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Count ways to remove pairs from a matrix such that remaining elements can be grouped in vertical or horizontal pairs
  • Last Updated : 22 Apr, 2021

Given an integer K and a square matrix mat[][] of size N * N with elements from the range[1, K], the task is to count ways to remove pairs of distinct matrix elements such that remaining elements can be arranged in pairs vertically or horizontally.

Examples:

Input: mat[][] = {{1, 2}, {3, 4}}, K = 4
Output: 4
Explanation:
Remove the row {1, 2}. Therefore, remaining elements are {3, 4}, which can be arranged in a horizontal group of 2.
Similarly, removing pairs (1, 3), (3, 4) and (2, 4) can also form such arrangements.

Input: mat[][] = {{1, 2, 3}, {2, 2, 2}, {1, 2, 2}}, K = 3
Output: 0

Approach: The idea is based on the below observations:



  • If N is odd, then no matter which pair is removed the remaining matrix can not be covered with the smaller matrices. This is because, upon removal of any pair of the matrix, the remaining elements will be odd, it is not possible to cover them using a 2×1 or 1×2 size matrix.
  • For other cases, covering the remaining matrix after removal of a pair (a, b) is only possible if the sum of row index, i and column index, j of a is even and that of b is odd or vice versa.

Follow the steps below to solve the problem:

  • If the size of the matrix is odd, then print 0 as there can be no possible arrangements.
  • Otherwise, check for the following:
    • Initialize a variable ans as 0 to store the number of required pairs.
    • Initialize an auxiliary matrix, dp[][] of size K×2, dp[i][0] indicating the number of occurrences of i in the matrix mat[][] whose sum of row index and column index is even, and dp[i][1] indicating the number of occurrences of i in mat[][] whose sum of row index and column index is odd.
    • Traverse the matrix mat[][] row-wise using row index i and column index j and if the sum of i and j is even, then increment dp[mat[i][j] – 1][0] by 1, else increment dp[mat[i][j] – 1][1] by 1.
    • Iterate over the range [0, K – 1] using the variable i and nested iterate in the range [i + 1, K – 1] using variable j:
      • Add the value of dp[i][0] * dp[j][1] to ans as the value of both block differ and (i + j) and the value of one is odd and (i + j) value of other is even.
      • Add the value of dp[i][1] * dp[j][0] to ans as the value of both block differ and (i + j) value of one is even and (i + j) value of other is odd
  • After the above steps, print the value of ans as the result.

Below is the implementation of the above approach :

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to count ways to remove pairs
// such that the remaining elements can
// be arranged in pairs vertically or horizontally
void numberofpairs(vector<vector<int> > v,
                   int k)
{
    // Store the size of matrix
    int n = v.size();
 
    // If N is odd, then no
    // such pair exists
    if (n % 2 == 1) {
        cout << 0;
        return;
    }
 
    // Store the number of
    // required pairs
    int ans = 0;
 
    // Initialize an auxiliary
    // matrix and fill it with 0s
    int dp[k][2];
 
    for (int i = 0; i < k; i++) {
        for (int j = 0; j < 2; j++) {
            dp[i][j] = 0;
        }
    }
 
    // Traverse the matrix v[][]
    for (int i = 0; i < n; i++) {
 
        for (int j = 0; j < n; j++) {
 
            // Check if i+j is odd or even
            if ((i + j) % 2 == 0)
 
                // Increment the value
                // dp[v[i][j]-1][0] by 1
                dp[v[i][j] - 1][0]++;
            else
 
                // Increment the value
                // dp[v[i][j]-1][1] by 1
                dp[v[i][j] - 1][1]++;
        }
    }
 
    // Iterate in range[0, k-1] using i
    for (int i = 0; i < k; i++) {
 
        // Iterate in range[i+1, k-1] using j
        for (int j = i + 1; j < k; j++) {
 
            // Update the ans
            ans += dp[i][0] * dp[j][1];
            ans += dp[i][1] * dp[j][0];
        }
    }
 
    // Print the answer
    cout << ans;
}
 
// Driver Code
int main()
{
    vector<vector<int> > mat = { { 1, 2 }, { 3, 4 } };
    int K = 4;
    numberofpairs(mat, K);
 
    return 0;
}

Java




// Java program for the above approach
import java.io.*;
class GFG{
 
// Function to count ways to remove pairs
// such that the remaining elements can
// be arranged in pairs vertically or horizontally
static void numberofpairs(int[][] v, int k)
{
   
    // Store the size of matrix
    int n = v.length;
 
    // If N is odd, then no
    // such pair exists
    if (n % 2 == 1) {
        System.out.println(0);
        return;
    }
 
    // Store the number of
    // required pairs
    int ans = 0;
 
    // Initialize an auxiliary
    // matrix and fill it with 0s
    int dp[][] = new int[k][2];
 
    for (int i = 0; i < k; i++) {
        for (int j = 0; j < 2; j++) {
            dp[i][j] = 0;
        }
    }
 
    // Traverse the matrix v[][]
    for (int i = 0; i < n; i++) {
 
        for (int j = 0; j < n; j++) {
 
            // Check if i+j is odd or even
            if ((i + j) % 2 == 0)
 
                // Increment the value
                // dp[v[i][j]-1][0] by 1
                dp[v[i][j] - 1][0]++;
            else
 
                // Increment the value
                // dp[v[i][j]-1][1] by 1
                dp[v[i][j] - 1][1]++;
        }
    }
 
    // Iterate in range[0, k-1] using i
    for (int i = 0; i < k; i++) {
 
        // Iterate in range[i+1, k-1] using j
        for (int j = i + 1; j < k; j++) {
 
            // Update the ans
            ans += dp[i][0] * dp[j][1];
            ans += dp[i][1] * dp[j][0];
        }
    }
 
    // Print the answer
    System.out.println(ans);
}
 
// Driver Code
public static void main(String[] args)
{
    int[][] mat = { { 1, 2 }, { 3, 4 } };
    int K = 4;
    numberofpairs(mat, K);
}
}
 
// This code is contributed by susmitakundogoaldanga.

Python3




# Python3 program for the above approach
 
# Function to count ways to remove pairs
# such that the remaining elements can
# be arranged in pairs vertically or horizontally
def numberofpairs(v, k) :
 
    # Store the size of matrix
    n = len(v)
  
    # If N is odd, then no
    # such pair exists
    if (n % 2 == 1) :
        print(0)
        return
  
    # Store the number of
    # required pairs
    ans = 0
  
    # Initialize an auxiliary
    # matrix and fill it with 0s
    dp = [[0 for i in range(2)] for j in range(k)]
  
    for i in range(k) :
        for j in range(2) :
            dp[i][j] = 0
  
    # Traverse the matrix v[][]
    for i in range(n) :
        for j in range(n) :
  
            # Check if i+j is odd or even
            if ((i + j) % 2 == 0) :
  
                # Increment the value
                # dp[v[i][j]-1][0] by 1
                dp[v[i][j] - 1][0] += 1
            else :
  
                # Increment the value
                # dp[v[i][j]-1][1] by 1
                dp[v[i][j] - 1][1] += 1
  
    # Iterate in range[0, k-1] using i
    for i in range(k) :
  
        # Iterate in range[i+1, k-1] using j
        for j in range(i + 1, k) :
  
            # Update the ans
            ans += dp[i][0] * dp[j][1]
            ans += dp[i][1] * dp[j][0]
  
    # Print the answer
    print(ans)
     
    # Driver code
mat = [ [ 1, 2 ], [ 3, 4 ] ]
K = 4
numberofpairs(mat, K)
 
# This code is contributed by divyeshrabdiya07.

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG {
     
    // Function to count ways to remove pairs
    // such that the remaining elements can
    // be arranged in pairs vertically or horizontally
    static void numberofpairs(List<List<int> > v,
                       int k)
    {
       
        // Store the size of matrix
        int n = v.Count;
      
        // If N is odd, then no
        // such pair exists
        if (n % 2 == 1) {
            Console.Write(0);
            return;
        }
      
        // Store the number of
        // required pairs
        int ans = 0;
      
        // Initialize an auxiliary
        // matrix and fill it with 0s
        int[,] dp = new int[k, 2];
      
        for (int i = 0; i < k; i++) {
            for (int j = 0; j < 2; j++) {
                dp[i, j] = 0;
            }
        }
      
        // Traverse the matrix v[][]
        for (int i = 0; i < n; i++) {
      
            for (int j = 0; j < n; j++) {
      
                // Check if i+j is odd or even
                if ((i + j) % 2 == 0)
      
                    // Increment the value
                    // dp[v[i][j]-1][0] by 1
                    dp[v[i][j] - 1, 0]++;
                else
      
                    // Increment the value
                    // dp[v[i][j]-1][1] by 1
                    dp[v[i][j] - 1, 1]++;
            }
        }
      
        // Iterate in range[0, k-1] using i
        for (int i = 0; i < k; i++) {
      
            // Iterate in range[i+1, k-1] using j
            for (int j = i + 1; j < k; j++) {
      
                // Update the ans
                ans += dp[i, 0] * dp[j, 1];
                ans += dp[i, 1] * dp[j, 0];
            }
        }
      
        // Print the answer
        Console.Write(ans);
    }
 
  // Driver code
  static void Main()
  {
    List<List<int> > mat = new List<List<int>>();
    mat.Add(new List<int>(new int[]{1, 2}));
    mat.Add(new List<int>(new int[]{3, 4}));
    int K = 4;
    numberofpairs(mat, K);
  }
}
 
// This code is contributed by divyesh072019.

Javascript




<script>
 
    // Javascript program for the above approach
     
    // Function to count ways to remove pairs
    // such that the remaining elements can
    // be arranged in pairs
    // vertically or horizontally
    function numberofpairs(v, k)
    {
 
        // Store the size of matrix
        let n = v.length;
 
        // If N is odd, then no
        // such pair exists
        if (n % 2 == 1) {
            document.write(0);
            return;
        }
 
        // Store the number of
        // required pairs
        let ans = 0;
 
        // Initialize an auxiliary
        // matrix and fill it with 0s
        let dp = new Array(k);
 
        for (let i = 0; i < k; i++) {
            dp[i] = new Array(2);
            for (let j = 0; j < 2; j++) {
                dp[i][j] = 0;
            }
        }
 
        // Traverse the matrix v[][]
        for (let i = 0; i < n; i++) {
 
            for (let j = 0; j < n; j++) {
 
                // Check if i+j is odd or even
                if ((i + j) % 2 == 0)
 
                    // Increment the value
                    // dp[v[i][j]-1][0] by 1
                    dp[v[i][j] - 1][0]++;
                else
 
                    // Increment the value
                    // dp[v[i][j]-1][1] by 1
                    dp[v[i][j] - 1][1]++;
            }
        }
 
        // Iterate in range[0, k-1] using i
        for (let i = 0; i < k; i++) {
 
            // Iterate in range[i+1, k-1] using j
            for (let j = i + 1; j < k; j++) {
 
                // Update the ans
                ans += dp[i][0] * dp[j][1];
                ans += dp[i][1] * dp[j][0];
            }
        }
 
        // Print the answer
        document.write(ans + "</br>");
    }
     
    let mat = [ [ 1, 2 ], [ 3, 4 ] ];
    let K = 4;
    numberofpairs(mat, K);
 
</script>

 
 

Output: 
4

 

Time Complexity: O(N2 + K2)
Auxiliary Space: O(K) 

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