Minimum Meeting Rooms

/*
 * Problem : Meeting rooms.
 * Input   : n, 2, then n rows.
 * Output  : Number of rooms.
 * Approach: Sorted starts and ends.
 * Time    : O(n log n)
 * Space   : O(n)
 * Java    : 8 or newer.
 */
import java.util.Scanner;

public class Main {
    // This method solves the task. It receives the input and returns the answer.
    static long solve(int[][] m) {
        // Step 1: Put all start times in one sorted array and all end times in another.
        int n = m.length;
        int[] starts = new int[n];
        int[] ends = new int[n];
        for (int i = 0; i < n; i++) {
            starts[i] = m[i][0];
            ends[i] = m[i][1];
        }
        java.util.Arrays.sort(starts);
        java.util.Arrays.sort(ends);
        // Step 2: Go through the starts. If the earliest unfinished meeting has ended, reuse its room. Otherwise open a new room.
        int rooms = 0;
        int endIndex = 0;
        for (int i = 0; i < n; i++) {
            if (starts[i] >= ends[endIndex]) {
                endIndex++;
            } else {
                rooms++;
            }
        }
        return rooms;
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int rows = scanner.nextInt();
        int cols = scanner.nextInt();
        int[][] m = new int[rows][cols];
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                m[i][j] = scanner.nextInt();
            }
        }
        System.out.println(solve(m));
    }
}