LeetCode 598: Range Addition II Solution in Python – A Step-by-Step Guide

What Is LeetCode 598: Range Addition II?

In LeetCode 598: Range Addition II, you’re given integers m and n defining an m x n grid initialized with zeros, and a list ops where each tuple (a, b) represents an operation to increment all cells in the top-left a x b subgrid by 1. Your task is to return the number of cells with the maximum value after all operations. For example, with m = 3, n = 3, ops = [(2,2), (3,3)], the result is 4 because the top-left 2x2 region reaches a max value of 2, appearing in 4 cells. This problem builds on LeetCode 370: Range Addition for range updates but focuses on counting max-value cells efficiently.

Problem Statement

• Input:
• m (int)—grid rows.
• n (int)—grid columns.
• ops (List[List[int]])—list of (a, b) operations.
• Output: Integer—number of cells with maximum value after operations.
• Rules: Each operation increments top-left a x b subgrid; grid starts at 0.

Constraints

• 1 <= m, n <= 10³
• 0 <= ops.length <= 10⁴
• 1 <= a <= m, 1 <= b <= n

Examples

• Input: m = 3, n = 3, ops = [[2,2],[3,3]]
• Output: 4
• Grid after ops: [[2,2,1], [2,2,1], [1,1,1]], max = 2, count = 4.
• Input: m = 3, n = 3, ops = [[2,2],[3,3],[3,3],[3,3]]
• Output: 4
• Max = 4, top-left 2x2 gets 4, count = 4.
• Input: m = 3, n = 3, ops = []
• Output: 9
• No ops, all 0, count = 9.

Understanding the Problem: Counting Max Cells

To solve LeetCode 598: Range Addition II in Python, we need to determine how many cells in an m x n grid reach the maximum value after applying range increment operations, handling up to 10⁴ operations efficiently. A brute-force approach simulating the grid (O(mnk)) could work but is slow for large grids and operation counts (k = ops length). Instead, a key insight—the maximum value occurs in the intersection of all operation ranges—leads to an O(k) solution by finding the smallest a and b, as only cells in this region get incremented by all operations. We’ll explore:

• Best Solution (Minimum Range Intersection): O(k) time, O(1) space—fast and optimal.
• Alternative Solution (Brute-Force Grid Simulation): O(mnk) time, O(m*n) space—thorough but slow.

Let’s dive into the intersection solution with a friendly breakdown!

Best Solution: Minimum Range Intersection

Why Intersection Wins

The minimum range intersection solution is the best for LeetCode 598 because it computes the count of maximum-value cells in O(k) time and O(1) space by recognizing that the maximum value equals the number of operations, occurring only in the overlapping region of all ops, which is the minimum a by minimum b. It’s like finding the common area everyone visits, counting how many stay—all in a quick, clever insight!

How It Works

Think of this as a grid-overlap detective:

• Step 1: Understand Max Value:
• Max value = number of operations (each op adds 1).
• Step 2: Find Intersection:
• Min a across all ops = rows always incremented.
• Min b across all ops = columns always incremented.
• Step 3: Compute Count:
• Count = min_a * min_b (cells in intersection).
• If no ops, count = m * n (all cells max at 0).
• Step 4: Return Result:
• Number of cells with max value.
• Why It Works:
• Only cells in smallest range get all increments.
• Intersection defines max overlap.

It’s like a max-cell counting maestro!

Step-by-Step Example

Example: m = 3, n = 3, ops = [(2,2),(3,3)]

• Step 1: Max value = 2 (2 ops).
• Step 2: Find intersection:
• Min a = min(2, 3) = 2 (rows).
• Min b = min(2, 3) = 2 (cols).
• Step 3: Count = 2 * 2 = 4.
• Grid: [[2,2,1], [2,2,1], [1,1,1]], top-left 2x2 = 2.
• Step 4: Return 4.
• Result: 4.

Example: m = 3, n = 3, ops = []

• Step 1: No ops, max = 0.
• Step 2: No min a or b, use full grid.
• Step 3: Count = 3 * 3 = 9 (all cells = 0).
• Result: 9.

Code with Detailed Line-by-Line Explanation

Here’s the Python code, explained for beginners:

Copy codefrom typing import List

class Solution:
    def maxCount(self, m: int, n: int, ops: List[List[int]]) -> int:
        # Step 1: If no operations, all cells are max (0)
        if not ops:
            return m * n

        # Step 2: Find minimum a and b across all operations
        min_a = m
        min_b = n
        for a, b in ops:
            min_a = min(min_a, a)
            min_b = min(min_b, b)

        # Step 3: Compute count of cells in intersection
        return min_a * min_b

• Lines 6-7: Handle no ops: all m*n cells are max (0).
• Lines 10-11: Initialize min_a, min_b to full grid size.
• Lines 12-14: Iterate ops, update min_a and min_b.
• Line 17: Return product of min dimensions.
• Time Complexity: O(k)—single pass over ops.
• Space Complexity: O(1)—constant space.

It’s like an intersection-counting wizard!

Alternative Solution: Brute-Force Grid Simulation

Why an Alternative Approach?

The brute-force grid simulation approach creates and updates the full grid, counting max-value cells afterward, running in O(mnk) time and O(m*n) space. It’s thorough but impractical for large grids or many operations, making it a good baseline for small inputs or understanding.

How It Works

Picture this as a grid-updating brute:

• Step 1: Initialize m x n grid with zeros.
• Step 2: Apply each operation, incrementing subgrid cells.
• Step 3: Find max value and count occurrences.
• Step 4: Return count.

It’s like a grid-painting tally!

Step-by-Step Example

Example: m = 3, n = 3, ops = [(2,2)]

• Step 1: Grid = [[0,0,0], [0,0,0], [0,0,0]].
• Step 2: Apply (2,2):
• Grid = [[1,1,0], [1,1,0], [0,0,0]].
• Step 3: Max = 1, count = 4.
• Result: 4.

Code for Brute-Force Approach

Copy codefrom typing import List

class Solution:
    def maxCount(self, m: int, n: int, ops: List[List[int]]) -> int:
        # Step 1: Initialize grid
        grid = [[0] * n for _ in range(m)]

        # Step 2: Apply operations
        for a, b in ops:
            for i in range(a):
                for j in range(b):
                    grid[i][j] += 1

        # Step 3: Find max value and count
        max_val = 0
        for i in range(m):
            for j in range(n):
                max_val = max(max_val, grid[i][j])

        count = 0
        for i in range(m):
            for j in range(n):
                if grid[i][j] == max_val:
                    count += 1

        # Step 4: Return count
        return count

• Line 6: Create m x n grid of zeros.
• Lines 9-12: Increment subgrid for each op.
• Lines 15-18: Find max value in grid.
• Lines 21-25: Count cells with max value.
• Time Complexity: O(mnk)—nested loops for ops and grid.
• Space Complexity: O(m*n)—grid storage.

It’s a brute-force grid counter!

Comparing the Two Solutions

• Intersection (Best):
• Pros: O(k), O(1), fast and insightful.
• Cons: Requires intersection insight.
• Brute-Force (Alternative):
• Pros: O(mnk), O(m*n), explicit simulation.
• Cons: Slow for large grids/ops.

Intersection wins for efficiency!

Additional Examples and Edge Cases

• No ops: Full grid count.
• Single op: Op area count.
• Overlapping ops: Smallest overlap count.

Intersection handles them all!

Complexity Recap

• Intersection: Time O(k), Space O(1).
• Brute-Force: Time O(mnk), Space O(m*n).

Intersection’s the speed champ!

Key Takeaways

• Intersection: Range finesse—learn at Python Basics!
• Brute-Force: Grid grind.
• Cells: Counting is fun.
• Python: Insight or brute, your pick!

Final Thoughts: Count Those Max Cells!

LeetCode 598: Range Addition II in Python is a clever grid challenge. Minimum range intersection is your fast track, while brute-force simulation offers a thorough dive. Want more? Try LeetCode 370: Range Addition or LeetCode 223: Rectangle Area. Ready to increment? Head to Solve LeetCode 598 on LeetCode and count those max cells today!