Find if a binary matrix exists with given row and column sums

Given an array Row[] of size R where Row[i] is the sum of elements of the i^th row and another array Column[] of size C where Column[i] is the sum of elements of the i^th column. The task is to check if it is possible to construct a binary matrix of R * C dimension which satisfies given row sums and column sums. A binary matrix is a matrix which is filled with only 0’s and 1’s.
Sum means the number of 1’s in particular row or column.

Examples:

Input: Row[] = {2, 2, 2, 2, 2}, Column[] = {5, 5, 0, 0}
Output: YES
Matrix is
{1, 1, 0, 0}
{1, 1, 0, 0}
{1, 1, 0, 0}
{1, 1, 0, 0}
{1, 1, 0, 0}

Input: Row[] = {0, 0, 3} Column[] = {3, 0, 0}
Output: NO

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

Key idea is that any cell in the matrix will contribute equally to both row and column sum, so sum of all the row sums must be equal to column sums.
Now, find the maximum of row sums, if this value is greater than the number of non zero column sums than matrix does not exist.
If the maximum of column sums is greater than the number of non zero row sums than matrix is not possible to construct.
If all the above 3 conditions is satisfied than matrix exists.

Below is the implementation of the above approach:

C++

// C++ implementation of the above approach 
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to check if matrix exists 
bool matrix_exist(int row[], int column[], int r, int c) 
{ 
    int row_sum = 0; 
    int column_sum = 0; 
    int row_max = -1; 
    int column_max = -1; 
    int row_non_zero = 0; 
    int column_non_zero = 0; 
  
    // Store sum of rowsums, max of row sum 
    // number of non zero row sums 
    for (int i = 0; i < r; i++) { 
        row_sum += row[i]; 
        row_max = max(row_max, row[i]); 
        if (row[i]) 
            row_non_zero++; 
    } 
  
    // Store sum of column sums, max of column sum 
    // number of non zero column sums 
    for (int i = 0; i < c; i++) { 
        column_sum += column[i]; 
        column_max = max(column_max, column[i]); 
        if (column[i]) 
            column_non_zero++; 
    } 
  
    // Check condition 1, 2, 3 
    if ((row_sum != column_sum) || 
        (row_max > column_non_zero) || 
        (column_max > row_non_zero)) 
        return false; 
  
    return true; 
} 
  
// Driver Code 
int main() 
{ 
    int row[] = { 2, 2, 2, 2, 2 }; 
    int column[] = { 5, 5, 0, 0 }; 
    int r = sizeof(row) / sizeof(row[0]); 
    int c = sizeof(column) / sizeof(column[0]); 
  
    if (matrix_exist(row, column, r, c)) 
        cout << "YES\n"; 
    else
        cout << "NO\n"; 
} 

Java

// Java implemenation of above approach 
import java.util.*; 
  
class GFG  
{ 
  
    // Function to check if matrix exists 
    static boolean matrix_exist(int row[], int column[], 
                                        int r, int c)  
    { 
        int row_sum = 0; 
        int column_sum = 0; 
        int row_max = -1; 
        int column_max = -1; 
        int row_non_zero = 0; 
        int column_non_zero = 0; 
  
        // Store sum of rowsums, max of row sum 
        // number of non zero row sums 
        for (int i = 0; i < r; i++)  
        { 
            row_sum += row[i]; 
            row_max = Math.max(row_max, row[i]); 
            if (row[i] > 0) 
            { 
                row_non_zero++; 
            } 
        } 
  
        // Store sum of column sums, max of column sum 
        // number of non zero column sums 
        for (int i = 0; i < c; i++) 
        { 
            column_sum += column[i]; 
            column_max = Math.max(column_max, column[i]); 
            if (column[i] > 0)  
            { 
                column_non_zero++; 
            } 
        } 
  
        // Check condition 1, 2, 3 
        if ((row_sum != column_sum) 
                || (row_max > column_non_zero) 
                || (column_max > row_non_zero)) 
        { 
            return false; 
        } 
  
        return true;  
    } 
  
// Driver Code 
public static void main(String[] args) 
{ 
    int row[] = { 2, 2, 2, 2, 2 }; 
    int column[] = { 5, 5, 0, 0 }; 
    int r = row.length; 
    int c = column.length; 
  
    if (matrix_exist(row, column, r, c)) 
        System.out.println("Yes"); 
    else
        System.out.println("No"); 
} 
} 
  
// This code has been contributed by 29AjayKumar 

Python

# Python3 implementation of the above approach 
  
# Function to check if matrix exists 
def matrix_exist(row, column, r, c) : 
      
    row_sum = 0
    column_sum = 0
    row_max = -1
    column_max = -1
    row_non_zero = 0
    column_non_zero = 0
  
    # Store sum of rowsums, max of row sum 
    # number of non zero row sum 
    for i in range (0, r): 
        row_sum += row[i] 
        row_max = max(row_max, row[i]) 
        if (row[i]) : 
            row_non_zero = row_non_zero + 1
      
  
    # Store sum of column sums, max of column sum 
    # number of non zero column sums 
    for i in range (0, c) : 
        column_sum += column[i] 
        column_max = max(column_max, column[i]) 
        if (column[i]) : 
            column_non_zero = column_non_zero + 1
      
  
    # Check condition 1, 2, 3 
    if ((row_sum != column_sum) or
        (row_max > column_non_zero) or
        (column_max > row_non_zero)) : 
        return False
  
    return True
  
  
# Driver Code 
row = [ 2, 2, 2, 2, 2 ] 
column = [ 5, 5, 0, 0 ] 
r = len(row) 
c = len(column) 
  
if (matrix_exist(row, column, r, c)) : 
    print("YES") 
else : 
    print("NO") 
  
# This code is contributed by ihritik 

C#

// C# implemenation of above approach 
using System; 
  
class GFG  
{ 
  
    // Function to check if matrix exists 
    static bool matrix_exist(int [] row, int []column, 
                                        int r, int c)  
    { 
        int row_sum = 0; 
        int column_sum = 0; 
        int row_max = -1; 
        int column_max = -1; 
        int row_non_zero = 0; 
        int column_non_zero = 0; 
  
        // Store sum of rowsums, max of row sum 
        // number of non zero row sums 
        for (int i = 0; i < r; i++)  
        { 
            row_sum += row[i]; 
            row_max = Math.Max(row_max, row[i]); 
            if (row[i] > 0) 
            { 
                row_non_zero++; 
            } 
        } 
  
        // Store sum of column sums, max of column sum 
        // number of non zero column sums 
        for (int i = 0; i < c; i++) 
        { 
            column_sum += column[i]; 
            column_max = Math.Max(column_max, column[i]); 
            if (column[i] > 0)  
            { 
                column_non_zero++; 
            } 
        } 
  
        // Check condition 1, 2, 3 
        if ((row_sum != column_sum) 
                || (row_max > column_non_zero) 
                || (column_max > row_non_zero)) 
        { 
            return false; 
        } 
  
        return true; 
    } 
      
    // Driver Code 
    public static void Main() 
    { 
        int [] row = new int [] { 2, 2, 2, 2, 2 }; 
        int []column = new int [] { 5, 5, 0, 0 }; 
        int r = row.Length; 
        int c = column.Length; 
      
        if (matrix_exist(row, column, r, c)) 
            Console.WriteLine("YES"); 
        else
            Console.WriteLine("NO"); 
    } 
} 
  
// This code is contributed by ihritik 

Output:

YES

Time Complexity : O(N)

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Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python

C#

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