# 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 = 2 + (10 * K) = 22 … 10 operations
mat = 4 + (9 * K) = 22 … 9 operations
mat = 22 … No operation
mat = 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the minimum ` `// number of operations required ` `int` `minOperations(``int` `n, ``int` `m, ``int` `k, ` `                  ``vector >& matrix) ` `{ ` `    ``// Create another array to ` `    ``// store the elements of matrix ` `    ``vector<``int``> arr(n * m, 0); ` ` `  `    ``int` `mod = matrix % 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 > matrix = { { 2, 4, 6 }, ` `                                    ``{ 8, 10, 12 }, ` `                                    ``{ 14, 16, 18 }, ` `                                    ``{ 20, 22, 24 } }; ` `    ``int` `n = matrix.size(); ` `    ``int` `m = matrix.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``]; ` `        ``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``]  ` `        ``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];  ` `            ``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  `

Output:

```36
```

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

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the minimum ` `// number of operations required ` `int` `minOperations(``int` `n, ``int` `m, ``int` `k, ` `                  ``vector >& 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) { ` `        ``mod = k - (``abs``(matrix) % k); ` `    ``} ` `    ``else` `{ ` `        ``mod = matrix % 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 > matrix = { { 2, 4, 6 }, ` `                                    ``{ 8, 10, 12 }, ` `                                    ``{ 14, 16, 18 }, ` `                                    ``{ 20, 22, 24 } }; ` `    ``int` `n = matrix.size(); ` `    ``int` `m = matrix.size(); ` `    ``int` `k = 2; ` `    ``cout << minOperations(n, m, k, matrix); ` `    ``return` `0; ` `} `

Output:

```36
```
Output:

```36
```

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