What is the time complexity of Reconstruct a 2-Row Binary Matrix?

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

What data structures are used to solve Reconstruct a 2-Row Binary Matrix?

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

What are the key insights for solving Reconstruct a 2-Row Binary Matrix?

Insight 1: The sum of 'upper' and 'lower' must equal the sum of all elements in 'colsum'; otherwise, no solution exists. Insight 2: Columns with a sum of 2 must have 1 in both the upper and lower rows. Columns with a sum of 0 must have 0 in both rows. Columns with a sum of 1 have flexibility: either the upper or lower row can be 1, and the other must be 0.

Which companies ask LeetCode 1253. Reconstruct a 2-Row Binary Matrix in interviews?

This problem is frequently asked at American Express, Grab.

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Reconstruct a 2-Row Binary Matrix - LeetCode 1253 Solution

Q: How to solve Reconstruct a 2-Row Binary Matrix?

The provided solution utilizes a greedy approach. First, it verifies that the sum of the 'upper' and 'lower' constraints equals the sum of the 'colsum' array. If not, an empty matrix is returned. Then, it iterates through the 'colsum' array. If a column's sum is 2, both corresponding entries in the result matrix are set to 1, and 'upper' and 'lower' are decremented. If at any point 'upper' or 'lower' become negative, an empty matrix is returned. Finally, it iterates through the 'colsum' array again, and for columns with a sum of 1, it greedily assigns 1 to the upper row if 'upper' is greater than 0, otherwise assigns 1 to the lower row.

Reconstruct a 2-Row Binary Matrix - Complete Solution Guide

Reconstruct a 2-Row Binary Matrix is LeetCode problem 1253, 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 the following details of a matrix with n columns and 2 rows : The matrix is a binary matrix, which means each element in the matrix can be 0 or 1 . The sum of elements of the 0-th(upper) row is given as upper . The sum of elements of the 1-st(lower) row is given as lower . The sum of elements in the i-th column(0-indexed) is colsum[i] , where colsum is given as an integer array with length n . Your task is to reconstruct the matrix with upper , lower and colsum . Return it as a 2-D integer

Detailed Explanation

The problem requires reconstructing a 2xN binary matrix given the sum of each row (upper and lower) and the sum of each column (colsum). Each element in the matrix can only be 0 or 1. The goal is to return *any* valid matrix that satisfies these constraints. If no valid matrix exists, return an empty 2D array. Constraints: colsum length <= 10^5, upper/lower <= colsum length, colsum[i] can be 0, 1, or 2.

Solution Approach

The provided solution utilizes a greedy approach. First, it verifies that the sum of the 'upper' and 'lower' constraints equals the sum of the 'colsum' array. If not, an empty matrix is returned. Then, it iterates through the 'colsum' array. If a column's sum is 2, both corresponding entries in the result matrix are set to 1, and 'upper' and 'lower' are decremented. If at any point 'upper' or 'lower' become negative, an empty matrix is returned. Finally, it iterates through the 'colsum' array again, and for columns with a sum of 1, it greedily assigns 1 to the upper row if 'upper' is greater than 0, otherwise assigns 1 to the lower row.

Step-by-Step Algorithm

Step 1: Check if the sum of 'upper' and 'lower' equals the sum of 'colsum'. If not, return an empty matrix (no solution possible).
Step 2: Initialize a 2xN matrix 'res' with all elements set to 0.
Step 3: Iterate through 'colsum'. For each column 'i':
Step 4: If colsum[i] == 2, set res[0][i] = 1 and res[1][i] = 1. Decrement 'upper' and 'lower'.
Step 5: If upper < 0 or lower < 0 after assigning all colsum == 2, return empty matrix (no solution).
Step 6: Iterate through 'colsum' again. For each column 'i':
Step 7: If colsum[i] == 1, if upper > 0, set res[0][i] = 1 and decrement 'upper'. Otherwise, set res[1][i] = 1 and decrement 'lower'.
Step 8: Return the reconstructed matrix 'res'.

Key Insights

Insight 1: The sum of 'upper' and 'lower' must equal the sum of all elements in 'colsum'; otherwise, no solution exists.
Insight 2: Columns with a sum of 2 must have 1 in both the upper and lower rows. Columns with a sum of 0 must have 0 in both rows. Columns with a sum of 1 have flexibility: either the upper or lower row can be 1, and the other must be 0.
Insight 3: A greedy approach can be used by prioritizing filling columns with sum 2 first and then greedily filling the remaining columns with sum 1 to satisfy 'upper' and 'lower' constraints.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Greedy, Matrix.

Companies

Asked at: American Express, Grab.

Reconstruct a 2-Row Binary Matrix | LeetCode 1253

Medium

Array Greedy Matrix

Time: O(n)Space: O(n)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.List;

class Solution {
    public List<List<Integer>> reconstructMatrix(int upper, int lower, int[] colsum) {
        int n = colsum.length;
        
        int sum = 0;
        for (int i = 0; i < n; i++) {
            sum += colsum[i];
        }
        
        if (upper + lower != sum) {
            return new ArrayList<>();
        }

        List<List<Integer>> res = new ArrayList<>();
        res.add(new ArrayList<>());
        res.add(new ArrayList<>());

        for (int i = 0; i < n; i++) {
            res.get(0).add(0);
            res.get(1).add(0);
        }
        
        for (int i = 0; i < n; i++) {
            if (colsum[i] == 2) {
                res.get(0).set(i, 1);
                res.get(1).set(i, 1);
                upper -= 1;
                lower -= 1;
            }
        }
        
        if (upper < 0 || lower < 0) {
            return new ArrayList<>();
        }
            
        for (int i = 0; i < n; i++) {
            if (colsum[i] == 1) {
                if (upper > 0) {
                    res.get(0).set(i, 1);
                    upper -= 1;
                } else {
                    res.get(1).set(i, 1);
                    lower -= 1;
                }
            }
        }
        
        return res;
    }
}

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Problem Info

Description

Problem 1253: Reconstruct a 2-Row Binary Matrix - Detailed explanation and solution approaches.

Topics

Array Greedy Matrix

Companies

American ExpressGrab

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:1253

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: Check if the sum of 'upper' and 'lower' equals the sum of 'colsum'. If not, return an empty matrix (no solution possible).
2
Step 2: Initialize a 2xN matrix 'res' with all elements set to 0.
3
Step 3: Iterate through 'colsum'. For each column 'i':
4
Step 4: If colsum[i] == 2, set res[0][i] = 1 and res[1][i] = 1. Decrement 'upper' and 'lower'.
5
Step 5: If upper < 0 or lower < 0 after assigning all colsum == 2, return empty matrix (no solution).
6
Step 6: Iterate through 'colsum' again. For each column 'i':
7
Step 7: If colsum[i] == 1, if upper > 0, set res[0][i] = 1 and decrement 'upper'. Otherwise, set res[1][i] = 1 and decrement 'lower'.
8
Step 8: Return the reconstructed matrix 'res'.

💡

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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is determined by the number of iterations through the 'colsum' array, which directly scales with the input size 'n'. The space complexity is determined by the size of the 'res' matrix, which also scales linearly with 'n'.

📖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

Backtracking

A backtracking approach could explore different combinations of assigning 1s to the upper and lower rows when 'colsum[i]' is 1. However, this is significantly less efficient.

Time: O(2^n) in the worst case, as there are n columns with colsum equal to 1, and each has 2 possibilities.Space: O(n) for the recursion stack.

Trade-offs: The backtracking approach is much slower and would likely result in a Time Limit Exceeded (TLE) error for larger input sizes. The greedy approach is more efficient and suitable for this problem.

✨

Key Insights

1Insight 1: The sum of 'upper' and 'lower' must equal the sum of all elements in 'colsum'; otherwise, no solution exists.
2Insight 2: Columns with a sum of 2 must have 1 in both the upper and lower rows. Columns with a sum of 0 must have 0 in both rows. Columns with a sum of 1 have flexibility: either the upper or lower row can be 1, and the other must be 0.
3Insight 3: A greedy approach can be used by prioritizing filling columns with sum 2 first and then greedily filling the remaining columns with sum 1 to satisfy 'upper' and 'lower' constraints.

📚

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: 'upper' + 'lower' != sum(colsum). The solution explicitly checks for this condition at the beginning and returns an empty list if it's true.
Edge case 2: 'upper' or 'lower' become negative after assigning 1s to columns where 'colsum[i]' is 2. The solution includes a check after the first loop to handle this case and return an empty list.
Edge case 3: 'upper' and 'lower' not being fully utilized. If after filling all 2's and 1's are not possible anymore because 'upper' or 'lower' are negative, then return empty.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not checking if 'upper' + 'lower' equals the sum of 'colsum' before proceeding. This leads to incorrect results.

✗Common mistake 2: Not handling the cases where 'upper' or 'lower' become negative during the assignment process, which can lead to invalid solutions.

✗Common mistake 3: Forgetting to initialize the result matrix to 0, as the algorithm relies on the initial values being 0.

✅

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: Resource allocation. Imagine assigning tasks to two workers ('upper' and 'lower') with limited capacity, where 'colsum' represents the workload for each task. The algorithm can help determine a feasible assignment strategy.

🏦Real-world scenario 2: Data encoding. Representing data in a binary format with constraints on the number of 1s in certain rows or columns. This could be used for error detection or data compression schemes.

💡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: November 21, 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.

What are common greedy problem patterns?

Activity selection (sort by end time), interval scheduling (sort by start/end), Huffman coding (always combine smallest), fractional knapsack (sort by value/weight ratio), and minimum spanning tree. Look for problems where local decisions don't affect future options.

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 "Reconstruct a 2-Row Binary Matrix"?

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

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