Find which element(x or y) has highest occurrence in the array

Given an array arr[] and the two elements x and y, the task is to find which element(x or y) has highest occurrence in the array. If the occurrences are the same, then return the smaller element.

Examples:

Input: arr[] = {1, 1, 2, 2, 3, 3, 4, 4, 4, 4, 5}; x = 4, y = 5; 
Output: 4 
Explanation: 
The occurrence of x in this array is = 4 times. 
The occurrence of y in this array is = 1 times. 
x has highest occurence than y, so we return 4. 

Input: arr[] = {1, 2, 3, 4, 5, 6, 7, 8}; x = 1, y = 7; 
Output: 1 
Explanation: 
The occurrence of x in this array is = 1 times. 
The occurrence of y in this array is = 1 times. 
both have same ocurrences, so we look for the smaller element. 
x = 1 is smaller than y = 7, so we return 1.

Approach:

1. Traverse the whole array.
2. Count the occurrences of x and y element.
3. Check whether the count of x and y occurrences are the same, If they are the same then return the smaller element among x and y.
4. Check which element(x or y) has the highest occurrence. If x has the highest occurrence then we return x, else we return y.

Below is the implementation of the above approach:

// Java program to find the element
// with the highest occurrence in the array

public class Cplusplus{
	
	// Function to find the element with the 
	// highest occurrence in the array.
	static int majorityElement(int[] a, int x, int y)
	{
		// Initialize the counter variable
		// x and y by zero.
		int count_x = 0;
		int count_y = 0;
		
		// Traverse the whole array.
		for(int i = 0; i < a.length; i++)
		{
			// Count the occurrence of x element.
			if(a[i] == x){
				count_x++;
			}
			
			// Count the occurrence of y element.
			else if(a[i] == y){
				count_y++;
			}
		}
		
		// If occurrences of x and y are same
		// then return smaller element among x and y.
		if(count_x == count_y){
			return Math.min(x,y);
		}
		
		// Find the element which has highest occurrence
		// highest occurrence, If x has highest occurrence
		// then return x, else return y.
		else if(count_x > count_y)
			return x;
		else
			return y;
    }
	
	// Driver Code.
	public static void main(String args[])
	{
		// Test Case 1:
		int arr1[] = {1, 1, 2, 2, 3, 3, 4, 4, 4, 4, 5};
		int x1 = 4;
		int y1 = 5;
		System.out.println(majorityElement(arr1,x1,y1));
		
		// Test Case 2:
		int arr2[] = {1, 2, 3, 4, 5, 6, 7, 8};
		int x2 = 1;
		int y2 = 7;
		System.out.println(majorityElement(arr2,x2,y2));
	}
}

Output:

4
1

Time Complexity: O(N).