Watch 10 video solutions for Range Addition II, a easy level problem involving Array, Math. This walkthrough by Coding Decoded has 2,994 views views. Want to try solving it yourself? Practice on FleetCode or read the detailed text solution.
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 <= nProblem Overview: You start with an m x n matrix filled with zeros. Each operation [a, b] increments every cell in the submatrix from (0,0) to (a-1,b-1). After applying all operations, the task is to count how many cells contain the maximum value in the matrix.
Approach 1: Minimize Rows and Columns (O(k) time, O(1) space)
The key observation is that every operation increments a rectangle starting from the top-left corner. The cells that end up with the largest value are the ones included in all operations. That overlapping region is determined by the smallest row boundary and the smallest column boundary across all operations. Iterate through the operations, track minRow = min(minRow, a) and minCol = min(minCol, b). The number of cells with the maximum value becomes minRow * minCol. This works because each operation overlaps in the top-left region, and the tightest bounds define the common intersection. Time complexity is O(k) where k is the number of operations, and space complexity is O(1). This approach mainly relies on simple iteration over an array of operations and basic math reasoning.
Approach 2: Direct Simulation (Less Efficient) (O(k * m * n) time, O(m * n) space)
A straightforward method is to simulate the process exactly as described. Create an m x n matrix initialized with zeros. For each operation [a, b], iterate through rows 0..a-1 and columns 0..b-1 and increment each cell. After processing all operations, scan the matrix to find the maximum value and count how many cells contain it. While easy to implement, this approach performs unnecessary repeated work since many cells are incremented multiple times across operations. The time complexity grows to O(k * m * n) with O(m * n) space, which becomes inefficient for larger matrices.
Recommended for interviews: The minimize rows and columns approach is the expected solution. It shows you recognize the overlapping rectangle pattern rather than simulating updates. Starting with the direct simulation demonstrates understanding of the problem mechanics, but identifying the intersection of all operations and reducing the solution to minRow * minCol demonstrates strong problem-solving skills and comfort with array traversal and simple mathematical reasoning.
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Minimize Rows and Columns | O(k) | O(1) | Best approach for interviews and production. Uses the intersection of all operations. |
| Direct Simulation | O(k * m * n) | O(m * n) | Useful for understanding the problem or when constraints are extremely small. |