Frequency Queries
Problem Statement :
You are given q queries. Each query is of the form two integers described below:
- 1 x: Insert x in your data structure.
- 2 y: Delete one occurence of y from your data structure, if present.
- 3 z: Check if any integer is present whose frequency is exactly z . If yes, print 1 else 0.
The queries are given in the form of a 2-D array querries of size q where queries[i][0] contains the operation, and queries[i][1] contains the data element. For example, you are given array
queries = [ (1,1), (2,2), (3,2), (1,1), (1,1) , (2,1), (3,2) ] . The results of each operation are:
Operation Array Output
(1,1) [1]
(2,2) [1]
(3,2) 0
(1,1) [1,1]
(1,1) [1,1,1]
(2,1) [1,1]
(3,2) 1
Return an array with the output: [0, 1] .
Function Description
Complete the freqQuery function in the editor below. It must return an array of integers where each element is a 1 if there is at least one element value with the queried number of occurrences in the current array, or 0 if there is not.
freqQuery has the following parameter(s):
queries: a 2-d array of integers
Input Format
The first line contains of an integer q, the number of queries.
Each of the next q lines contains two integers denoting the 2-d array queries.
Constraints
1 <= q <= 10^5
1 <= x, y, z <= 10^9
1 <= queries[i][1] <= 10^9
Output Format
Return an integer array consisting of all the outputs of queries of type 3.
Solution :
Solution in Python :
In Python3 :
#!/bin/python3
import os
from collections import defaultdict
def freqQuery(queries):
elementFreq = defaultdict(int)
freqCount = defaultdict(int)
ans = []
for i, j in queries:
if i == 1:
if freqCount[elementFreq[j]]:
freqCount[elementFreq[j]] -= 1
elementFreq[j] += 1
freqCount[elementFreq[j]] += 1
elif i == 2:
if elementFreq[j]:
freqCount[elementFreq[j]] -= 1
elementFreq[j] -= 1
freqCount[elementFreq[j]] += 1
else:
# operation 3
if j in freqCount and freqCount[j]:
ans.append(1)
else:
ans.append(0)
return ans
if __name__ == '__main__':
fptr = open(os.environ['OUTPUT_PATH'], 'w')
q = int(input())
queries = []
for _ in range(q):
queries.append(map(int, input().rstrip().split()))
ans = freqQuery(queries)
fptr.write('\n'.join(map(str, ans)))
fptr.write('\n')
fptr.close()
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