Not able to get problem

#1

sir i am not able to get the problem basically what the problem is saying

#2

sir plzz tell me i am not getting the problem

Top most frequent element using heap
#3

Hey @singh.pankhil40
We basically need to print top k numbers sorted by frequency when input stream has included k distinct elements, else need to print all distinct elements sorted by frequency.
i can explain you better with examples :-
Input : arr[] = {5, 2, 1, 3, 2}
k = 4
Output : 5 2 5 1 2 5 1 2 3 5 2 1 3 5
Explanation:

  1. After reading 5, there is only one element 5 whose frequency is max till now.
    so print 5.
  2. After reading 2, we will have two elements 2 and 5 with the same frequency.
    As 2, is smaller than 5 but their frequency is the same so we will print 2 5.
  3. After reading 1, we will have 3 elements 1, 2 and 5 with the same frequency,
    so print 1 2 5.
  4. Similarly after reading 3, print 1 2 3 5
  5. After reading last element 2 since 2 has already occurred so we have now a
    frequency of 2 as 2. So we keep 2 at the top and then rest of the element
    with the same frequency in sorted order. So print, 2 1 3 5.

Input : arr[] = {5, 2, 1, 3, 4}
k = 4
Output : 5 2 5 1 2 5 1 2 3 5 1 2 3 4
Explanation:

  1. After reading 5, there is only one element 5 whose frequency is max till now.
    so print 5.
  2. After reading 2, we will have two elements 2 and 5 with the same frequency.
    As 2, is smaller than 5 but their frequency is the same so we will print 2 5.
  3. After reading 1, we will have 3 elements 1, 2 and 5 with the same frequency,
    so print 1 2 5.
    Similarly after reading 3, print 1 2 3 5
  4. After reading last element 4, All the elements have same frequency
    So print, 1 2 3 4.
#4

can you explain me with the help of this example
n=5,k=2;
5 1 3 5 2

#5

its output is
5 1 5 1 3 5 1 5 1

#6

So we first get 5 its freq is 1 and since its the only element we print it :5
Next we get 1 its freq is 1 but its smaller than 5 so it comes before it and since k is 2 we print : 1 5
Next we get 3 now all 1 ,3 ,5 have same freq and 1,3 comes first on order so we print :1 3
Next comes 5 now 5 freq is 2 and 1,3 freq is 1 so while printing we print : 5(most freq) 1(second most freq but smaller than others having same freq)
Next comes 2 its freq is 1 so again we print :5 1