Range Addition II – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

You are given an m x n matrix M initialized with all 0‘s and an array of operations ops, where ops[i] = [ai, bi] means M[x][y] should be incremented by one for all 0 <= x < ai and 0 <= y < bi.

Count and return the number of maximum integers in the matrix after performing all the operations.

Example 1:

Input: m = 3, n = 3, ops = [[2,2],[3,3]]
Output: 4
Explanation: 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 3:

Input: m = 3, n = 3, ops = []
Output: 9

Constraints:

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

C++ Range Addition II LeetCode Solution

class Solution {
public:
    int maxCount(int m, int n, vector<vector<int>>& ops) {
        for (int i = 0; i < ops.size(); ++i) {
            m = min(m, ops[i][0]);
            n = min(n, ops[i][1]);
        }
        return m*n;
    }
};

Java Range Addition II LeetCode Solution

public class Solution {
    public int maxCount(int m, int n, int[][] ops) {
        if (ops == null || ops.length == 0) {
            return m * n;
        }
        
        int row = Integer.MAX_VALUE, col = Integer.MAX_VALUE;
        for(int[] op : ops) {
            row = Math.min(row, op[0]);
            col = Math.min(col, op[1]);
        }
        
        return row * col;
    }
}
 

Python 3 Range Addition II LeetCode Solution

public class Solution {
    public int maxCount(int m, int n, int[][] ops) {
        if (ops == null || ops.length == 0) {
            return m * n;
        }
        
        int row = Integer.MAX_VALUE, col = Integer.MAX_VALUE;
        for(int[] op : ops) {
            row = Math.min(row, op[0]);
            col = Math.min(col, op[1]);
        }
        
        return row * col;
    }
}

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