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

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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);
}

Java

// Java program for the above approach
import java.util.*;
 
public class Main {
 
  // Custom comparator to sort vector
  // of bitsets
  static class BitsetComparator
    implements Comparator<BitSet> {
    public int compare(BitSet p1, BitSet p2)
    {
      return p1.toString().compareTo(p2.toString());
    }
  }
 
  // Function to convert the given
  // matrix into the target matrix
  static int minCost(List<List<Integer> > a,
                     List<List<Integer> > t)
  {
    // Number of rows and columns
    int n = a.size();
    int m = a.get(0).size();
 
    List<BitSet> mat = new ArrayList<>(n);
    List<BitSet> tar = new ArrayList<>(n);
 
    // Iterate over rows
    for (int i = 0; i < n; i++) {
      StringBuilder sb = new StringBuilder();
      for (int j = 0; j < m; j++) {
        sb.append(a.get(i).get(j));
      }
      mat.add(
        i, BitSet.valueOf(new long[] {
          Long.parseLong(sb.toString(), 2) }));
    }
 
    // Iterate over rows
    for (int i = 0; i < n; i++) {
      StringBuilder sb = new StringBuilder();
      for (int j = 0; j < m; j++) {
        sb.append(t.get(i).get(j));
      }
      tar.add(
        i, BitSet.valueOf(new long[] {
          Long.parseLong(sb.toString(), 2) }));
    }
 
    // Sort the matrix
    Collections.sort(tar, new BitsetComparator());
 
    int ans = Integer.MAX_VALUE;
 
    // Check all possible rows as the
    // first row of target
    for (int i = 0; i < n; i++) {
 
      List<BitSet> copy = new ArrayList<>(mat);
 
      // Get the flip pattern
      BitSet flip = (BitSet)copy.get(i).clone();
      flip.xor(tar.get(0));
 
      for (int j = 0; j < copy.size(); j++) {
        copy.get(j).xor(flip);
      }
 
      Collections.sort(copy, new BitsetComparator());
 
      // Number of flip operations
      // is the count of set bits in flip
      if (copy.equals(tar)) {
        ans = Math.min(ans, flip.cardinality());
      }
    }
 
    // If it is not possible
    if (ans == Integer.MAX_VALUE) {
      return -1;
    }
 
    // Return the answer
    return ans;
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    // Given matrices
    List<List<Integer> > matrix = Arrays.asList(
      Arrays.asList(0, 0), Arrays.asList(1, 0),
      Arrays.asList(1, 1));
 
    List<List<Integer> > target = Arrays.asList(
      Arrays.asList(0, 1), Arrays.asList(1, 0),
      Arrays.asList(1, 1));
 
    // Function Call
    System.out.println(minCost(matrix, target));
  }
}
 
// This code is contributed by rutikbhosale

Python3

# Python code for the above approach
import sys
 
# Function to convert bitset to string
def bitset_to_str(bs, size):
    return ''.join(str(bs >> i & 1) for i in range(size - 1, -1, -1))
 
# Custom comparator to sort vector of bitsets
def cmp(p1, p2):
    return bitset_to_str(p1, 105) < bitset_to_str(p2, 105)
 
# Function to convert the given matrix into the target matrix
def minCost(a, t):
    # Number of rows and columns
    n = len(a)
    m = len(a[0])
 
    mat = []
    tar = []
 
    # Iterate over rows
    for i in range(n):
        s = ''
 
        for j in range(m):
            s += str(a[i][j])
        mat.append(int(s, 2))
 
    # Iterate over rows
    for i in range(n):
        s = ''
 
        for j in range(m):
            s += str(t[i][j])
        tar.append(int(s, 2))
 
    # Sort the matrix
    tar.sort(key=lambda x: bitset_to_str(x, 105))
 
    ans = sys.maxsize
 
    # Check all possible rows as the first row of target
    for i in range(n):
        copy = mat.copy()
 
        # Get the flip pattern
        flip = copy[i] ^ tar[0]
 
        for j in range(len(copy)):
            copy[j] ^= flip
 
        copy.sort(key=lambda x: bitset_to_str(x, 105))
 
        # Number of flip operations is the count of set bits in flip
        if copy == tar:
            ans = min(ans, bin(flip).count('1'))
 
    # If it is not possible
    if ans == sys.maxsize:
        return -1
 
    # Return the answer
    return ans
 
# Driver Code
matrix = [[0, 0], [1, 0], [1, 1]]
target = [[0, 1], [1, 0], [1, 1]]
 
print(minCost(matrix, target))
 
# This code is contributed by sdeadityasharma

Javascript

// JavaScript code for the above approach
 
// Function to convert bitset to string
function bitset_to_str(bs, size) {
    let result = '';
    for (let i = size - 1; i >= 0; i--) {
        result += (bs >> i) & 1;
    }
    return result;
}
 
// Custom comparator to sort array of bitsets
function cmp(p1, p2) {
    return bitset_to_str(p1, 105) < bitset_to_str(p2, 105) ? -1 : 1;
}
 
// Function to convert the given matrix into the target matrix
function minCost(a, t) {
    // Number of rows and columns
    let n = a.length;
    let m = a[0].length;
    let mat = [];
    let tar = [];
 
    // Iterate over rows
    for (let i = 0; i < n; i++) {
        let s = '';
 
        for (let j = 0; j < m; j++) {
            s += a[i][j];
        }
        mat.push(parseInt(s, 2));
    }
 
    // Iterate over rows
    for (let i = 0; i < n; i++) {
        let s = '';
 
        for (let j = 0; j < m; j++) {
            s += t[i][j];
        }
        tar.push(parseInt(s, 2));
    }
 
    // Sort the matrix
    tar.sort(cmp);
 
    let ans = Number.MAX_SAFE_INTEGER;
 
    // Check all possible rows as the first row of target
    for (let i = 0; i < n; i++) {
        let copy = [...mat];
 
        // Get the flip pattern
        let flip = copy[i] ^ tar[0];
 
        for (let j = 0; j < copy.length; j++) {
            copy[j] ^= flip;
        }
 
        copy.sort(cmp);
 
        // Number of flip operations is the count of set bits in flip
        if (JSON.stringify(copy) === JSON.stringify(tar)) {
            ans = Math.min(ans, (flip).toString(2).match(/1/g).length);
        }
    }
 
    // If it is not possible
    if (ans == Number.MAX_SAFE_INTEGER) {
        return -1;
    }
 
    // Return the answer
    return ans;
}
 
// Driver Code
let matrix = [
    [0, 0],
    [1, 0],
    [1, 1]
];
let target = [
    [0, 1],
    [1, 0],
    [1, 1]
];
 
console.log(minCost(matrix, target));
 
// Contributed by adityasharmadev01

C#

// C# code for the above approach
 
 
using System;
using System.Collections.Generic;
using System.Linq;
 
public class Program
{
    // Function to convert bitset to string
    public static string BitsetToStr(int bs, int size)
    {
        string result = "";
        for (int i = size - 1; i >= 0; i--)
        {
            result += ((bs >> i) & 1);
        }
        return result;
    }
 
    // Custom comparator to sort array of bitsets
    public static int Cmp(int p1, int p2)
    {
        return BitsetToStr(p1, 105).CompareTo(BitsetToStr(p2, 105));
    }
 
    // Function to convert the given matrix into the target matrix
    public static int MinCost(int[][] a, int[][] t)
    {
        // Number of rows and columns
        int n = a.Length;
        int m = a[0].Length;
        List<int> mat = new List<int>();
        List<int> tar = new List<int>();
         
        // Iterate over rows
        for (int i = 0; i < n; i++)
        {
            string s = "";
            for (int j = 0; j < m; j++)
            {
                s += a[i][j];
            }
            mat.Add(Convert.ToInt32(s, 2));
        }
         
        // Iterate over rows
        for (int i = 0; i < n; i++)
        {
            string s = "";
            for (int j = 0; j < m; j++)
            {
                s += t[i][j];
            }
            tar.Add(Convert.ToInt32(s, 2));
        }
        // Sort the matrix
        tar.Sort(Cmp);
        int ans = int.MaxValue;
         
        // Check all possible rows as the first row of target
        for (int i = 0; i < n; i++)
        {
            List<int> copy = new List<int>(mat);
            int flip = copy[i] ^ tar[0];
            for (int j = 0; j < copy.Count; j++)
            {
                copy[j] ^= flip;
            }
            copy.Sort(Cmp);
             
             
            // Number of flip operations is the count of set bits in flip
            if (copy.SequenceEqual(tar))
            {
                ans = Math.Min(ans, Convert.ToString(flip, 2).Count(x => x == '1'));
            }
        }
         
        // If it is not possible
        if (ans == int.MaxValue)
        {
            return -1;
        }
        return ans;
    }
 
    // Driver Code
    public static void Main()
    {
        int[][] matrix = new int[][] {
            new int[] { 0, 0 },
            new int[] { 1, 0 },
            new int[] { 1, 1 }
        };
        int[][] target = new int[][] {
            new int[] { 0, 1 },
            new int[] { 1, 0 },
            new int[] { 1, 1 }
        };
        Console.WriteLine(MinCost(matrix, target));
    }
}
 
// This code is contributed by Shivhack999

Output:

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

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