Find All Elements in an Array which appears more than N/K times, N is Array Size and k is a Number.

Objective: Given an array of size of N and number k. Find all elements in an Array which appears more than N/K times.

Input: Array [] and number k.

Example:

int[] arrA = { 2, 2, 4, 4, 3, 5, 3, 4, 4, 6, 4, 3, 3, 8 };

K = 4

N/k = 14/4 = 3

Output will be [3,4] they appear 5, 4 times respectively.

Approach:

Naive Approach: Take two for loops , check every element with all other elements, Time Complexity –   O(n2) time.

Better Approach: Tetris Game technique- O(Nk)

  • We will create a class structure which will store an element and its count.
class Elements{
   intelement;
   intcount;
   public Elements(int element, int count){
   this.element = element;
   this.count =count;
  }
}
  • Create array etms[] contains k-1 objects of class Elements with element =0 and count = 0.
  • So idea is, navigate all the elements of given array.
  • Check if element exist in etms[] if not, insert it with the count 1 and if exist then increase its count.
  • Also check if etms[] gets full when inserting an element, if it is not, follow the previous step. If it is full then reduce the count of every existing element in the etms[]. (Just think of a Tetris game, when row gets full, it gets deleted and size of the Tetris reduced by 1) see the picture below.
  • Once all the elements of array gets over, check every element of etms[] with array and print those elements which has N/K times.
Find All Elements in an Array which appears more than NbyK times

Find All Elements in an Array which appears more than NbyK times

Complete Code:


Output:

4 appears more than n/4 times, Actual count is 5
3 appears more than n/4 times, Actual count is 4

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If you find anything incorrect or you feel that there is any better approach to solve the above problem, please write comment.
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  • cabhishek

    Why not use hashmap to store the count ?

    • tutorialhorizon

      You can use the Hashmap , if you are storing all the elements in hashmap then it will take O(n) space. you can apply the same logic which is mentioned in the post with hash map size k-1, then space complexity will be O(k-1)

  • Amarnath

    Good logic.
    I have a question, in the above chart’s last step 8 has 1, 4 has 3, 3 has 2. As per logic we will look at the count of all the elements and the elements which are having >= N/K = 14/4 = 3 will be the result. Here the count of element 4 is having count >= 3. Hence this should be the answer. But why is 3 also part of the result? Since element 3 has count only 2, it should not be added in the result right?

  • Ankita

    Thank you so much for the diagram !! It made understanding this algo a lot easier 🙂
    Also, the code is very clean.

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