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  1. Prepare
  2. Algorithms
  3. Strings
  4. Sherlock and the Valid String
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Sherlock and the Valid String

  • lylia_luong
     + 0 comments

    Python solution with Collections.Counter. Time: O(N), Space: O(1)

    "from collections import Counter

def isValid(s):
    counter_s = Counter(s)
    frequency = Counter(counter_s.values())

    if len(frequency) == 1:
        return ""YES""

    if len(frequency) == 2:
        key1, key2 = frequency.keys()
        val1, val2 = frequency[key1], frequency[key2]

        # Case A: one char has freq=1 and it appears once
        if (key1 == 1 and val1 == 1) or (key2 == 1 and val2 == 1):
            return ""YES""

        # Case B: frequencies differ by 1 and higher one appears once
        if abs(key1 - key2) == 1 and (frequency[max(key1, key2)] == 1):
            return ""YES""

    return ""NO""
"