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 * 1040 <= ops.length <= 104ops[i].length == 21 <= ai <= m1 <= bi <= nTo solve this problem efficiently, observe that each operation affects a submatrix starting from the top-left corner. Thus, the final result of the operations is determined by the smallest intersected submatrix affected by all operations. Find the minimum value of "a" and "b" from all operations in ops. This will give you the dimensions of the area where the maximum integers will exist after applying all operations.
The key insight is that the size of the submatrix affected by all operations determines the count of maximum numbers in the final matrix.
This solution iterates over all operations provided and finds the minimum values of rows and columns from the operations. These minimums define the dimensions of the submatrix that will have the maximum value after all operations. We simply calculate the area of this submatrix to get the count of maximum integers.
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Time Complexity: O(k), where k is the number of operations.
Space Complexity: O(1), as we are using only a few extra variables.
This approach simulates the operations directly on the matrix. For each operation, we iterate through the specified submatrix and increment the values. Finally, we determine the number of times the maximum value occurs in the matrix. Although this method is less efficient due to its higher computational cost, it reinforces understanding of the problem.
This solution creates a matrix of size m x n and sets all values to zero. For each operation, it increments the values within the specified submatrix. After processing all operations, it finds the maximum value in the matrix and counts its occurrences.
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Time Complexity: O(m * n * k), where m and n are the dimensions of the matrix, and k is the number of operations.
Space Complexity: O(m * n), for the matrix.
| Approach | Complexity |
|---|---|
| Approach 1: Minimize Rows and Columns | Time Complexity: O(k), where k is the number of operations. Space Complexity: O(1), as we are using only a few extra variables. |
| Approach 2: Direct Simulation (Less Efficient) | Time Complexity: O(m * n * k), where m and n are the dimensions of the matrix, and k is the number of operations. Space Complexity: O(m * n), for the matrix. |
Range Sum Query 2D - Immutable - Leetcode 304 - Python • NeetCode • 50,535 views views
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