**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()
```

## View More Similar Problems

## Array and simple queries

Given two numbers N and M. N indicates the number of elements in the array A[](1-indexed) and M indicates number of queries. You need to perform two types of queries on the array A[] . You are given queries. Queries can be of two types, type 1 and type 2. Type 1 queries are represented as 1 i j : Modify the given array by removing elements from i to j and adding them to the front. Ty

View Solution →## Median Updates

The median M of numbers is defined as the middle number after sorting them in order if M is odd. Or it is the average of the middle two numbers if M is even. You start with an empty number list. Then, you can add numbers to the list, or remove existing numbers from it. After each add or remove operation, output the median. Input: The first line is an integer, N , that indicates the number o

View Solution →## Maximum Element

You have an empty sequence, and you will be given N queries. Each query is one of these three types: 1 x -Push the element x into the stack. 2 -Delete the element present at the top of the stack. 3 -Print the maximum element in the stack. Input Format The first line of input contains an integer, N . The next N lines each contain an above mentioned query. (It is guaranteed that each

View Solution →## Balanced Brackets

A bracket is considered to be any one of the following characters: (, ), {, }, [, or ]. Two brackets are considered to be a matched pair if the an opening bracket (i.e., (, [, or {) occurs to the left of a closing bracket (i.e., ), ], or }) of the exact same type. There are three types of matched pairs of brackets: [], {}, and (). A matching pair of brackets is not balanced if the set of bra

View Solution →## Equal Stacks

ou have three stacks of cylinders where each cylinder has the same diameter, but they may vary in height. You can change the height of a stack by removing and discarding its topmost cylinder any number of times. Find the maximum possible height of the stacks such that all of the stacks are exactly the same height. This means you must remove zero or more cylinders from the top of zero or more of

View Solution →## Game of Two Stacks

Alexa has two stacks of non-negative integers, stack A = [a0, a1, . . . , an-1 ] and stack B = [b0, b1, . . . , b m-1] where index 0 denotes the top of the stack. Alexa challenges Nick to play the following game: In each move, Nick can remove one integer from the top of either stack A or stack B. Nick keeps a running sum of the integers he removes from the two stacks. Nick is disqualified f

View Solution →