What is the time complexity of Maximum Side Length of a Square with Sum Less than or Equal to Threshold?

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

What data structures are used to solve Maximum Side Length of a Square with Sum Less than or Equal to Threshold?

This problem uses Array, Binary Search, Matrix, Prefix Sum. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Maximum Side Length of a Square with Sum Less than or Equal to Threshold?

Insight 1: A brute-force approach of checking all possible squares would be inefficient. Using prefix sums allows us to calculate the sum of any submatrix (including squares) in O(1) time. Insight 2: The maximum side length 'k' can be increased incrementally. If a square of side 'k' satisfies the condition, we should try to see if a square of side 'k+1' also satisfies the condition, starting from the current largest side.

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Maximum Side Length of a Square with Sum Less than or Equal to Threshold - LeetCode 1292 Solution

Q: What is the space complexity of Maximum Side Length of a Square with Sum Less than or Equal to Threshold?

The space complexity is O(m*n).

Q: Which companies ask LeetCode 1292. Maximum Side Length of a Square with Sum Less than or Equal to Threshold in interviews?

This problem is frequently asked at IMC.

Maximum Side Length of a Square with Sum Less than or Equal to Threshold - Complete Solution Guide

Maximum Side Length of a Square with Sum Less than or Equal to Threshold is LeetCode problem 1292, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given a m x n matrix mat and an integer threshold , return the maximum side-length of a square with a sum less than or equal to threshold or return 0 if there is no such square . Example 1: Input: mat = [[1,1,3,2,4,3,2],[1,1,3,2,4,3,2],[1,1,3,2,4,3,2]], threshold = 4 Output: 2 Explanation: The maximum side length of square with sum less than 4 is 2 as shown. Example 2: Input: mat = [[2,2,2,2,2],[2,2,2,2,2],[2,2,2,2,2],[2,2,2,2,2],[2,2,2,2,2]], threshold = 1 Output: 0 Constraints: m == mat.length

Detailed Explanation

The problem requires us to find the maximum side length of a square within a given m x n matrix such that the sum of all elements within that square is less than or equal to a given threshold. If no such square exists, we should return 0. The input consists of the matrix 'mat' and the integer 'threshold'. The output is an integer representing the maximum side length of the qualifying square.

Solution Approach

The solution employs a prefix sum technique to efficiently compute the sum of elements within any square submatrix. First, a prefix sum matrix is constructed. Then, it iterates through each possible top-left corner of a square within the matrix. For each corner, it attempts to increase the square's side length, checking if the sum of elements within the square exceeds the threshold. The largest valid side length encountered is the final result.

Step-by-Step Algorithm

Step 1: Create a prefix sum matrix 'prefix' of size (m+1) x (n+1) initialized with zeros. The prefix sum at prefix[i][j] will store the sum of all elements in the submatrix mat[0...i-1][0...j-1].
Step 2: Populate the prefix sum matrix 'prefix' using the formula: prefix[r][c] = mat[r-1][c-1] + prefix[r-1][c] + prefix[r][c-1] - prefix[r-1][c-1]. This formula efficiently calculates the sum of the submatrix up to row 'r' and column 'c'.
Step 3: Initialize 'max_len' to 0, which will store the maximum valid side length found so far.
Step 4: Iterate through each possible top-left corner (r, c) of a square within the original matrix 'mat', starting from (1,1) to (m, n).
Step 5: For each corner (r, c), increment the potential side length 'k' from max_len + 1 and check if a square of side 'k' can be formed (i.e., r >= k and c >= k).
Step 6: If a square of side 'k' can be formed, calculate the sum of its elements using the prefix sum matrix: current_sum = prefix[r][c] - prefix[r-k][c] - prefix[r][c-k] + prefix[r-k][c-k].
Step 7: If current_sum is less than or equal to the 'threshold', update 'max_len' to 'k', as we've found a larger valid square.
Step 8: After iterating through all possible corners, return the final value of 'max_len'.

Key Insights

Insight 1: A brute-force approach of checking all possible squares would be inefficient. Using prefix sums allows us to calculate the sum of any submatrix (including squares) in O(1) time.
Insight 2: The maximum side length 'k' can be increased incrementally. If a square of side 'k' satisfies the condition, we should try to see if a square of side 'k+1' also satisfies the condition, starting from the current largest side.
Insight 3: Efficiently calculating the sum of each potential square using the prefix sum technique is crucial for optimizing the algorithm's runtime.

Complexity Analysis

Time Complexity: O(m*n)

Space Complexity: O(m*n)

Topics

This problem involves: Array, Binary Search, Matrix, Prefix Sum.

Companies

Asked at: IMC.

Maximum Side Length of a Square with Sum Less than or Equal to Threshold | LeetCode 1292

Medium

Array Binary Search Matrix Prefix Sum

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

Solution Code

Available in:

class Solution {
    public int maxSideLength(int[][] mat, int threshold) {
        int m = mat.length;
        int n = mat[0].length;

        int[][] prefix = new int[m + 1][n + 1];
        for (int r = 1; r <= m; r++) {
            for (int c = 1; c <= n; c++) {
                prefix[r][c] = mat[r-1][c-1] + prefix[r-1][c] + prefix[r][c-1] - prefix[r-1][c-1];
            }
        }

        int max_len = 0;
        for (int r = 1; r <= m; r++) {
            for (int c = 1; c <= n; c++) {
                int k = max_len + 1;
                if (r >= k && c >= k) {
                    int current_sum = prefix[r][c] - prefix[r-k][c] - prefix[r][c-k] + prefix[r-k][c-k];
                    if (current_sum <= threshold) {
                        max_len = k;
                    }
                }
            }
        }

        return max_len;
    }
}

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Problem Info

Description

Problem 1292: Maximum Side Length of a Square with Sum Less than or Equal to Threshold - Detailed explanation and solution approaches.

Topics

Array Binary Search Matrix Prefix Sum

Companies

IMC

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1292

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

8 steps

1
Step 1: Create a prefix sum matrix 'prefix' of size (m+1) x (n+1) initialized with zeros. The prefix sum at prefix[i][j] will store the sum of all elements in the submatrix mat[0...i-1][0...j-1].
2
Step 2: Populate the prefix sum matrix 'prefix' using the formula: prefix[r][c] = mat[r-1][c-1] + prefix[r-1][c] + prefix[r][c-1] - prefix[r-1][c-1]. This formula efficiently calculates the sum of the submatrix up to row 'r' and column 'c'.
3
Step 3: Initialize 'max_len' to 0, which will store the maximum valid side length found so far.
4
Step 4: Iterate through each possible top-left corner (r, c) of a square within the original matrix 'mat', starting from (1,1) to (m, n).
5
Step 5: For each corner (r, c), increment the potential side length 'k' from max_len + 1 and check if a square of side 'k' can be formed (i.e., r >= k and c >= k).
6
Step 6: If a square of side 'k' can be formed, calculate the sum of its elements using the prefix sum matrix: current_sum = prefix[r][c] - prefix[r-k][c] - prefix[r][c-k] + prefix[r-k][c-k].
7
Step 7: If current_sum is less than or equal to the 'threshold', update 'max_len' to 'k', as we've found a larger valid square.
8
Step 8: After iterating through all possible corners, return the final value of 'max_len'.

💡

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 O(m*n) because we iterate through each element of the matrix once to build the prefix sum matrix and then, in the worst case, iterate through the entire matrix again while increasing the side length of the square. The space complexity is O(m*n) because we store the prefix sum matrix of size (m+1) x (n+1).

📖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

Binary Search

For each possible square's top-left corner, we can binary search for the maximum side length. We could use the prefix sum to validate if a given side length is valid (sum <= threshold).

Time: O(m*n*log(min(m, n)))Space: O(m*n)

Trade-offs: This approach trades constant time increments for logarithmic time increments when searching for the maximum side length. If the threshold is usually small (meaning the maximum square is relatively small), the linear increment approach is faster. If the threshold is large, the binary search approach can avoid checking a large number of invalid square sizes.

✨

Key Insights

1Insight 1: A brute-force approach of checking all possible squares would be inefficient. Using prefix sums allows us to calculate the sum of any submatrix (including squares) in O(1) time.
2Insight 2: The maximum side length 'k' can be increased incrementally. If a square of side 'k' satisfies the condition, we should try to see if a square of side 'k+1' also satisfies the condition, starting from the current largest side.
3Insight 3: Efficiently calculating the sum of each potential square using the prefix sum technique is crucial for optimizing the algorithm's runtime.

📚

Key Terms Explained

For Students

Array

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

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Edge Cases

Edge case 1: If the matrix is empty (m=0 or n=0), the prefix sum calculation will not work. However, based on the constraints, m and n are always greater than or equal to 1, so there is no need to handle this explicitly.
Edge case 2: If all the elements of the matrix are very large, the sum of any square might exceed the threshold immediately. The algorithm correctly returns 0 in this case because max_len will never be updated.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to subtract prefix[r-1][c-1] when calculating the sum of a submatrix using the prefix sum matrix. This is crucial to avoid double-counting the overlapping region.

✗Common mistake 2: Incorrectly indexing into the original matrix 'mat' when calculating the prefix sum. Remember that 'prefix' matrix is (m+1) x (n+1) and is indexed from 1, while 'mat' is m x n and is indexed from 0. A common error is accessing mat[r][c] instead of mat[r-1][c-1].

✅

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 - identifying regions of interest with a certain average intensity (threshold). The matrix represents pixel intensities and the square represents the region.

🏦Real-world scenario 2: Data analysis - finding clusters of data points that meet certain criteria (threshold) in a spatial dataset. The matrix represents the spatial data and the square represents a potential cluster.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ 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: July 3, 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 14

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

@CodeNinjaJul 7

❓

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 avoid off-by-one errors in binary search?

Use the pattern: left = 0, right = n-1, while (left <= right), mid = left + (right - left) / 2. Decide if you need to find exact match, first occurrence, or last occurrence, and adjust your conditions accordingly. Test with arrays of size 0, 1, and 2.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

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 "Maximum Side Length of a Square with Sum Less than or Equal to Threshold"?

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

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