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
// CPP program to compute number of sets // in a binary matrix. #include <bits/stdc++.h> using namespace std; const int m = 3; // no of columns const int n = 2; // no of rows // function to calculate the number of // non empty sets of cell long long countSets(int a[n][m]) { // stores the final answer long res = 0; // traverses row-wise 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; } // traverses column wise 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; } // at the end subtract n*m as no of // single sets have been added twice. return res - (n * m); } // driver program to test the above function. int main() { int a[][3] = { { 1, 0, 1 }, { 0, 1, 0 } }; cout << countSets(a); return 0; } |
Java
// Java program to compute number of sets // in a binary matrix. import java.util.*; class GFG { static final int m = 3; // no of columns static final int n = 2; // no of rows // function to calculate the number of // non empty sets of cell static long countSets(int a[][]) { // stores the final answer long res = 0; // traverses row-wise 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; } // traverses column wise 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; } // at the end subtract n*m as no of // single sets have been added twice. return res - (n * m); } // Driver code public static void main(String[] args) { int a[][] = {{1, 0, 1}, {0, 1, 0}}; System.out.print(countSets(a)); } } // This code is contributed by Anant Agarwal. |
Python3
# Python3 program to compute number of sets # in a binary matrix. m = 3 # no of columns n = 2 # no of rows # function to calculate the number of # non empty sets of cell def countSets(a): # stores the final answer res = 0 # traverses row-wise 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 # traverses column wise 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 # at the end subtract n*m as no of # single sets have been added twice. return res - (n*m) # Driver program to test the above function. a = [[1, 0, 1],[0, 1, 0]] print(countSets(a)) # This code is contributed by shubhamsingh10 |
C#
// C# program to compute number of // sets in a binary matrix. using System; class GFG { static int m = 3; // no of columns static int n = 2; // no of rows // function to calculate the number of // non empty sets of cell static long countSets(int [,]a) { // stores the final answer long res = 0; // Traverses row-wise 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; } // Traverses column wise 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; } // at the end subtract n*m as no of // single sets have been added twice. return res - (n * m); } // Driver code public static void Main() { int [,]a = {{1, 0, 1}, {0, 1, 0}}; Console.WriteLine(countSets(a)); } } // This code is contributed by vt_m. |
PHP
<?php // PHP program to compute // number of sets // in a binary matrix. // no of columns $m = 3; // no of rows $n = 2; // function to calculate the number // of non empty sets of cell function countSets($a) { global $m, $n; // stores the final answer $res = 0; // traverses row-wise for ($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, $v) - 1; } // traverses column wise for ($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; } // at the end subtract // n*m as no of single // sets have been added // twice. return $res-($n*$m); } // Driver Code $a = array(array(1, 0, 1), array(0, 1, 0)); echo countSets($a); // This code is contributed by anuj_67. ?> |
Javascript
<script> // javascript program to compute number of sets // in a binary matrix. var m = 3; // no of columns var n = 2; // no of rows // function to calculate the number of // non empty sets of cell function countSets(a) { // stores the final answer var res = 0; // traverses row-wise 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; } // traverses column wise 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; } // at the end subtract n*m as no of // single sets have been added twice. return res - (n * m); } // Driver code var a = [ [ 1, 0, 1 ], [ 0, 1, 0 ] ]; document.write(countSets(a)); // This code is contributed by Rajput-Ji </script> |
Output
8
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.
This article is contributed by Raj. If you like neveropen and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the neveropen main page and help other Geeks.
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
