598. Range Addition II ¶ Time: $O(N)$, where $N = |\texttt{ops}|$ Space: C++JavaPython 1 2 3 4 5 6 7 8 9 10 11 12 13 14class Solution { public: int maxCount(int m, int n, vector<vector<int>>& ops) { int minY = m; int minX = n; for (const vector<int>& op : ops) { minY = min(minY, op[0]); minX = min(minX, op[1]); } return minX * minY; } }; 1 2 3 4 5 6 7 8 9 10 11 12 13class Solution { public int maxCount(int m, int n, int[][] ops) { int minY = m; int minX = n; for (int[] op : ops) { minY = Math.min(minY, op[0]); minX = Math.min(minX, op[1]); } return minX * minY; } } 1 2 3 4 5 6 7 8 9 10class Solution: def maxCount(self, m: int, n: int, ops: list[list[int]]) -> int: minY = m minX = n for y, x in ops: minY = min(minY, y) minX = min(minX, x) return minX * minY