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  1. Frequency Queries
  2. Discussions

Frequency Queries

  • c_m_s_gan
     + 1 comment

    Not shorter but this is a passable solution without the use of Counter from Collections:

    def freqQuery(queries):

    a1 = dict()  #Keep track of the number of times each queried number occurs in the array
    a2 = dict() #[0]*len(queries)  #Keep track of how many numbers occur once, twice, etc.
    
    out = []

    for (op,num) in queries:

        if  (op == 1):
            if not(num in a1):
                a1[num] = 0

            if not(a1[num] in a2):
                a2[a1[num]] = 1

            a2[a1[num]] -= 1
            
            a1[num] += 1

            if not(a1[num] in a2):
                a2[a1[num]] = 0

            a2[a1[num]] += 1
        
        if (op == 2) & (num in a1):

            a2[a1[num]] -= 1
            a1[num] -= 1
            a2[a1[num]] += 1

            if a1[num] <= 0:
                a1.pop(num)
            

        if (op == 3) & (num in a2):
            out.append(int(a2[num]>0))
            #out.append(a2[num])

        elif (op==3):
            out.append(0)

    return out