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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