What is the time complexity of Stamping the Grid?

The optimal solution has a time complexity of O(m*n).

What is the space complexity of Stamping the Grid?

The space complexity is O(m*n).

What data structures are used to solve Stamping the Grid?

This problem uses Array, Greedy, Matrix, Prefix Sum. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Stamping the Grid?

Insight 1: Using prefix sums allows for efficient calculation of the number of 1s within any subgrid. Insight 2: Pre-calculating all possible locations where a stamp can be placed without covering a 1 cell is crucial.

Which companies ask LeetCode 2132. Stamping the Grid in interviews?

This problem is frequently asked at Rubrik.

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Stamping the Grid - LeetCode 2132 Solution

Stamping the Grid - Complete Solution Guide

Stamping the Grid is LeetCode problem 2132, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given an m x n binary matrix grid where each cell is either 0 (empty) or 1 (occupied). You are then given stamps of size stampHeight x stampWidth . We want to fit the stamps such that they follow the given restrictions and requirements : Cover all the empty cells. Do not cover any of the occupied cells. We can put as many stamps as we want. Stamps can overlap with each other. Stamps are not allowed to be rotated . Stamps must stay completely inside the grid. Return true if it is possible

Detailed Explanation

The problem asks whether it's possible to cover all '0' cells in a given binary grid using overlapping stamps of a fixed size (stampHeight x stampWidth) without covering any '1' cells or exceeding the grid boundaries. The stamps cannot be rotated.

Solution Approach

The solution uses a three-step approach: 1) Compute prefix sums on the original grid to efficiently determine subgrid sums (number of 1s). 2) Identify all possible locations for the stamp. A stamp can be placed at a location (r, c) if the sum of the corresponding stampHeight x stampWidth subgrid is 0 (meaning no 1s). 3) Create a prefix sum matrix based on the possible stamp locations. Iterate through the grid; if a cell is 0, verify that at least one valid stamp covers it. This verification involves checking if the count of 'possible stamps' in the relevant area (defined by the stamp dimensions relative to the current 0 cell) is greater than 0.

Step-by-Step Algorithm

Step 1: Calculate the prefix sum matrix for the input grid. This allows for O(1) calculation of the sum of any rectangular subgrid.
Step 2: Iterate through the grid, checking for locations where a stamp can be placed. A stamp can be placed starting at (r, c) if the subgrid of size stampHeight x stampWidth starting at (r, c) contains only 0s. Mark these locations in the 'possible_stamps' matrix.
Step 3: Calculate the prefix sum matrix for the 'possible_stamps' matrix. This allows for O(1) calculation of the number of possible stamp locations within any rectangular subgrid.
Step 4: Iterate through the grid again. For each 0 cell at (r, c), determine the range of possible starting locations for a stamp that covers this cell (r - stampHeight + 1 to r, and c - stampWidth + 1 to c).
Step 5: Use the 'stamp_prefix_sum' to count the number of possible stamp locations within the range calculated in the previous step. If the count is 0, it means no stamp can cover the current 0 cell, so return false.
Step 6: If all 0 cells can be covered by at least one stamp, return true.

Key Insights

Insight 1: Using prefix sums allows for efficient calculation of the number of 1s within any subgrid.
Insight 2: Pre-calculating all possible locations where a stamp can be placed without covering a 1 cell is crucial.
Insight 3: Applying another prefix sum technique on the possible stamp locations enables quick verification if every 0 cell is covered by at least one valid stamp placement.
Insight 4: The key idea is to first find valid stamp locations and then efficiently check if these stamps cover all empty cells. Trying to directly place stamps on each empty cell would be inefficient due to potential overlaps and the need to check for invalid placements.

Complexity Analysis

Time Complexity: O(m*n)

Space Complexity: O(m*n)

Topics

This problem involves: Array, Greedy, Matrix, Prefix Sum.

Companies

Asked at: Rubrik.

Stamping the Grid | LeetCode 2132

Hard

Array Greedy Matrix Prefix Sum

Time: O(m*n)Space: O(m*n)

Solution Code

Available in:

class Solution {
    public boolean possibleToStamp(int[][] grid, int stampHeight, int stampWidth) {
        int m = grid.length;
        int n = grid[0].length;
        int h = stampHeight;
        int w = stampWidth;

        int[][] prefixSum = new int[m + 1][n + 1];
        for (int r = 0; r < m; r++) {
            for (int c = 0; c < n; c++) {
                prefixSum[r + 1][c + 1] = grid[r][c] + prefixSum[r][c + 1] + prefixSum[r + 1][c] - prefixSum[r][c];
            }
        }

        // Helper function to calculate the sum of elements within a rectangle
        function get_rect_sum(r1, c1, r2, c2)
        int getRectSum(int r1, int c1, int r2, int c2) {
            return prefixSum[r2 + 1][c2 + 1] - prefixSum[r1][c2 + 1] - prefixSum[r2 + 1][c1] + prefixSum[r1][c1];
        }

        int[][] possibleStamps = new int[m][n];
        for (int r = 0; r <= m - h; r++) {
            for (int c = 0; c <= n - w; c++) {
                if (getRectSum(r, c, r + h - 1, c + w - 1) == 0) {
                    possibleStamps[r][c] = 1;
                }
            }
        }

        int[][] stampPrefixSum = new int[m + 1][n + 1];
        for (int r = 0; r < m; r++) {
            for (int c = 0; c < n; c++) {
                stampPrefixSum[r + 1][c + 1] = possibleStamps[r][c] + stampPrefixSum[r][c + 1] + stampPrefixSum[r + 1][c] - stampPrefixSum[r][c];
            }
        }

        // Helper function to count the number of possible stamps within a rectangle
        int countPossibleStamps(int r1, int c1, int r2, int c2) {
            r1 = Math.max(0, r1);
            c1 = Math.max(0, c1);
            if (r1 > r2 || c1 > c2) {
                return 0;
            }
            return stampPrefixSum[r2 + 1][c2 + 1] - stampPrefixSum[r1][c2 + 1] - stampPrefixSum[r2 + 1][c1] + stampPrefixSum[r1][c1];
        }

        for (int r = 0; r < m; r++) {
            for (int c = 0; c < n; c++) {
                if (grid[r][c] == 0) {
                    int trStart = r - h + 1;
                    int tcStart = c - w + 1;
                    int trEnd = r;
                    int tcEnd = c;
                    if (countPossibleStamps(trStart, tcStart, trEnd, tcEnd) == 0) {
                        return false;
                    }
                }
            }
        }

        return true;
    }
}

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Problem Info

Description

Problem 2132: Stamping the Grid - Detailed explanation and solution approaches.

Topics

Array Greedy Matrix Prefix Sum

Companies

Rubrik

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2132

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

6 steps

1
Step 1: Calculate the prefix sum matrix for the input grid. This allows for O(1) calculation of the sum of any rectangular subgrid.
2
Step 2: Iterate through the grid, checking for locations where a stamp can be placed. A stamp can be placed starting at (r, c) if the subgrid of size stampHeight x stampWidth starting at (r, c) contains only 0s. Mark these locations in the 'possible_stamps' matrix.
3
Step 3: Calculate the prefix sum matrix for the 'possible_stamps' matrix. This allows for O(1) calculation of the number of possible stamp locations within any rectangular subgrid.
4
Step 4: Iterate through the grid again. For each 0 cell at (r, c), determine the range of possible starting locations for a stamp that covers this cell (r - stampHeight + 1 to r, and c - stampWidth + 1 to c).
5
Step 5: Use the 'stamp_prefix_sum' to count the number of possible stamp locations within the range calculated in the previous step. If the count is 0, it means no stamp can cover the current 0 cell, so return false.
6
Step 6: If all 0 cells can be covered by at least one stamp, return true.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(m*n)

💾

Space Complexity

O(m*n)

📝

Why?

The time complexity is dominated by the nested loops used to iterate through the grid and the possible stamp locations. The space complexity arises from the auxiliary matrices used to store prefix sums and possible stamp locations.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Dynamic Programming

Instead of prefix sums, a dynamic programming approach could be used to determine if a region can be stamped. However, it would still require iterating through the grid multiple times, likely resulting in the same time complexity.

Time: O(m*n)Space: O(m*n)

Trade-offs: The prefix sum approach is generally cleaner and easier to implement for this specific problem. DP might be beneficial if there were significant overlapping subproblems, but that is not the case here.

Brute Force

For each '0' cell, iterate through all possible stamp placements that could cover it and verify that each placement is valid. This would mean recalculating rectangle sums repeatedly.

Time: O(m*n * stampHeight * stampWidth). For each zero cell, we'd need to consider up to stampHeight * stampWidth possible stamp placements and check them. This makes it very slow.Space: O(1). No extra space is needed, aside from constant variables.

Trade-offs: This approach avoids extra space, but is much slower due to repeated calculations. Highly inefficient for larger grids.

✨

Key Insights

1Insight 1: Using prefix sums allows for efficient calculation of the number of 1s within any subgrid.
2Insight 2: Pre-calculating all possible locations where a stamp can be placed without covering a 1 cell is crucial.
3Insight 3: Applying another prefix sum technique on the possible stamp locations enables quick verification if every 0 cell is covered by at least one valid stamp placement.
4Insight 4: The key idea is to first find valid stamp locations and then efficiently check if these stamps cover all empty cells. Trying to directly place stamps on each empty cell would be inefficient due to potential overlaps and the need to check for invalid placements.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: The grid contains no 0s. The algorithm will correctly return true because there are no empty cells to cover.
Edge case 2: The stampHeight or stampWidth is larger than the grid dimensions. The loops will correctly not place any stamps out of bounds. The prefix sum logic handles boundary conditions by initializing the first row and column to 0.
Edge case 3: The grid contains only 0's and the stamp dimensions are 1x1. The algorithm will correctly cover all the 0's with many possible stamp locations.
Edge case 4: A single isolated 0. The algorithm correctly checks if it can be covered by a stamp and returns false if it cannot.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating the prefix sum or using it improperly, leading to incorrect subgrid sums.

✗Common mistake 2: Not properly handling the boundary conditions when placing the stamps or calculating the prefix sums, causing out-of-bounds errors.

✗Common mistake 3: Forgetting to account for overlapping stamps when checking if all empty cells are covered.

✗Common mistake 4: Assuming that the grid is always square, which is not guaranteed by the problem constraints.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Image processing, where the grid represents an image, 0s are pixels to be 'filled', and the stamp represents a brush. Determining if an area can be filled with a certain brush size without covering other specified regions.

🏦Real-world scenario 2: Resource allocation in a grid-based system. The grid represents available land, 0s are available spots, and the stamp is a building of a certain size. Verifying if resources can be allocated to all available spots given size constraints.

🎮Real-world scenario 3: Manufacturing defect detection - Identifying if defects (represented as empty cells) can be covered/repaired by a tool (stamp) of certain dimension.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 31, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I verify if a greedy approach is correct?

Prove the greedy choice property: the locally optimal choice leads to globally optimal solution. Use exchange argument: show that swapping any element with the greedy choice doesn't improve the solution. If you can't prove it, test extensively or consider DP.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Stamping the Grid"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether greedy techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?