Given a 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.
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)
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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