Flip Columns For Maximum Number Of Equal Rows - Leetcode Solution

Problem Description:

The problem "Flip Columns For Maximum Number Of Equal Rows" on LeetCode is a string manipulation problem where you are given a binary matrix of rows M and columns N. You are required to flip some columns of the binary matrix such that the number of rows having all the elements as 0 or all the elements as 1 is maximized. The objective is to find the maximum number of rows that can be made the same by at most flipping K columns of the matrix.

Detailed Solution:

Approach:

• Iterate over each column of the binary matrix from left to right. Let column C1 be the first column and column Cn be the last column.
• Take any two columns Ci and Cj, where i<j. Swap the columns and count the maximum number of rows that are the same in the modified matrix. You can use a hash table to store the frequency of the rows that are the same.
• Swap columns Ci and Cj again to bring it back to its original state.
• Continue the above steps until all the column permutations have been tried.
• Return the maximum number of rows that are the same.

Algorithm:

• Initialize the max_count variable to 0.
• Iterate over each column i from 0 to N-1.
  • Iterate over each column j from i+1 to N.
    • Swap the columns i and j of the matrix.
    • Create a hash table to store the frequency of rows with all elements as 0 or 1.
    • Iterate over each row k from 0 to M-1.
      • If the row is not present in the hash table, add it along with its frequency as 1.
      • Else, update the frequency of the row in the hash table.
    • Get the maximum frequency of a row from the hash table.
    • Update the max_count variable with the maximum frequency.
    • Swap back the columns i and j of the matrix.
• Return max_count.

Time Complexity:

The time complexity of the above algorithm is O(N^2*M). This is because in the worst case, we need to iterate over all the column permutations (N choose 2) and then iterate over all the rows to find the frequency of rows with all elements as 0 or 1.

Space Complexity:

The space complexity of the above algorithm is O(M*N) because we need to store the input binary matrix and the hash table to store the frequencies of the rows.

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