Minimum operations of given type to make all elements of a matrix equal

Given an integer K and a matrix of N rows and M columns, the task is to find the minimum number of operations required to make all the elements of the matrix equal. In a single operation, K can be added to or subtracted from any element of the matrix. Print -1 if it is impossible to do so.

Examples:

Input: mat[][] = {{2, 4}, {22, 24}}, K = 2
Output: 20
mat[0][0] = 2 + (10 * K) = 22 … 10 operations
mat[0][1] = 4 + (9 * K) = 22 … 9 operations
mat[1][0] = 22 … No operation
mat[1][1] = 24 – K = 22 … 1 operations
10 + 9 + 1 = 20



Input: mat[][] = {
{3, 63, 42},
{18, 12, 12},
{15, 21, 18},
{33, 84, 24}},
K = 3
Output: 63

Approach: Since we are only allowed to add or subtract K from any element, we can easily infer that mod of all the elements with K should be equal because x % K = (x + K) % K = (x – K) % K.
If that is not the case simply print -1. Else, sort all the elements of the matrix in non-deceasing order and find the median of the sorted elements. The minimum number of steps would occur if we convert all the elements equal to the median. Calculate these steps and print the result.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the minimum
// number of operations required
int minOperations(int n, int m, int k,
                  vector<vector<int> >& matrix)
{
    // Create another array to
    // store the elements of matrix
    vector<int> arr(n * m, 0);
  
    int mod = matrix[0][0] % k;
  
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < m; ++j) {
            arr[i * m + j] = matrix[i][j];
  
            // If not possible
            if (matrix[i][j] % k != mod) {
                return -1;
            }
        }
    }
  
    // Sort the array to get median
    sort(arr.begin(), arr.end());
  
    int median = arr[(n * m) / 2];
  
    // To count the minimum operations
    int minOperations = 0;
    for (int i = 0; i < n * m; ++i) 
        minOperations += abs(arr[i] - median) / k;
  
    // If there are even elements, then there 
    // are two medians. We consider the best
    // of two as answer.
    if ((n * m) % 2 == 0)
    {
       int median2 = arr[(n * m) / 2];
       int minOperations2 = 0;
       for (int i = 0; i < n * m; ++i) 
          minOperations2 += abs(arr[i] - median2) / k;
  
       minOperations = min(minOperations, minOperations2);
    }
  
    // Return minimum operations required
    return minOperations;
}
  
// Driver code
int main()
{
    vector<vector<int> > matrix = { { 2, 4, 6 },
                                    { 8, 10, 12 },
                                    { 14, 16, 18 },
                                    { 20, 22, 24 } };
    int n = matrix.size();
    int m = matrix[0].size();
    int k = 2;
    cout << minOperations(n, m, k, matrix);
    return 0;
}

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Java

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// Java implementation of the approach
import java.util.*;
  
class GFG
{
    // Function to return the minimum
    // number of operations required
    static int minOperations(int n, int m,
                        int k, int matrix[][])
    {
        // Create another array to
        // store the elements of matrix
        int [] arr = new int[n * m];
      
        int mod = matrix[0][0] % k;
      
        for (int i = 0; i < n; ++i)
        {
            for (int j = 0; j < m; ++j) 
            {
                arr[i * m + j] = matrix[i][j];
      
                // If not possible
                if (matrix[i][j] % k != mod) 
                {
                    return -1;
                }
            }
        }
      
        // Sort the array to get median
        Arrays.sort(arr);
      
        int median = arr[(n * m) / 2];
      
        // To count the minimum operations
        int minOperations = 0;
        for (int i = 0; i < n * m; ++i) 
            minOperations += Math.abs(arr[i] - median) / k;
      
        // If there are even elements, then there 
        // are two medians. We consider the best
        // of two as answer.
        if ((n * m) % 2 == 0)
        {
        int median2 = arr[(n * m) / 2];
        int minOperations2 = 0;
        for (int i = 0; i < n * m; ++i) 
            minOperations2 += Math.abs(arr[i] - median2) / k;
  
        minOperations = Math.min(minOperations, minOperations2);
        }
      
        // Return minimum operations required
        return minOperations;
    }
      
    // Driver code
    public static void main(String []args)
    {
        int matrix [][] = { { 2, 4, 6 },
                            { 8, 10, 12 },
                            { 14, 16, 18 },
                            { 20, 22, 24 } };
                              
        int n = matrix.length;
        int m = matrix[0].length;
        int k = 2;
        System.out.println(minOperations(n, m, k, matrix));
    }
}
  
// This code is contributed by ihritik

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Python3

# Python3 implementation of the approach



# Function to return the minimum
# number of operations required
def minOperations(n, m, k, matrix):

# Create another array to store the
# elements of matrix
arr = [0] * (n * m)

mod = matrix[0][0] % k

for i in range(0, n):
for j in range(0, m):
arr[i * m + j] = matrix[i][j]

# If not possible
if matrix[i][j] % k != mod:
return -1

# Sort the array to get median
arr.sort()

median = arr[(n * m) // 2]

# To count the minimum operations
minOperations = 0
for i in range(0, n * m):
minOperations += abs(arr[i] – median) // k

# If there are even elements, then there
# are two medians. We consider the best
# of two as answer.
if (n * m) % 2 == 0:

median2 = arr[(n * m) // 2]
minOperations2 = 0
for i in range(0, n * m):
minOperations2 += abs(arr[i] – median2) // k

minOperations = min(minOperations,
minOperations2)

# Return minimum operations required
return minOperations

# Driver code
if __name__ == “__main__”:

matrix = [[2, 4, 6],
[8, 10, 12],
[14, 16, 18],
[20, 22, 24]]

n = len(matrix)
m = len(matrix[0])
k = 2
print(minOperations(n, m, k, matrix))

# This code is contributed by Rituraj Jain

C#

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// C# implementation of the approach 
using System;
  
class GFG 
    // Function to return the minimum 
    // number of operations required 
    static int minOperations(int n, int m, 
                        int k, int [,]matrix) 
    
          
        // Create another array to 
        // store the elements of matrix 
        int []arr = new int[n * m]; 
      
        int mod = matrix[0, 0] % k; 
      
        for (int i = 0; i < n; ++i) 
        
            for (int j = 0; j < m; ++j) 
            
                arr[i * m + j] = matrix[i,j]; 
      
                // If not possible 
                if (matrix[i,j] % k != mod) 
                
                    return -1; 
                
            
        
      
        // Sort the array to get median 
        Array.Sort(arr); 
      
        int median = arr[(n * m) / 2]; 
      
        // To count the minimum operations 
        int minOperations = 0; 
        for (int i = 0; i < n * m; ++i) 
            minOperations += Math.Abs(arr[i] - median) / k; 
      
        // If there are even elements, then there 
        // are two medians. We consider the best 
        // of two as answer. 
        if ((n * m) % 2 == 0) 
        
            int median2 = arr[(n * m) / 2]; 
            int minOperations2 = 0; 
            for (int i = 0; i < n * m; ++i) 
                minOperations2 += Math.Abs(arr[i] - median2) / k; 
  
            minOperations = Math.Min(minOperations, minOperations2); 
        
      
        // Return minimum operations required 
        return minOperations; 
    
      
    // Driver code 
    public static void Main() 
    
        int [,]matrix = { { 2, 4, 6 }, 
                            { 8, 10, 12 }, 
                            { 14, 16, 18 }, 
                            { 20, 22, 24 } }; 
                              
        int n = matrix.GetLength(0); 
        int m = matrix.GetLength(1); 
        int k = 2; 
        Console.WriteLine(minOperations(n, m, k, matrix)); 
    
  
// This code is contributed by Ryuga 

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Output:

36


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