C++で行列のすべての要素を等しくするための特定の型の最小演算

問題の説明

整数KとMxNの行列が与えられた場合、タスクは、行列のすべての要素を等しくするために必要な操作の最小数を見つけることです。 1回の操作で、行列の任意の要素にKを加算または減算できます。

例

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

アルゴリズム

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

例

#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;
}

上記のプログラムをコンパイルして実行する場合。次の出力を生成します

出力

Minimum required operations = 27