Range Addition II - Math - Easy - LeetCode
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Range Addition II - Math - Easy - LeetCode

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  • 1The problem involves a matrix initialized with zeros and multiple update operations that increment values in specified submatrices.
  • 2After applying all operations, the task is to determine the maximum integer in the matrix and count its occurrences.
  • 3The solution efficiently calculates the result using the minimum dimensions affected by the operations, with a time complexity of O(n).

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"The problem involves a matrix initialized with zeros and multiple update operations that increment values in specified submatrices."

Range Addition II - Math - Easy - LeetCode

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.

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.

public class Solution {
 public int MaxCount(int m, int n, int[][] ops) {
 int x = m;
 int y = n;
 for(int i=0;i<ops.Length;i++){
 x = Math.Min(ops[i][0],x);
 y = Math.Min(ops[i][1],y);
 }
 return x*y;
 }
}

Time Complexity: O(n)

Space Complexity: O(1)

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sumitc91

Published on 1 October 2020 · 1 min read · 221 words

Part of AskGif Blog · coding

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