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

Sherlock and the Valid String

  • illogicalsleepi1
     + 0 comments

    here f1==1 is for edge case - where only single alphabet is in string while all other have same frequency

    def isValid(s):
    lst=list(s)
    set_s=set(s)
    dct={}
    for i in set_s:
        dct[i]=lst.count(i)
    freqs=list(dct.values())
    unique=set(freqs)
    if len(unique)==1:
        return 'YES'
    elif len(unique)==2:
        f1,f2=sorted(unique)
        count_f1=freqs.count(f1)
        count_f2=freqs.count(f2)
        if count_f1==1 and (abs(f2-f1)==1 or f1==1):
            return 'YES'
        elif count_f2==1 and abs(f2-f1)==1:
            return 'YES'     
    return 'NO'