Given an m * n matrix M initialized with all 0‘s and several update operations.
Operations are represented by a 2D array, and each operation is represented by an array with two positive integers a and bwhich means M[i][j] should be Added by one for all 0 <= i < a and 0 <= j < b.
You need to count and return the number of maximum integers in the matrix after performing all the operations.
Example 1:
Input: m = 3, n = 3 operations = [[2,2],[3,3]] Output: 4 Explanation: Initially, M = [[0, 0, 0], [0, 0, 0], [0, 0, 0]] After performing [2,2], M = [[1, 1, 0], [1, 1, 0], [0, 0, 0]] After performing [3,3], M = [[2, 2, 1], [2, 2, 1], [1, 1, 1]] So the maximum integer in M is 2, and there are four of it in M. So return 4.
Note:
- The range of m and n is [1,40000].
- The range of a is [1,m], and the range of b is [1,n].
- The range of operations size won’t exceed 10,000.
This problem looks intimidating, but after looking at it on paper, I found it to be very simple – it's about finding the smallest range (note that it's a range; I initially thought of it as finding the smallest block).
public class Solution { public int maxCount(int m, int n, int[][] ops) { if(ops.length == 0) return m*n; int r = ops.length-1; int minrow = Integer.MAX_VALUE; int mincol = Integer.MAX_VALUE; while(r>=0){ if(ops[r][0] < minrow){ minrow = ops[r][0]; } if(ops[r][1] < mincol){ mincol = ops[r][1]; } r--; } return mincol * minrow; } } This websiteOriginal articleAll follow "Attribution-NonCommercial-ShareAlike 4.0 License (CC BY-NC-SA 4.0)Please retain the following annotations when sharing or adapting:
Original author:Jake Tao,source:「LeetCode – 598. Range Addition II」
