Given n × m binary matrix, count the number of sets where a set can be formed one or more same values in a row or column.
Examples:
Input: 1 0 1
0 1 0
Output: 8
Explanation: There are six one-element sets
(three 1s and three 0s). There are two two-
element sets, the first one consists of the
first and the third cells of the first row.
The second one consists of the first and the
third cells of the second row.
Input: 1 0
1 1
Output: 6
The number of non-empty subsets of x elements is 2x – 1. We traverse every row and calculate numbers of 1’s and 0’s cells. For every u zeros and v ones, total sets is 2u – 1 + 2v – 1. We then traverse all columns and compute same values and compute overall sum. We finally subtract m x n from the overall sum as single elements are considered twice.
Implementation:
CPP
#include <bits/stdc++.h>
using namespace std;
const int m = 3;
const int n = 2;
long long countSets( int a[n][m])
{
long res = 0;
for ( int i = 0; i < n; i++) {
int u = 0, v = 0;
for ( int j = 0; j < m; j++) {
if (a[i][j] == 1)
u++;
else
v++;
}
res += pow (2, u) - 1 + pow (2, v) - 1;
}
for ( int i = 0; i < m; i++) {
int u = 0, v = 0;
for ( int j = 0; j < n; j++) {
if (a[j][i] == 1)
u++;
else
v++;
}
res += pow (2, u) - 1 + pow (2, v) - 1;
}
return res - (n * m);
}
int main()
{
int a[][3] = { { 1, 0, 1 }, { 0, 1, 0 } };
cout << countSets(a);
return 0;
}
|
Java
import java.util.*;
class GFG {
static final int m = 3 ;
static final int n = 2 ;
static long countSets( int a[][]) {
long res = 0 ;
for ( int i = 0 ; i < n; i++) {
int u = 0 , v = 0 ;
for ( int j = 0 ; j < m; j++) {
if (a[i][j] == 1 )
u++;
else
v++;
}
res += Math.pow( 2 , u) - 1 + Math.pow( 2 , v) - 1 ;
}
for ( int i = 0 ; i < m; i++) {
int u = 0 , v = 0 ;
for ( int j = 0 ; j < n; j++) {
if (a[j][i] == 1 )
u++;
else
v++;
}
res += Math.pow( 2 , u) - 1 + Math.pow( 2 , v) - 1 ;
}
return res - (n * m);
}
public static void main(String[] args) {
int a[][] = {{ 1 , 0 , 1 }, { 0 , 1 , 0 }};
System.out.print(countSets(a));
}
}
|
Python3
m = 3
n = 2
def countSets(a):
res = 0
for i in range (n):
u = 0
v = 0
for j in range (m):
if a[i][j]:
u + = 1
else :
v + = 1
res + = pow ( 2 , u) - 1 + pow ( 2 , v) - 1
for i in range (m):
u = 0
v = 0
for j in range (n):
if a[j][i]:
u + = 1
else :
v + = 1
res + = pow ( 2 , u) - 1 + pow ( 2 , v) - 1
return res - (n * m)
a = [[ 1 , 0 , 1 ],[ 0 , 1 , 0 ]]
print (countSets(a))
|
C#
using System;
class GFG {
static int m = 3;
static int n = 2;
static long countSets( int [,]a)
{
long res = 0;
for ( int i = 0; i < n; i++)
{
int u = 0, v = 0;
for ( int j = 0; j < m; j++)
{
if (a[i,j] == 1)
u++;
else
v++;
}
res += ( long )(Math.Pow(2, u) - 1
+ Math.Pow(2, v)) - 1;
}
for ( int i = 0; i < m; i++)
{
int u = 0, v = 0;
for ( int j = 0; j < n; j++)
{
if (a[j,i] == 1)
u++;
else
v++;
}
res += ( long )(Math.Pow(2, u) - 1
+ Math.Pow(2, v)) - 1;
}
return res - (n * m);
}
public static void Main()
{
int [,]a = {{1, 0, 1}, {0, 1, 0}};
Console.WriteLine(countSets(a));
}
}
|
PHP
<?php
$m = 3;
$n = 2;
function countSets( $a )
{
global $m , $n ;
$res = 0;
for ( $i = 0; $i < $n ; $i ++)
{
$u = 0; $v = 0;
for ( $j = 0; $j < $m ; $j ++)
$a [ $i ][ $j ] ? $u ++ : $v ++;
$res += pow(2, $u ) - 1 + pow(2, $v ) - 1;
}
for ( $i = 0; $i < $m ; $i ++)
{
$u = 0; $v = 0;
for ( $j = 0; $j < $n ; $j ++)
$a [ $j ][ $i ] ? $u ++ : $v ++;
$res += pow(2, $u ) - 1 +
pow(2, $v ) - 1;
}
return $res -( $n * $m );
}
$a = array ( array (1, 0, 1),
array (0, 1, 0));
echo countSets( $a );
?>
|
Javascript
<script>
var m = 3;
var n = 2;
function countSets(a) {
var res = 0;
for (i = 0; i < n; i++) {
var u = 0, v = 0;
for (j = 0; j < m; j++) {
if (a[i][j] == 1)
u++;
else
v++;
}
res += Math.pow(2, u) - 1 + Math.pow(2, v) - 1;
}
for (i = 0; i < m; i++) {
var u = 0, v = 0;
for (j = 0; j < n; j++) {
if (a[j][i] == 1)
u++;
else
v++;
}
res += Math.pow(2, u) - 1 + Math.pow(2, v) - 1;
}
return res - (n * m);
}
var a = [ [ 1, 0, 1 ], [ 0, 1, 0 ] ];
document.write(countSets(a));
</script>
|
Time Complexity: O(N*M), as we are using nested loops to traverse N*M times.
Auxiliary Space: O(1), as we are not using any extra space.
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