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++
// C++ program to count unique cells in
// a matrix
#include <bits/stdc++.h>
using namespace std;
const int MAX = 100;
// Returns true if mat[i][j] is unique
bool isUnique(int mat[][MAX], int i, int j,
int n, int m)
{
// checking in row calculating sumrow
// will be moving column wise
int sumrow = 0;
for (int k = 0; k < m; k++) {
sumrow += mat[i][k];
if (sumrow > 1)
return false;
}
// checking in column calculating sumcol
// will be moving row wise
int sumcol = 0;
for (int k = 0; k < n; k++) {
sumcol += mat[k][j];
if (sumcol > 1)
return false;
}
return true;
}
int countUnique(int mat[][MAX], 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] &&
isUnique(mat, i, j, n, m))
uniquecount++;
return uniquecount;
}
// Driver code
int main()
{
int mat[][MAX] = {{0, 1, 0, 0},
{0, 0, 1, 0},
{1, 0, 0, 1}};
cout << countUnique(mat, 3, 4);
return 0;
}
// Efficient Java program to count unique
// cells in a binary matrix
import java.io.*;
class GFG {
static final int MAX = 100;
// Returns true if mat[i][j] is unique
static boolean isUnique(int mat[][], int i, int j,
int n, int m)
{
// checking in row calculating sumrow
// will be moving column wise
int sumrow = 0;
for (int k = 0; k < m; k++) {
sumrow += mat[i][k];
if (sumrow > 1)
return false;
}
// checking in column calculating sumcol
// will be moving row wise
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;
}
// Driver code
static public 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));
}
}
// This code is contributed by Rajput-Ji
# Python3 program to count unique cells in
# a matrix
MAX = 100
# Returns true if mat[i][j] is unique
def isUnique(mat, i, j, n, m):
# checking in row calculating sumrow
# will be moving column wise
sumrow = 0
for k in range(m):
sumrow += mat[i][k]
if (sumrow > 1):
return False
# checking in column calculating sumcol
# will be moving row wise
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
# Driver code
mat = [[0, 1, 0, 0],
[0, 0, 1, 0],
[1, 0, 0, 1]]
print(countUnique(mat, 3, 4))
# This code is contributed by mohit kumar 29
// Efficient C# program to count unique
// cells in a binary matrix
using System;
public class GFG {
static readonly int MAX = 100;
// Returns true if mat[i][j] is unique
static bool isUnique(int [,]mat, int i, int j,
int n, int m)
{
// checking in row calculating sumrow
// will be moving column wise
int sumrow = 0;
for (int k = 0; k < m; k++) {
sumrow += mat[i,k];
if (sumrow > 1)
return false;
}
// checking in column calculating sumcol
// will be moving row wise
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;
}
// Driver code
static public void Main() {
int [,]mat = {{0, 1, 0, 0},
{0, 0, 1, 0},
{1, 0, 0, 1}};
Console.Write(countUnique(mat, 3, 4));
}
}
// This code is contributed by Rajput-Ji
<?php
// PHP program to count
// unique cells in a matrix
$MAX = 100;
// Returns true if
// mat[i][j] is unique
function isUnique($mat, $i,
$j, $n, $m)
{
global $MAX;
// checking in row calculating
// sumrow will be moving column wise
$sumrow = 0;
for ($k = 0; $k < $m; $k++)
{
$sumrow += $mat[$i][$k];
if ($sumrow > 1)
return false;
}
// checking in column
// calculating sumcol
// will be moving row wise
$sumcol = 0;
for ($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;
}
// Driver code
$mat = array(array(0, 1, 0, 0),
array(0, 0, 1, 0),
array(1, 0, 0, 1));
echo countUnique($mat, 3, 4);
// This code is contributed by ajit
?>
<script>
// Efficient Javascript program to count
// unique cells in a binary matrix
let MAX = 100;
// Returns true if mat[i][j] is unique
function isUnique(mat, i, j, n, m)
{
// Checking in row calculating sumrow
// will be moving column wise
let sumrow = 0;
for(let k = 0; k < m; k++)
{
sumrow += mat[i][k];
if (sumrow > 1)
return false;
}
// Checking in column calculating sumcol
// will be moving row wise
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;
}
// Driver code
let mat = [ [ 0, 1, 0, 0 ],
[ 0, 0, 1, 0 ],
[ 1, 0, 0, 1 ] ];
document.write(countUnique(mat, 3, 4));
// This code is contributed by decode2207
</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++
// Efficient C++ program to count unique
// cells in a binary matrix
#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));
// Count number of 1s in each row
// and in each column
for (int i = 0; i < n; i++)
for (int j = 0; j < m; j++)
if (mat[i][j])
{
rowsum[i]++;
colsum[j]++;
}
// Using above count arrays, find
// cells
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;
}
// Driver code
int main()
{
int mat[][MAX] = {{0, 1, 0, 0},
{0, 0, 1, 0},
{1, 0, 0, 1}};
cout << countUnique(mat, 3, 4);
return 0;
}
// Efficient Java program to count unique
// cells in a binary matrix
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];
// Count number of 1s in each row
// and in each column
for (int i = 0; i < n; i++)
for (int j = 0; j < m; j++)
if (mat[i][j] != 0)
{
rowsum[i]++;
colsum[j]++;
}
// Using above count arrays, find
// cells
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;
}
// Driver code
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));
}
}
// This code is contributed by Rajput-Ji
# Efficient Python3 program to count unique
# cells in a binary matrix
MAX = 100;
def countUnique(mat, n, m):
rowsum = [0] * n;
colsum = [0] * m;
# Count number of 1s in each row
# and in each column
for i in range(n):
for j in range(m):
if (mat[i][j] != 0):
rowsum[i] += 1;
colsum[j] += 1;
# Using above count arrays,
# find cells
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;
# Driver code
if __name__ == '__main__':
mat = [[ 0, 1, 0, 0 ],
[ 0, 0, 1, 0 ],
[ 1, 0, 0, 1 ]];
print(countUnique(mat, 3, 4));
# This code is contributed by 29AjayKumar
// Efficient C# program to count unique
// cells in a binary matrix
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];
// Count number of 1s in each row
// and in each column
for (int i = 0; i < n; i++)
for (int j = 0; j < m; j++)
if (mat[i, j] != 0)
{
rowsum[i]++;
colsum[j]++;
}
// Using above count arrays, find
// cells
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;
}
// Driver code
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));
}
}
// This code is contributed by Rajput-Ji
<script>
// Efficient Javascript program to count unique cells in a binary matrix
let MAX = 100;
function countUnique(mat, n, m)
{
let rowsum = new Array(n);
rowsum.fill(0);
let colsum = new Array(m);
colsum.fill(0);
// Count number of 1s in each row
// and in each column
for (let i = 0; i < n; i++)
for (let j = 0; j < m; j++)
if (mat[i][j] != 0)
{
rowsum[i]++;
colsum[j]++;
}
// Using above count arrays, find
// cells
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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