Matrix Analyzer

Example

Input : 5
           2 3 1 7 3
           4 1 9 6 8
           5 5 2 4 4
           6 5 2 6 7
           8 4 2 2 2
Output : 4

 

My answer:

import java.util.Scanner;

class MatrixAnalyzer {
	public static void main(String[] args) {
		MatrixAnalyzer t = new MatrixAnalyzer();
		Scanner kb = new Scanner(System.in);
		int n = kb.nextInt();
		int[][] a = new int[n][n];
		for(int i = 0; i < n; i++) {
			for(int j = 0; j < n; j++) {
				a[i][j] = kb.nextInt();
			}	
		}
		System.out.println(t.solution(n, a));
	}

	int solution(int n, int[][] a) {
		int[][] b = new int[n][n];
		for(int j = 0; j < n; j++) {
			for(int i = 0; i < n; i++) {
				int tmp = a[i][j];
				int cnt = n - 1;
				while(cnt >= 0) {
					if(tmp == a[cnt][j]) {
						b[i][j] += 1;
					}
					cnt--;
				}
			}
		}

		int max = 0, answer = 0;

		for(int i = 0; i < n; i++) {
			int sum = 0;
			for(int j = 0; j < n; j++) {
				sum += b[i][j];
			}
			if(max < sum) {
				answer = i + 1;
				max = sum;
			}
		}
		return answer;
	}
}

 

1. Read the size of the matrix (n) from the user.
2. Read the elements of the matrix from the user and store them in a 2D array a.
3. Create a new 2D array b to store the count of repeated elements in each column.
4. Iterate over each column of the matrix (a) and for each element, compare it with the elements in the same column. Increment the count in the corresponding position in the b array.
5. Find the row with the maximum sum of counts in the b array. The row index (1-based) is considered the answer.