Top k frequent number

#1


not passes any test cases

#2

Sample Input

1
5 2
5 1 3 5 2

Sample Output

5 1 5 1 3 5 1 5 1

But your code prints
3 5

In case you are not understanding the output format:

5 comes : print( { 5 } )
1 comes : print( { 1,5 } )
3 comes : print( { 1,3 } ) , all three have same frequency so taking smaller 2.
5 comes : print( { 5,1 } ) , now 5 has max frequency.
2 comes : print( { 5,1 } )

Final output : 5, 1 5, 1 3, 5 1, 5 1.

#3

Help me with time complexity. Using Hashmap solves it in O(n*k) then why use heap.

#4

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