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 = 20Input: 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> |
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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> |
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