_598.java

package com.fishercoder.solutions;

public class _598 {

    public static class Solution1 {
        /**
         * Since the incrementing starts from zero to op[0] and op[1], we only need to find the range that has the most overlaps.
         * Thus we keep finding the minimum of both x and y.
         */
        public int maxCount(int m, int n, int[][] ops) {
            int x = m;
            int y = n;
            for (int[] op : ops) {
                x = Math.min(x, op[0]);
                y = Math.min(y, op[1]);
            }
            return x * y;
        }
    }
}