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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
    • renato314159 + 0 comments

      So I rewrote your solution with some comments to try and figure out your logic. Can you check to see if I've understood it?

      def freqQuery(queries):
    arr = dict() # val:count of said values    
    freq = dict() #number of counts, ie 1:numbers that appear only
    out = []

for (oper, val) in queries:
    if (oper == 1):
        #insertion
        if not(val in arr):
            arr[val] = 0 #if value isn't in arr dict, then add value to dict and give it a 0 bc we can't assign a 1 just yet
        if not(arr[val] in freq):
            freq[arr[val]] = 1 #if freq dictionary doesn't already have arr[val]'s number, then we automatically assign by 1
        freq[arr[val]] -= 1 #subtracting arr[val]'s number by one because it's always kept at 1
        arr[val] += 1 #adding 1 to arr[val]
        if not(arr[val] in freq):
            freq[arr[val]] = 0  #if arr[val]'s cnt isn't in freq, then we assign it 0
        freq[arr[val]] += 1 #otherwise we increase it by 1

    if (oper == 2) and (val in arr):
        #deletion
        freq[arr[val]] -= 1 #same as above, we subtract 1 from it to keep it at 1
        arr[val] -= 1
        freq[arr[val]] += 1 #same as above, we add 1 from it to keept it at one

        if arr[val] <= 0:
            arr.pop(val)

    if (oper == 3) and (val in freq):
        out.append(int(freq[val]>0))
    elif(oper == 3):
        out.append(0)
return out