What is the time complexity of Check if Every Row and Column Contains All Numbers?

The optimal solution has a time complexity of O(n^2).

What data structures are used to solve Check if Every Row and Column Contains All Numbers?

This problem uses Array, Hash Table, Matrix. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Check if Every Row and Column Contains All Numbers?

Insight 1: We need to check both rows and columns independently for completeness. A simple nested loop isn't sufficient to efficiently handle both checks simultaneously. Insight 2: Using a set (or boolean array) to track the presence of numbers in each row and column is efficient. Sets provide fast membership checking (O(1) on average). Boolean arrays offer a slightly more space-efficient option in this specific case.

Which companies ask LeetCode 2133. Check if Every Row and Column Contains All Numbers in interviews?

This problem is frequently asked at Indeed, Instacart, Karat.

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Check if Every Row and Column Contains All Numbers - LeetCode 2133 Solution

Check if Every Row and Column Contains All Numbers - Complete Solution Guide

Check if Every Row and Column Contains All Numbers is LeetCode problem 2133, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

An n x n matrix is valid if every row and every column contains all the integers from 1 to n ( inclusive ). Given an n x n integer matrix matrix , return true if the matrix is valid . Otherwise, return false . Example 1: Input: matrix = [[1,2,3],[3,1,2],[2,3,1]] Output: true Explanation: In this case, n = 3, and every row and column contains the numbers 1, 2, and 3. Hence, we return true. Example 2: Input: matrix = [[1,1,1],[1,2,3],[1,2,3]] Output: false Explanation: In this case, n = 3, but the

Detailed Explanation

The problem asks us to determine if an n x n matrix is valid. A valid matrix means that every row and every column contains all the integers from 1 to n, without any duplicates or missing numbers. The input is a square integer matrix, and the output is a boolean value: `true` if the matrix is valid, and `false` otherwise.

Solution Approach

The provided solutions utilize a set (Python) or boolean array (Java, C++, C) to efficiently check each row and column. For each row and column, it iterates through the elements, adding them to the set/marking them in the boolean array. If a duplicate is encountered, or if the number is outside the valid range (1 to n), the function immediately returns `false`. After processing all elements of a row/column, it verifies if the set/array contains all numbers from 1 to n. If not, it returns `false`. If all rows and columns pass these checks, the function returns `true`.

Step-by-Step Algorithm

Step 1: Iterate through each row of the matrix.
Step 2: For each row, create a set (or boolean array) to store the numbers encountered in that row. Iterate through the elements of the row, adding each element to the set (or marking it in the array). If a duplicate is found or a number is out of range, immediately return `false`.
Step 3: After processing all elements of the row, check if the set (or array) contains all numbers from 1 to n. If not, immediately return `false`.
Step 4: Repeat steps 1-3 for each column of the matrix. This involves iterating through columns and using another set (or boolean array) for each column.
Step 5: If all rows and columns pass the checks, return `true`.

Key Insights

Insight 1: We need to check both rows and columns independently for completeness. A simple nested loop isn't sufficient to efficiently handle both checks simultaneously.
Insight 2: Using a set (or boolean array) to track the presence of numbers in each row and column is efficient. Sets provide fast membership checking (O(1) on average). Boolean arrays offer a slightly more space-efficient option in this specific case.
Insight 3: We can optimize slightly by returning `false` immediately if we find a number outside the range [1, n] or a duplicate within a row or column. This avoids unnecessary further checks.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table, Matrix.

Companies

Asked at: Indeed, Instacart, Karat.

Check if Every Row and Column Contains All Numbers | LeetCode 2133

Easy

Array Hash Table Matrix

Time: O(n^2)Space: O(n)

Solution Code

Available in:

class Solution {
    public boolean checkValid(int[][] matrix) {
        int n = matrix.length;

        for (int i = 0; i < n; i++) {
            boolean[] rowCheck = new boolean[n + 1];
            boolean[] colCheck = new boolean[n + 1];

            for (int j = 0; j < n; j++) {
                if (matrix[i][j] < 1 || matrix[i][j] > n) return false;
                if (rowCheck[matrix[i][j]]) return false;
                rowCheck[matrix[i][j]] = true;

                if (matrix[j][i] < 1 || matrix[j][i] > n) return false;
                if (colCheck[matrix[j][i]]) return false;
                colCheck[matrix[j][i]] = true;
            }
        }

        return true;
    }
}

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Problem Info

Description

Problem 2133: Check if Every Row and Column Contains All Numbers - Detailed explanation and solution approaches.

Topics

Array Hash Table Matrix

Companies

IndeedInstacartKarat

Stats

Difficulty:Easy

Languages:4

Companies:3

Problem #:2133

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

5 steps

1
Step 1: Iterate through each row of the matrix.
2
Step 2: For each row, create a set (or boolean array) to store the numbers encountered in that row. Iterate through the elements of the row, adding each element to the set (or marking it in the array). If a duplicate is found or a number is out of range, immediately return `false`.
3
Step 3: After processing all elements of the row, check if the set (or array) contains all numbers from 1 to n. If not, immediately return `false`.
4
Step 4: Repeat steps 1-3 for each column of the matrix. This involves iterating through columns and using another set (or boolean array) for each column.
5
Step 5: If all rows and columns pass the checks, 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(n^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The nested loops iterate through each cell of the n x n matrix, resulting in O(n^2) time complexity. The space complexity is O(n) because we use a set (or boolean array) of size n to track the numbers in each row and column. The space used for the sets/arrays is proportional to n, and the other variables' space is constant.

📖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 only boolean arrays

Instead of sets, use only boolean arrays in Python as well. This will provide a small space optimization but might lose the slight speed advantage of set lookups in Python.

Time: O(n^2)Space: O(n)

Trade-offs: This approach is slightly more space-efficient but could be marginally slower in languages like Python where set lookups are very optimized.

✨

Key Insights

1Insight 1: We need to check both rows and columns independently for completeness. A simple nested loop isn't sufficient to efficiently handle both checks simultaneously.
2Insight 2: Using a set (or boolean array) to track the presence of numbers in each row and column is efficient. Sets provide fast membership checking (O(1) on average). Boolean arrays offer a slightly more space-efficient option in this specific case.
3Insight 3: We can optimize slightly by returning `false` immediately if we find a number outside the range [1, n] or a duplicate within a row or column. This avoids unnecessary further checks.

📚

Key Terms Explained

For Students

Array

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

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Edge Cases

Edge case 1: An empty matrix (n=0). The code should ideally handle this explicitly (although the constraints prevent it), by returning `true` or raising an exception depending on the requirements. The current implementations don't explicitly handle this.
Edge case 2: A matrix with numbers outside the range [1,n]. The code correctly handles this by returning `false` immediately.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to check both rows and columns. Many beginners might only check the rows or only the columns, leading to incorrect results.

✗Common mistake 2: Inefficiently checking for duplicates. Using nested loops to check for duplicates within each row/column leads to O(n^3) time complexity.

✅

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: Sudoku solvers. Checking the validity of a partially filled Sudoku grid is very similar to this problem. The algorithm can be used to ensure that the current placement of numbers is valid.

🏦Real-world scenario 2: Data validation in spreadsheets or databases. Ensuring that data entries in a table adhere to certain constraints (all values in a column belong to a specific set) uses a similar approach.

💡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

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ 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: November 1, 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

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

@CodeNinjaJul 13

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

@TechLeadJul 6

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jun 29

❓

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.

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

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 "Check if Every Row and Column Contains All Numbers"?

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

🤖 Need help with this problem?