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Given a N x 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.
PHP
<?php
// PHP program to compute number of sets
// in a binary matrix.
// Number of columns
$m = 3;
// Number 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);
?>
Time Complexity: O(n * m)
Space Complexity: O(1) as no extra space has been taken.
Please refer complete article on Counting sets of 1s and 0s in a binary matrix for more details!
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