**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(n^{2}) time.

**Better Approach: Tetris Game technique- O(Nk)**

- We will create a class structure which will store an element and its count.

**class** Elements{
**int**element;
**int**count;
**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

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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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