What is the main idea behind solving Range Addition II efficiently?

The main idea is to find the overlapping region affected by all operations using min_ai and min_bi, then count its area for maximum integers.

Why not simulate each increment in the matrix?

Simulating every increment is inefficient for large matrices; the problem only requires counting maximums using overlap analysis.

How do I handle an empty ops array?

If ops is empty, the entire m x n matrix retains the initial value 0, so return m * n.

Does the solution scale for large m and n?

Yes, by only tracking minimum ai and bi across ops, the solution avoids full matrix storage and updates.

Which pattern is essential in this problem?

The array plus math pattern is key: computing minimum bounds from operations avoids unnecessary simulation and identifies maximum regions.

#598

Easy

auto_awesome数组·数学

LeetCode 题解工作台

区间加法 II

给你一个 m x n 的矩阵 M 和一个操作数组 op 。矩阵初始化时所有的单元格都为 0 。 ops[i] = [ai, bi] 意味着当所有的 0 和 0 时， M[x][y] 应该加 1。在执行完所有操作后，计算并返回矩阵中最大整数的个数。示例 1: 输入: m = 3, n = …

数组数学

题目描述

给你一个 m x n 的矩阵 M 和一个操作数组 op 。矩阵初始化时所有的单元格都为 0 。ops[i] = [ai, bi] 意味着当所有的 0 <= x < ai 和 0 <= y < bi 时， M[x][y] 应该加 1。

在 执行完所有操作后 ，计算并返回 矩阵中最大整数的个数 。

示例 1:

输入: m = 3, n = 3，ops = [[2,2],[3,3]]
输出: 4
解释: M 中最大的整数是 2, 而且 M 中有4个值为2的元素。因此返回 4。

示例 2:

输入: m = 3, n = 3, ops = [[2,2],[3,3],[3,3],[3,3],[2,2],[3,3],[3,3],[3,3],[2,2],[3,3],[3,3],[3,3]]
输出: 4

示例 3:

输入: m = 3, n = 3, ops = []
输出: 9

提示:

1 <= m, n <= 4 * 10⁴
0 <= ops.length <= 10⁴
ops[i].length == 2
1 <= a_i <= m
1 <= b_i <= n

lightbulb

解题思路

方法一：脑筋急转弯

我们注意到，所有操作子矩阵的交集就是最终的最大整数所在的子矩阵，并且每个操作子矩阵都是从左上角 $(0, 0)$ 开始的，因此，我们遍历所有操作子矩阵，求出行数和列数的最小值，最后返回这两个值的乘积即可。

注意，如果操作数组为空，那么矩阵中的最大整数个数就是 $m \times n$ 。

时间复杂度 $O(k)$ ，其中 $k$ 是操作数组 $\textit{ops}$ 的长度。空间复杂度 $O(1)$ 。

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class Solution:
    def maxCount(self, m: int, n: int, ops: List[List[int]]) -> int:
        for a, b in ops:
            m = min(m, a)
            n = min(n, b)
        return m * n

speed

复杂度分析

指标	值
时间	complexity is O(k) for scanning all operations to find the minimum ai and bi, where k = ops.length. Space complexity is O(1) as no additional matrix storage is needed.
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
The candidate immediately identifies the overlap of all operations without full matrix simulation.
question_mark
They recognize that only the minimum ai and bi determine the maximum value region.
question_mark
They handle the empty ops edge case and large matrix sizes efficiently.

warning

常见陷阱

外企场景

error
Attempting to simulate each increment in the full matrix, which is unnecessary and inefficient.
error
Failing to handle an empty ops array, incorrectly returning zero instead of m * n.
error
Mixing up row and column limits when computing the overlap rectangle, leading to wrong counts.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Return the sum of all maximum integers instead of the count, which involves similar overlap reasoning.
arrow_right_alt
Operations allow decrements as well, requiring careful tracking of minimum and maximum overlaps.
arrow_right_alt
Compute the maximum value itself instead of the count, combining overlap detection with cumulative increments.

help

常见问题

外企场景

继续练习

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