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
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
