Check whether row or column swaps produce maximum size binary sub-matrix with all 1s
Given a binary matrix, the task is to find whether row swaps or column swaps give maximum size sub-matrix with all 1’s. In a row swap, we are allowed to swap any two rows. In a column swap, we are allowed to swap any two columns. Output “Row Swap” or “Column Swap” and the maximum size.
Input : 1 1 1 1 0 1 Output : Column Swap 4 By swapping column 1 and column 2(0-based indexing), index (0, 0) to (1, 1) makes the largest binary sub-matrix. Input : 0 0 0 1 1 0 1 1 0 0 0 0 1 1 0 Output : Row Swap 6 Input : 1 1 0 0 0 0 0 0 0 1 1 0 1 1 0 0 0 0 1 1 0 Output : Row Swap 8
The idea is to find both row swap and column swap maximum size binary submatrix and compare.
To find the maximum-sized binary sub-matrix with row swaps allowed, make a 2-D array, say dp[i][j]. Each value of dp[i][j] contains the number of consecutive 1s on right side of (i,j) in i-th row. Now, store each column in the 1-D temporary array one by one, say b and sort, and find maximum b[i] * (n – i), since b[i] is indicating the sub-matrix width and (n – i) is sub-matrix height.
Similarly, to find the maximum size binary sub-matrix with column swap allowed, find dp[i][j], where each value contains the number of consecutive 1 below the (i, j) in j-th column. Similarly, store each row in the 1-D temporary array one by one, say b and sort. Find maximum b[i] * (m – i), since b[i] is indicating the submatrix height and (n – i) is submatrix width.
Below is the implementation of this approach:
Row Swap 6
Time Complexity: O(R*C* log(C))
Auxiliary Space: O(R*C)
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