Find if a binary matrix exists with given row and column sums
Last Updated :
01 Nov, 2022
Given an array Row[] of size R where Row[i] is the sum of elements of the ith row and another array Column[] of size C where Column[i] is the sum of elements of the ith 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
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++
#include <bits/stdc++.h>
using namespace std;
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;
for ( int i = 0; i < r; i++) {
row_sum += row[i];
row_max = max(row_max, row[i]);
if (row[i])
row_non_zero++;
}
for ( int i = 0; i < c; i++) {
column_sum += column[i];
column_max = max(column_max, column[i]);
if (column[i])
column_non_zero++;
}
if ((row_sum != column_sum) ||
(row_max > column_non_zero) ||
(column_max > row_non_zero))
return false ;
return true ;
}
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
import java.util.*;
class GFG
{
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 ;
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++;
}
}
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++;
}
}
if ((row_sum != column_sum)
|| (row_max > column_non_zero)
|| (column_max > row_non_zero))
{
return false ;
}
return true ;
}
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" );
}
}
|
Python3
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
for i in range (r):
row_sum + = row[i]
row_max = max (row_max, row[i])
if (row[i]):
row_non_zero = row_non_zero + 1
for i in range (c):
column_sum = column_sum + column[i]
column_max = max (column_max, column[i])
if (column[i]):
column_non_zero = column_non_zero + 1
if ((row_sum ! = column_sum)
or (row_max > column_non_zero)
or (column_max > row_non_zero)):
return False
return True
if __name__ = = '__main__' :
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" )
|
C#
using System;
public class GFG{
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;
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++;
}
}
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++;
}
}
if ((row_sum != column_sum)
|| (row_max > column_non_zero)
|| (column_max > row_non_zero))
{
return false ;
}
return true ;
}
static public void Main ()
{
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))
Console.Write( "YES" );
else
Console.Write( "NO" );
}
}
|
Javascript
<script>
function matrix_exist(row, column, r, c)
{
var row_sum = 0;
var column_sum = 0;
var row_max = -1;
var column_max = -1;
var row_non_zero = 0;
var column_non_zero = 0;
for ( var i = 0; i < r; i++) {
row_sum += row[i];
row_max = Math.max(row_max, row[i]);
if (row[i])
row_non_zero++;
}
for ( var i = 0; i < c; i++) {
column_sum += column[i];
column_max = Math.max(column_max, column[i]);
if (column[i])
column_non_zero++;
}
if ((row_sum != column_sum) ||
(row_max > column_non_zero) ||
(column_max > row_non_zero))
return false ;
return true ;
}
var row = [2, 2, 2, 2, 2];
var column = [5, 5, 0, 0];
var r = row.length;
var c = column.length;
if (matrix_exist(row, column, r, c))
document.write( "YES" );
else
document.write( "NO" );
</script>
|
Time Complexity : O(N)
Auxiliary Space: O(1)
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