]
    if K > 0:
        _, pv = c[K - 1]
        if pv == ckv:
            return "AMBIGUOUS"
    if K < len(c) - 1:
        _, pv = c[K + 1]
        if pv == ckv:
            return "AMBIGUOUS"
    return ck

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