1class Solution {2public int numberOfWeakCharacters(int[][] properties) {3int[] maxH = new int[100002];4int count = 0;5for (int[] point : properties) {6maxH[point[0]] = Math.max(point[1], maxH[point[0]]);7}8for (int i = 100000; i >= 0; i--) {9maxH[i] = Math.max(maxH[i + 1], maxH[i]);10}1112for (int[] point : properties) {13if (point[1] < maxH[point[0] + 1]) count++;14}15return count;16}17}