TOP k most frequent no in a stream

#1

can you please help me in building logic of this question?

#2

hey @CODER_JATIN let me explain you this question with an approach, initialise a k priority queue window that store top k frequent number.
Whenever we gets a new number we will updated its frequency and then check it if that can be a part of our window or not.
We will find the element which has least frequncy . if this frequency is less than our current element frequency’ then we will remove that least frequent element and add then add our current element in our window. For eg: See this test case.
For input:
n=5,k=2;
arr[] = 5 1 3 5 2

Explanation
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

#3

bhaiya, ye mene pehle hi padh lia , but mujhe smjh ni aaya , pls achhe se smjhado thoda

#4

yeh video dekho


same question hai
#5

bhaiya , but how can we for running stream of numbers ?

#6

running stream ko kch ni bs array smjho. fr toh possible ho jayega na…

#7

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