#598

Easy

LeetCode Problem Workspace

Range Addition II

Range Addition II requires counting maximum values in a matrix after incremental operations using array and math reasoning patterns.

array math

Problem Statement

You are given an m x n matrix initially filled with zeros and an array of operations ops, where each operation ops[i] = [ai, bi] means incrementing all elements M[x][y] by one for 0 <= x < ai and 0 <= y < bi. Your task is to apply all operations and determine the number of integers that reach the maximum value after all increments.

For example, given m = 3, n = 3, and ops = [[2,2],[3,3]], after applying the operations, the maximum integer in the matrix appears four times. Implement an efficient method that counts these maximum integers without simulating every increment individually.

Examples

Example 1

Input: m = 3, n = 3, ops = [[2,2],[3,3]]

Output: 4

The maximum integer in M is 2, and there are four of it in M. So return 4.

Example 2

Input: 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]]

Output: 4

Example details omitted.

Example 3

Input: m = 3, n = 3, ops = []

Output: 9

Example details omitted.

Constraints

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

Solution Approach

Identify Overlap Region

Calculate the minimum ai and bi across all operations to determine the overlapping rectangle that receives every increment. Only this region will contain the maximum value.

Compute Maximum Count

Multiply the dimensions of the overlap rectangle (min_ai * min_bi) to directly obtain the count of maximum integers, avoiding full matrix updates.

Handle Edge Cases

If ops is empty, the entire matrix retains the initial value 0, so return m * n. Ensure the method handles large m and n efficiently using only the min calculation.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time 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.

What Interviewers Usually Probe

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

Common Pitfalls or Variants

Common pitfalls

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

Follow-up variants

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

FAQ

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.

terminal

Solution

Solution 1: Brain Teaser

We notice that the intersection of all operation submatrices is the submatrix where the final maximum integer is located, and each operation submatrix starts from the top-left corner $(0, 0)$. Therefore, we traverse all operation submatrices to find the minimum number of rows and columns. Finally, we return the product of these two values.

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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

Continue Practicing

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