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++
// 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 implementation 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 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 sums
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
# Store sum of column sums, max of column sum
# number of non zero column sums
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
# 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
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")
# this code is contributed by nirajgusain5
// C# implementation of above approach
using System;
public 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
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");
}
}
// This code has been contributed by shubhamsingh10
<script>
// Javascript implementation of the above approach
// Function to check if matrix exists
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;
// Store sum of rowsums, max of row sum
// number of non zero row sums
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++;
}
// Store sum of column sums, max of column sum
// number of non zero column sums
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++;
}
// 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
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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