Unique cells in a binary matrix
Last Updated :
12 Dec, 2022
Given a matrix of size n × m consisting of 0’s and 1’s. We need to find the number of unique cells with value 1 such that the corresponding entire row and the entire column do not have another 1. Return the number of unique cells.
Examples:
Input : mat[][] = {0, 1, 0, 0
0, 0, 1, 0
1, 0, 0, 1}
Answer : 2
The two 1s that are unique
in their rows and columns
are highlighted.
Input : mat[][] = {
{0, 0, 0, 0, 0, 0, 1}
{0, 1, 0, 0, 0, 0, 0}
{0, 0, 0, 0, 0, 1, 0}
{1, 0, 0, 0, 0, 0, 0}
{0, 0, 1, 0, 0, 0, 1}
Output : 3
Method 1- Brute Force Approach:
In this approach, we are going to check for each cell with value 1 whether the corresponding rows satisfy our requirement. We will check in the corresponding rows and columns of each cell with the value 1.
Implementation:
C++
Java
Python3
MAX = 100
def isUnique(mat, i, j, n, m):
sumrow = 0
for k in range (m):
sumrow + = mat[i][k]
if (sumrow > 1 ):
return False
sumcol = 0
for k in range (n):
sumcol + = mat[k][j]
if (sumcol > 1 ):
return False
return True
def countUnique(mat, n, m):
uniquecount = 0
for i in range (n):
for j in range (m):
if (mat[i][j] and isUnique(mat, i, j, n, m)):
uniquecount + = 1
return uniquecount
mat = [[ 0 , 1 , 0 , 0 ],
[ 0 , 0 , 1 , 0 ],
[ 1 , 0 , 0 , 1 ]]
print (countUnique(mat, 3 , 4 ))
|
C#
using System;
public class GFG {
static readonly int MAX = 100;
static bool isUnique( int [,]mat, int i, int j,
int n, int m)
{
int sumrow = 0;
for ( int k = 0; k < m; k++) {
sumrow += mat[i,k];
if (sumrow > 1)
return false ;
}
int sumcol = 0;
for ( int k = 0; k < n; k++) {
sumcol += mat[k,j];
if (sumcol > 1)
return false ;
}
return true ;
}
static int countUnique( int [,]mat, int n, int m)
{
int uniquecount = 0;
for ( int i = 0; i < n; i++)
for ( int j = 0; j < m; j++)
if (mat[i,j]!=0 &&
isUnique(mat, i, j, n, m))
uniquecount++;
return uniquecount;
}
static public void Main() {
int [,]mat = {{0, 1, 0, 0},
{0, 0, 1, 0},
{1, 0, 0, 1}};
Console.Write(countUnique(mat, 3, 4));
}
}
|
PHP
<?php
$MAX = 100;
function isUnique( $mat , $i ,
$j , $n , $m )
{
global $MAX ;
$sumrow = 0;
for ( $k = 0; $k < $m ; $k ++)
{
$sumrow += $mat [ $i ][ $k ];
if ( $sumrow > 1)
return false;
}
$sumcol = 0;
for ( $k = 0; $k < $n ; $k ++)
{
$sumcol += $mat [ $k ][ $j ];
if ( $sumcol > 1)
return false;
}
return true;
}
function countUnique( $mat , $n , $m )
{
$uniquecount = 0;
for ( $i = 0; $i < $n ; $i ++)
for ( $j = 0; $j < $m ; $j ++)
if ( $mat [ $i ][ $j ] &&
isUnique( $mat , $i ,
$j , $n , $m ))
$uniquecount ++;
return $uniquecount ;
}
$mat = array ( array (0, 1, 0, 0),
array (0, 0, 1, 0),
array (1, 0, 0, 1));
echo countUnique( $mat , 3, 4);
?>
|
Javascript
<script>
let MAX = 100;
function isUnique(mat, i, j, n, m)
{
let sumrow = 0;
for (let k = 0; k < m; k++)
{
sumrow += mat[i][k];
if (sumrow > 1)
return false ;
}
let sumcol = 0;
for (let k = 0; k < n; k++)
{
sumcol += mat[k][j];
if (sumcol > 1)
return false ;
}
return true ;
}
function countUnique(mat, n, m)
{
let uniquecount = 0;
for (let i = 0; i < n; i++)
for (let j = 0; j < m; j++)
if (mat[i][j] != 0 &&
isUnique(mat, i, j, n, m))
uniquecount++;
return uniquecount;
}
let mat = [ [ 0, 1, 0, 0 ],
[ 0, 0, 1, 0 ],
[ 1, 0, 0, 1 ] ];
document.write(countUnique(mat, 3, 4));
</script>
|
Time Complexity: O((n*m)*(n+m))
Auxiliary Space: O(1), since no extra space has been taken.
This goes to the order of cubic due to check condition for every corresponding row and column
Method 2- O(n*m) Approach:
In this approach, we are going to use extra space for rowsum array and colsum array and then check for each cell with value 1 whether the corresponding rowsum array and colsum array values are 1.
Implementation:
C++
#include <bits/stdc++.h>
using namespace std;
const int MAX = 100;
int countUnique( int mat[][MAX], int n, int m)
{
int rowsum[n], colsum[m];
memset (colsum, 0, sizeof (colsum));
memset (rowsum, 0, sizeof (rowsum));
for ( int i = 0; i < n; i++)
for ( int j = 0; j < m; j++)
if (mat[i][j])
{
rowsum[i]++;
colsum[j]++;
}
int uniquecount = 0;
for ( int i = 0; i < n; i++)
for ( int j = 0; j < m; j++)
if (mat[i][j] &&
rowsum[i] == 1 &&
colsum[j] == 1)
uniquecount++;
return uniquecount;
}
int main()
{
int mat[][MAX] = {{0, 1, 0, 0},
{0, 0, 1, 0},
{1, 0, 0, 1}};
cout << countUnique(mat, 3, 4);
return 0;
}
|
Java
import java.util.*;
class GFG
{
static int MAX = 100 ;
static int countUnique( int mat[][], int n, int m)
{
int []rowsum = new int [n];
int []colsum = new int [m];
for ( int i = 0 ; i < n; i++)
for ( int j = 0 ; j < m; j++)
if (mat[i][j] != 0 )
{
rowsum[i]++;
colsum[j]++;
}
int uniquecount = 0 ;
for ( int i = 0 ; i < n; i++)
for ( int j = 0 ; j < m; j++)
if (mat[i][j] != 0 &&
rowsum[i] == 1 &&
colsum[j] == 1 )
uniquecount++;
return uniquecount;
}
public static void main(String[] args)
{
int mat[][] = {{ 0 , 1 , 0 , 0 },
{ 0 , 0 , 1 , 0 },
{ 1 , 0 , 0 , 1 }};
System.out.print(countUnique(mat, 3 , 4 ));
}
}
|
Python3
MAX = 100 ;
def countUnique(mat, n, m):
rowsum = [ 0 ] * n;
colsum = [ 0 ] * m;
for i in range (n):
for j in range (m):
if (mat[i][j] ! = 0 ):
rowsum[i] + = 1 ;
colsum[j] + = 1 ;
uniquecount = 0 ;
for i in range (n):
for j in range (m):
if (mat[i][j] ! = 0 and
rowsum[i] = = 1 and
colsum[j] = = 1 ):
uniquecount + = 1 ;
return uniquecount;
if __name__ = = '__main__' :
mat = [[ 0 , 1 , 0 , 0 ],
[ 0 , 0 , 1 , 0 ],
[ 1 , 0 , 0 , 1 ]];
print (countUnique(mat, 3 , 4 ));
|
C#
using System;
class GFG
{
static int MAX = 100;
static int countUnique( int [,]mat,
int n, int m)
{
int []rowsum = new int [n];
int []colsum = new int [m];
for ( int i = 0; i < n; i++)
for ( int j = 0; j < m; j++)
if (mat[i, j] != 0)
{
rowsum[i]++;
colsum[j]++;
}
int uniquecount = 0;
for ( int i = 0; i < n; i++)
for ( int j = 0; j < m; j++)
if (mat[i, j] != 0 &&
rowsum[i] == 1 &&
colsum[j] == 1)
uniquecount++;
return uniquecount;
}
public static void Main(String[] args)
{
int [,]mat = {{0, 1, 0, 0},
{0, 0, 1, 0},
{1, 0, 0, 1}};
Console.Write(countUnique(mat, 3, 4));
}
}
|
Javascript
<script>
let MAX = 100;
function countUnique(mat, n, m)
{
let rowsum = new Array(n);
rowsum.fill(0);
let colsum = new Array(m);
colsum.fill(0);
for (let i = 0; i < n; i++)
for (let j = 0; j < m; j++)
if (mat[i][j] != 0)
{
rowsum[i]++;
colsum[j]++;
}
let uniquecount = 0;
for (let i = 0; i < n; i++)
for (let j = 0; j < m; j++)
if (mat[i][j] != 0 &&
rowsum[i] == 1 &&
colsum[j] == 1)
uniquecount++;
return uniquecount;
}
let mat = [[0, 1, 0, 0],
[0, 0, 1, 0],
[1, 0, 0, 1]];
document.write(countUnique(mat, 3, 4));
</script>
|
Time Complexity : O(n*m)
Auxiliary Space: O(n+m)
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