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

Problem statement

Given an integer K and a matrix of M x N, 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.

Example

If input matrix is:
{
{2, 4},
{20, 40}
} and K = 2 then total 27 operations required as follows;
Matrix[0][0] = 2 + (K * 9) = 20 = 9 operations
Matrix[0][1] = 4 + (k * 8) = 20 = 8 operations
Matrix[1][0] = 20 + (k * 10) = 40 = 10 operations

Algorithm

1. 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. If it’s not, then return -1
2. sort all the elements of the matrix in non-deceasing order and find the median of the sorted elements
3. The minimum number of steps would occur if we convert all the elements equal to the median

Example

Live Demo

#include <bits/stdc++.h>
using namespace std;
int getMinOperations(int n, int m, int k, vector<vector<int> >& 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 (matrix[i][j] % k != mod) {
return -1;
}
}
}
sort(arr.begin(), arr.end());
int median = arr[(n * m) / 2];
int minOperations = 0;
for (int i = 0; i < n * m; ++i)
minOperations += abs(arr[i] - median) / k;
if ((n * m) % 2 == 0) {
int newMedian = arr[(n * m) / 2];
int newMinOperations = 0;
for (int i = 0; i < n * m; ++i)
newMinOperations += abs(arr[i] - newMedian) / k;
minOperations = min(minOperations, newMinOperations);
}
return minOperations;
}
int main() {
vector<vector<int> > matrix = {
{ 2, 4},
{ 20, 40},
};
int n = matrix.size();
int m = matrix[0].size();
int k = 2;
cout << "Minimum required operations = " <<
getMinOperations(n, m, k, matrix) << endl;
return 0;
}

When you compile and execute above program. It generates following output

Output

Minimum required operations = 27

Updated on: 23-Dec-2019

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