What is the time complexity of Count Square Submatrices with All Ones?

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

What data structures are used to solve Count Square Submatrices with All Ones?

This problem uses Array, Dynamic Programming, Matrix. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Count Square Submatrices with All Ones?

Insight 1: The size of the largest square submatrix ending at a given cell (i, j) is determined by the minimum size of the squares ending at its top (i-1, j), left (i, j-1), and top-left (i-1, j-1) neighbors, plus 1 (if the current cell is 1). Insight 2: Dynamic programming can be effectively used to store the size of the largest square submatrix ending at each cell. The original matrix itself can be used as the DP table, optimizing space.

Which companies ask LeetCode 1277. Count Square Submatrices with All Ones in interviews?

This problem is frequently asked at Tinkoff.

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Count Square Submatrices with All Ones - LeetCode 1277 Solution

Count Square Submatrices with All Ones - Complete Solution Guide

Count Square Submatrices with All Ones is LeetCode problem 1277, 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 * n matrix of ones and zeros, return how many square submatrices have all ones. Example 1: Input: matrix = [ [0,1,1,1], [1,1,1,1], [0,1,1,1] ] Output: 15 Explanation: There are 10 squares of side 1. There are 4 squares of side 2. There is 1 square of side 3. Total number of squares = 10 + 4 + 1 = 15 . Example 2: Input: matrix = [ [1,0,1], [1,1,0], [1,1,0] ] Output: 7 Explanation: There are 6 squares of side 1. There is 1 square of side 2. Total number of squares = 6 + 1 = 7 . Constrain

Detailed Explanation

The problem asks us to find the total number of square submatrices containing only ones within a given m x n matrix of ones and zeros. A square submatrix is a contiguous square block of the matrix. The goal is to count all such squares, regardless of their size, that are composed entirely of ones.

Solution Approach

The solution employs dynamic programming to efficiently count the number of square submatrices with all ones. The provided code modifies the input matrix in place, using it as a DP table. For each cell (i, j) containing a 1, the code checks its top, left, and top-left neighbors. If any of these neighbors are zeros, then the maximum possible square has size 1. Otherwise, the code updates the value of matrix[i][j] to be 1 + the minimum of its three neighbors. The value at matrix[i][j] then represents the size of the largest square submatrix ending at that cell. The sum of all the values in the modified matrix gives the total count of square submatrices with all ones.

Step-by-Step Algorithm

Step 1: Initialize `total_squares` to 0. This variable will store the count of square submatrices.
Step 2: Iterate through each cell `matrix[i][j]` in the matrix.
Step 3: If `matrix[i][j]` is 1, check if it's not in the first row or column (i.e., `i > 0` and `j > 0`).
Step 4: If the cell is not in the first row or column, update its value with `matrix[i][j] = 1 + min(matrix[i-1][j], matrix[i][j-1], matrix[i-1][j-1])`. This represents the size of the largest square ending at this cell.
Step 5: Add the value of `matrix[i][j]` to `total_squares`. This accumulates the count of squares.
Step 6: After iterating through all the cells, return `total_squares`.

Key Insights

Insight 1: The size of the largest square submatrix ending at a given cell (i, j) is determined by the minimum size of the squares ending at its top (i-1, j), left (i, j-1), and top-left (i-1, j-1) neighbors, plus 1 (if the current cell is 1).
Insight 2: Dynamic programming can be effectively used to store the size of the largest square submatrix ending at each cell. The original matrix itself can be used as the DP table, optimizing space.
Insight 3: The count of squares can be computed by summing up the sizes of the largest squares ending at each cell within the matrix. This represents the total number of squares found.

Complexity Analysis

Time Complexity: O(m*n)

Space Complexity: O(1)

Topics

This problem involves: Array, Dynamic Programming, Matrix.

Companies

Asked at: Tinkoff.

Count Square Submatrices with All Ones | LeetCode 1277

Medium

Array Dynamic Programming Matrix

Time: O(m*n)Space: O(1)

Solution Code

Available in:

class Solution {
    public int countSquares(int[][] matrix) {
        int m = matrix.length;
        int n = matrix[0].length;
        int total_squares = 0;

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (matrix[i][j] == 1) {
                    if (i > 0 && j > 0) {
                        matrix[i][j] = 1 + Math.min(matrix[i-1][j], Math.min(matrix[i][j-1], matrix[i-1][j-1]));
                    }
                    total_squares += matrix[i][j];
                }
            }
        }

        return total_squares;
    }
}

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Problem Info

Description

Problem 1277: Count Square Submatrices with All Ones - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Matrix

Companies

Tinkoff

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1277

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: Initialize `total_squares` to 0. This variable will store the count of square submatrices.
2
Step 2: Iterate through each cell `matrix[i][j]` in the matrix.
3
Step 3: If `matrix[i][j]` is 1, check if it's not in the first row or column (i.e., `i > 0` and `j > 0`).
4
Step 4: If the cell is not in the first row or column, update its value with `matrix[i][j] = 1 + min(matrix[i-1][j], matrix[i][j-1], matrix[i-1][j-1])`. This represents the size of the largest square ending at this cell.
5
Step 5: Add the value of `matrix[i][j]` to `total_squares`. This accumulates the count of squares.
6
Step 6: After iterating through all the cells, return `total_squares`.

💡

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(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is determined by the nested loops that iterate over all cells in the matrix. The in-place modification of the matrix ensures constant space complexity because no auxiliary data structures that scale with the size of the input are used.

📖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

Using a separate DP table

Instead of modifying the input matrix in place, a separate DP table of the same dimensions can be created to store the sizes of the largest square submatrices. This preserves the original matrix.

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

Trade-offs: Using a separate DP table offers the advantage of not modifying the input matrix but increases the space complexity to O(m*n). Use this approach if preserving the original matrix is a requirement.

✨

Key Insights

1Insight 1: The size of the largest square submatrix ending at a given cell (i, j) is determined by the minimum size of the squares ending at its top (i-1, j), left (i, j-1), and top-left (i-1, j-1) neighbors, plus 1 (if the current cell is 1).
2Insight 2: Dynamic programming can be effectively used to store the size of the largest square submatrix ending at each cell. The original matrix itself can be used as the DP table, optimizing space.
3Insight 3: The count of squares can be computed by summing up the sizes of the largest squares ending at each cell within the matrix. This represents the total number of squares found.

📚

Key Terms Explained

For Students

Array

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

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: Input matrix with only one cell. The algorithm correctly handles this case, returning 1 if the cell value is 1, and 0 otherwise.
Edge case 2: Input matrix containing only zeros. The algorithm correctly returns 0, as there are no square submatrices of ones.
Edge case 3: Matrix with only first row or first column having ones. The solution iterates over the matrix and increments only the elements containing 1 and so works fine.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to check if a cell contains 1 before performing the DP update. This leads to incorrect results if a cell containing 0 is considered as the bottom-right corner of a square submatrix.

✗Common mistake 2: Not handling the cases where i = 0 or j = 0 correctly. If a cell is on the first row or column, it can only form a square of size 1 if its value is 1.

✗Common mistake 3: Incorrectly calculating the minimum of the neighboring cells. Ensure that the `min` function is used correctly with all three neighbors.

✅

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 patterns or features within images, where '1' might represent a pixel of interest. For example, finding contiguous regions of a specific color or texture.

🏦Real-world scenario 2: Resource allocation - determining the largest contiguous block of available resources (represented by '1') in a grid-like structure, such as assigning seats in a theater or allocating memory blocks in a system.

💡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: June 18, 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

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Count Square Submatrices with All Ones"?

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 dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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