1. Description

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 b, which 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.

2. Runtime Distribution

3. Submission Details

4. Example

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.

5. Code

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public int maxCount(int m, int n, int[][] ops) {
	if (m <= 0 || n <= 0 || ops == null) {
		return 0;
	}

	int minRow = m;
	int minCol = n;
	for (int i = 0; i < ops.length; i++) {
		if (minRow > ops[i][0]) {
			minRow = ops[i][0];
		}
		if (minCol > ops[i][1]) {
			minCol = ops[i][1];
		}
	}

	return minRow * minCol;
}

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6.Test

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public class LeetCode0598 {
	public int maxCount(int m, int n, int[][] ops) {
		if (m <= 0 || n <= 0 || ops == null) {
			return 0;
		}

		int minRow = m;
		int minCol = n;
		for (int i = 0; i < ops.length; i++) {
			if (minRow > ops[i][0]) {
				minRow = ops[i][0];
			}
			if (minCol > ops[i][1]) {
				minCol = ops[i][1];
			}
		}

		return minRow * minCol;
	}

	public static void main(String[] args) {
		LeetCode0598 leetcode = new LeetCode0598();
		int ops[][] = { { 2, 2 }, { 3, 3 } };

		System.out.println(leetcode.maxCount(3, 3, ops));
	}
}

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