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Leetcode 598. Range Addition II

Problem Statement

Given an m x n matrix M initialized with all 0’s and several update operations.

Operations are represented by a 2D array, where each operation is represented by an array with two positive integers a and b, which means M[i][j] should be incremented by 1 for all 0 <= i < a and 0 <= j < b.

You need to return the count of the maximum integer in the matrix after performing all the operations.

Example:

Input: m = 3, n = 3, ops = [[2,2],[3,3]]
Output: 4
Explanation: 
Initially, the M = 
[[0, 0, 0],
 [0, 0, 0],
 [0, 0, 0]]

After performing [2,2], the matrix M is:
[[1, 1, 0],
 [1, 1, 0],
 [0, 0, 0]]

After performing [3,3], the matrix M is:
[[2, 2, 1],
 [2, 2, 1],
 [1, 1, 1]]

The largest integers in M are 2, and there are four of them in M. So return 4.

Constraints:

Clarifying Questions

  1. Can the operations have overlapping regions? If so, how should they be treated?
  2. What should the output be if there are no operations?
  3. Are all matrix elements initialized to zero?

Code

Assuming the answers:

  1. Yes, overlapping regions increment the values in the overlapping areas.
  2. If there are no operations, the entire matrix remains zero, and the maximum is zero.
  3. Yes, all matrix elements are initialized to zero.

Strategy

Instead of directly simulating the operations on the matrix, which could become computationally expensive (especially for large values of m and n), we can determine the region that is incremented the most number of times.

Each operation updates a submatrix from (0,0) to (a,b). The submatrix with the most updates will be the smallest (a, b) across all operations. Thus, all we need to do is find the minimum a and b across all operations.

  1. Initialize min_a to m and min_b to n.
  2. Iterate over each operation and update min_a and min_b with the minimum values of a and b respectively.
  3. The area of the maximum integer will be the intersection of these minimum values, i.e., min_a * min_b.

Time Complexity

The time complexity for iterating over the operations is O(k) where k is the number of operations. Space complexity is O(1) as we only maintain a few variables.

Code

#include <vector>
#include <algorithm>

class Solution {
public:
    int maxCount(int m, int n, std::vector<std::vector<int>>& ops) {
        int minA = m;
        int minB = n;
        
        for (const auto& op : ops) {
            minA = std::min(minA, op[0]);
            minB = std::min(minB, op[1]);
        }
        
        return minA * minB;
    }
};

This code effectively captures the essence of determining the maximum impacted area after all operations and computes it in an efficient manner.

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