Minimize cost to convert a given matrix to another by flipping columns and reordering rows

Given two binary matrices mat[][] and target[][] of dimensions N * M, the task is to find the minimum cost to convert the matrix mat[][] into target[][] using the following operations:

Flip a particular column in mat[][] such that all 1s become 0s and vice-versa. The cost of this operation is 1.
Reorder the rows of mat[][]. The cost of this operation is 0.

If its not possible to convert the matrix mat[][] to target[][], then print “-1”.

Examples:

Input: mat[][] = {{0, 0}, {1, 0}, {1, 1}}, target[][] = {{0, 1}, {1, 0}, {1, 1}}
Output: 1
Explanation:
Step 1: Reorder 2^nd and 3^rd rows to modify mat[][] to {{0, 0}, {1, 1}, {1, 0}}. Cost = 0
Step 2: Flip the second column . mat[][] modifies to {{0, 1}, {1, 0}, {1, 1}}, which is equal to target[][]. Cost = 1

Input: mat[][] = {{0, 0, 0}, {1, 0, 1}, {1, 1, 0}}, target[][] = {{0, 0, 1}, {1, 0, 1}, {1, 1, 1}}
Output: -1

Approach: The idea is to convert the rows of the given matrices into bitsets and then find the minimum cost of operation to make the matrices equal. Below are the steps:

First, convert the rows of mat[][] to binary numbers using bitset.
After completing the above step, find all possible rows which can be the first row of the target matrix.
The number of flips required to convert the row of mat[][] to tar[][] is the count of set bits in the Bitwise XOR value of the bitsets.
Compute Bitwise XOR of every row with the flip pattern and check if the new matrix, when sorted, is equal to the sorted target[][] matrix or not.
If they are the same, then store the number of flips.
Calculate all such count of flips in the above steps and return the minimum among them.

Below is the implementation of the above approach:

C++

// C++ program for the above approach 
  
#include <bits/stdc++.h> 
using namespace std; 
  
// Custom comparator to sort vector 
// of bitsets 
bool static cmp(bitset<105>& p1, 
                bitset<105>& p2) 
{ 
    return p1.to_string() < p2.to_string(); 
} 
  
// Function to convert the given 
// matrix into the target matrix 
int minCost(vector<vector<int> >& a, 
            vector<vector<int> >& t) 
{ 
  
    // Number of rows and columns 
    int n = a.size(); 
    int m = a[0].size(); 
  
    vector<bitset<105> > mat(n), tar(n); 
  
    // Iterate over rows 
    for (int i = 0; i < n; i++) { 
        string s; 
  
        for (int j = 0; j < m; j++) { 
            s += char(a[i][j] + '0'); 
        } 
        mat[i] = bitset<105>(s); 
    } 
  
    // Iterate over rows 
    for (int i = 0; i < n; i++) { 
        string s; 
  
        for (int j = 0; j < m; j++) { 
            s += char(t[i][j] + '0'); 
        } 
        tar[i] = bitset<105>(s); 
    } 
  
    // Sort the matrix 
    sort(tar.begin(), tar.end(), cmp); 
  
    int ans = INT_MAX; 
  
    // Check all possible rows as the 
    // first row of target 
    for (int i = 0; i < n; i++) { 
  
        vector<bitset<105> > copy(mat); 
  
        // Get the flip pattern 
        auto flip = copy[i] ^ tar[0]; 
  
        for (auto& row : copy) 
            row ^= flip; 
  
        sort(copy.begin(), copy.end(), cmp); 
  
        // Number of flip operations 
        // is the count of set bits in flip 
        if (copy == tar) 
            ans = min(ans, (int)flip.count()); 
    } 
  
    // If it is not possible 
    if (ans == INT_MAX) 
        return -1; 
  
    // Return the answer 
    return ans; 
} 
  
// Driver Code 
int main() 
{ 
    // Given matrices 
    vector<vector<int> > matrix{ { 0, 0 }, 
                                 { 1, 0 }, 
                                 { 1, 1 } }; 
  
    vector<vector<int> > target{ { 0, 1 }, 
                                 { 1, 0 }, 
                                 { 1, 1 } }; 
  
    // Function Call 
    cout << minCost(matrix, target); 
} 

Output:

Time Complexity: O(N * M)
Auxiliary Space: O(N * M)

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My Personal Notes

C++

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