Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

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

  • Difficulty Level : Easy
  • Last Updated : 15 Nov, 2021

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:

Want to learn from the best curated videos and practice problems, check out the C++ Foundation Course for Basic to Advanced C++ and C++ STL Course for the language and STL. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.

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 the 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. Otherwise, sort all the elements of the matrix in non-decreasing order and find the median of the sorted elements. The minimum number of steps would occur if we convert all the elements to equal to the median. Calculate these steps and print the result.

Below is the implementation of the above approach: 

C++




// 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) - 1];
       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;
}

Java




// 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 ) - 1];
        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

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  ) - 1]
        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#




// 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 ) - 1];
            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

Javascript




<script>
    // Javascript implementation of the approach
     
    // Function to return the minimum
    // number of operations required
    function minOperations(n, m, k, matrix)
    {
        // Create another array to
        // store the elements of matrix
        let arr = new Array(n * m);
        arr.fill(0);
      
        let mod = matrix[0][0] % k;
      
        for (let i = 0; i < n; ++i)
        {
            for (let 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
        arr.sort(function(a, b){return a - b});
      
        let median = arr[parseInt((n * m) / 2, 10)];
      
        // To count the minimum operations
        let minOperations = 0;
        for (let i = 0; i < n * m; ++i)
            minOperations += parseInt(Math.abs(arr[i] - median) / k, 10);
      
        // If there are even elements, then there
        // are two medians. We consider the best
        // of two as answer.
        if ((n * m) % 2 == 0)
        {
          let median2 = arr[parseInt((n * m) / 2, 10)];
          let minOperations2 = 0;
          for (let i = 0; i < n * m; ++i)
              minOperations2 += parseInt(Math.abs(arr[i] - median2) / k, 10);
 
          minOperations = Math.min(minOperations, minOperations2);
        }
      
        // Return minimum operations required
        return minOperations;
    }
     
    // Driver code
    let matrix = [ [ 2, 4, 6 ],
                   [ 8, 10, 12 ],
                   [ 14, 16, 18 ],
                   [ 20, 22, 24 ] ];
 
    let n = 4;
    let m = 3;
    let k = 2;
    document.write(minOperations(n, m, k, matrix));
     
    // This code is contributed by mukesh07.
</script>
Output: 
36

 

Below is an implementation that handles negative numbers also in the input matrix:  

C++




// 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;
 
    int mod;
     
    // will not work for negative elements, so ..
    // adding this
    if (matrix[0][0] < 0) {
        mod = k - (abs(matrix[0][0]) % k);
    }
    else {
        mod = matrix[0][0] % k;
    }
     
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < m; ++j) {
            arr.push_back(matrix[i][j]);
 
            // adding this to handle negative elements too .
            int val = matrix[i][j];
            if (val < 0) {
                int res = k - (abs(val) % k);
                if (res != mod) {
                    return -1;
                }
            }
            else {
                int foo = matrix[i][j];
                if (foo % 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) {
 
        // changed here as in case of even elements there will be 2 medians
        int median2 = arr[((n * m) / 2) - 1];
 
        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;
}

Java




// Java implementation of the approach
import java.util.*;
class GFG
{
 
  // Function to return the minimum
  // number of operations required
  public static int minOperations(int n, int m, int k,
                                  int matrix[][])
  {
    // Create another array to
    // store the elements of matrix
    Vector<Integer> arr = new Vector<>(); 
 
    int mod;
 
    // will not work for negative elements, so ..
    // adding this
    if (matrix[0][0] < 0)
    {
      mod = k - (Math.abs(matrix[0][0]) % k);
    }
    else
    {
      mod = matrix[0][0] % k;
    }
 
    for (int i = 0; i < n; ++i)
    {
      for (int j = 0; j < m; ++j)
      {
        arr.add(matrix[i][j]);
 
        // adding this to handle
        // negative elements too .
        int val = matrix[i][j];
        if (val < 0)
        {
          int res = k - (Math.abs(val) % k);
          if (res != mod)
          {
            return -1;
          }
        }
        else
        {
          int foo = matrix[i][j];
          if (foo % k != mod)
          {
            return -1;
          }
        }
      }
    }
 
    // Sort the array to get median
    Collections.sort(arr);     
    int median = arr.get((n * m) / 2);
 
    // To count the minimum operations
    int minOperations = 0;
    for (int i = 0; i < n * m; ++i)
      minOperations += Math.abs(arr.get(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)
    {
 
      // changed here as in case of
      // even elements there will be 2 medians
      int median2 = arr.get(((n * m) / 2) - 1);
 
      int minOperations2 = 0;
      for (int i = 0; i < n * m; ++i)
        minOperations2 += Math.abs(arr.get(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 divyesh072019

Python3




# Python3 implementation of the
# above 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 = []
 
    # will not work for negative
    # elements, so .. adding this
    if (matrix[0][0] < 0):
        mod = k - (abs(matrix[0][0]) % k)
    else:
        mod = matrix[0][0] % k
 
    for i in range(n):
        for j in range(m):
            arr.append(matrix[i][j])
 
            # adding this to handle
            # negative elements too .
            val = matrix[i][j]
             
            if (val < 0):
                res = k - (abs(val) % k)
                if (res != mod):
                    return -1
            else:
                foo = matrix[i][j]
                if (foo % 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(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):
 
        # changed here as in case of
        # even elements there will be
        # 2 medians
        median2 = arr[((n * m) //
                        2) - 1]
 
        minOperations2 = 0
        for i in range(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 Chitranayal

C#




// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG {
     
    // Function to return the minimum
    // number of operations required
    static int minOperations(int n, int m, int k,
                      List<List<int>> matrix)
    {
        // Create another array to
        // store the elements of matrix
        List<int> arr = new List<int>();
      
        int mod;
          
        // will not work for negative elements, so ..
        // adding this
        if (matrix[0][0] < 0) {
            mod = k - (Math.Abs(matrix[0][0]) % k);
        }
        else {
            mod = matrix[0][0] % k;
        }
          
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < m; ++j) {
                arr.Add(matrix[i][j]);
      
                // adding this to handle negative elements too .
                int val = matrix[i][j];
                if (val < 0) {
                    int res = k - (Math.Abs(val) % k);
                    if (res != mod) {
                        return -1;
                    }
                }
                else {
                    int foo = matrix[i][j];
                    if (foo % k != mod) {
                        return -1;
                    }
                }
            }
        }
      
        // Sort the array to get median
        arr.Sort();
      
        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) {
      
            // changed here as in case of
            // even elements there will be 2 medians
            int median2 = arr[((n * m) / 2) - 1];
      
            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;
    }
 
  static void Main() {
    List<List<int>> matrix = new List<List<int>>{
        new List<int> {2, 4, 6},
        new List<int> {8, 10, 12},
        new List<int> {14, 16, 18},
        new List<int> {20, 22, 24},
    };
    int n = matrix.Count;
    int m = matrix[0].Count;
    int k = 2;
    Console.Write(minOperations(n, m, k, matrix));
  }
}
 
// This code is contributed by divyeshrabadiya07.

Javascript




<script>
// Javascript implementation of the approach
 
// Function to return the minimum
// number of operations required
function minOperations(n,m,k,matrix)
{
    // Create another array to
    // store the elements of matrix
    let arr = [];
  
    let mod;
  
    // will not work for negative elements, so ..
    // adding this
    if (matrix[0][0] < 0)
    {
      mod = k - (Math.abs(matrix[0][0]) % k);
    }
    else
    {
      mod = matrix[0][0] % k;
    }
  
    for (let i = 0; i < n; ++i)
    {
      for (let j = 0; j < m; ++j)
      {
        arr.push(matrix[i][j]);
  
        // adding this to handle
        // negative elements too .
        let val = matrix[i][j];
        if (val < 0)
        {
          let res = k - (Math.abs(val) % k);
          if (res != mod)
          {
            return -1;
          }
        }
        else
        {
          let foo = matrix[i][j];
          if (foo % k != mod)
          {
            return -1;
          }
        }
      }
    }
  
    // Sort the array to get median
    arr.sort(function(a,b){return a-b;});    
    let median = arr[(n * m) / 2];
  
    // To count the minimum operations
    let minOperations = 0;
    for (let 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)
    {
  
      // changed here as in case of
      // even elements there will be 2 medians
      let median2 = arr[((n * m) / 2) - 1];
  
      let minOperations2 = 0;
      for (let 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
let matrix = [[2, 4, 6],
              [8, 10, 12],
              [14, 16, 18],
              [20, 22, 24]];
               
let n = matrix.length;
let m = matrix[0].length;
let k = 2;
document.write(minOperations(n, m, k, matrix));
 
// This code is contributed by rag2127
</script>
Output: 
36

 

 




My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!