Leetcode 598. Range Addition II

Problem Explanation:

This problem is essentially asking us to perform range addition on a 2D matrix with the provided operations and then count the integer that occurs most frequently after all the operations are performed.

An operation is represented as an array [a, b], which means that for every cell within the range of 0 to a in rows and 0 to b in columns, we add 1 to the value in the cell.

After performing all the operations, we need to find the maximum integer value in the matrix and return the count of that integer in the matrix as the output.

The key to simplifying the problem lies in observing that the cells (rows and columns) that receive the maximum sums are ones that are in the overlap of the ranges specified in all the operations. Therefore, rather than applying the operation to all cells in the matrix, we can simply find the smallest operation range (defined as the cell at position [min_a, min_b]) because it will be the one that is included in all operations. The number of cells in this range (which is min_a * min_b) correspond to the maximum integer in the matrix.

Solution:

Python

1
2python
3class Solution:
4    def maxCount(self, m: int, n: int, ops: List[List[int]]) -> int:
5        min_rows = m
6        min_cols = n
7        
8        for op in ops:
9            min_rows = min(min_rows, op[0])
10            min_cols = min(min_cols, op[1])
11        
12        return min_rows * min_cols

Java

1
2java
3class Solution {
4    public int maxCount(int m, int n, int[][] ops) {
5        int minRows = m;
6        int minCols = n;
7        
8        for(int[] op : ops) {
9            minRows = Math.min(minRows, op[0]);
10            minCols = Math.min(minCols, op[1]);
11        }
12        
13        return minRows * minCols;
14    }
15}

JavaScript

1
2javascript
3var maxCount = function(m, n, ops) {
4    let minRows = m;
5    let minCols = n;
6    
7    for(let op of ops) {
8        minRows = Math.min(minRows, op[0]);
9        minCols = Math.min(minCols, op[1]);
10    }
11    
12    return minRows * minCols;
13};

C++

1
2cpp
3class Solution {
4public:
5    int maxCount(int m, int n, vector<vector<int>>& ops) {
6        int minRows = m;
7        int minCols = n;
8        
9        for(auto & op : ops) {
10            minRows = min(minRows, op[0]);
11            minCols = min(minCols, op[1]);
12        }
13        
14        return minRows * minCols;
15    }
16};

C#

1
2csharp
3public class Solution {
4    public int MaxCount(int m, int n, int[][] ops) {
5        int minRows = m;
6        int minCols = n;
7        
8        foreach(int[] op in ops) {
9            minRows = Math.Min(minRows, op[0]);
10            minCols = Math.Min(minCols, op[1]);
11        }
12        
13        return minRows * minCols;
14    }
15}

In all the solutions above, we first initialize minRows and minCols with m and n respectively. Then we iterate through each operation (op) in ops and update minRows and minCols with the minimum of their current values and the rows and columns specified in the operation. After looping through all the operations, we return the product of minRows and minCols as the result. This is because the smallest range specified by the operations is the range that will be included in all operations and thus will end up with the maximum counts in the matrix.Discussing Time Complexity:

In Python, Java, JavaScript, C++, and C#, the time complexity of this solution is O(n), where n is the number of operations, because we are merely iterating through all operations once.

Discussing Space Complexity:

The space complexity of the solution is O(1), which means that the amount of memory used does not change with the size of the input array. This is because we are using a constant amount of space to store minRows and minCols.

Regardless of the size of the input, we use only these two variables. Hence, while the input size could potentially increase, our memory usage remains constant, making the space complexity O(1).

Finally, this solution is optimal as the problem could not be solved in a better time or space complexity class. We only need to check the smallest range of each operation instead of executing each operation. Thus, this algorithm provides a significant reduction in time complexity compared to other possible solutions. It's a perfect demonstration of how problem understanding and careful analysis can lead to efficient solutions.


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