给定类型的最小运算，以使矩阵的所有元素在C ++中相等
问题陈述
给定一个整数K和一个MxN的矩阵，任务是找到使矩阵的所有元素相等所需的最小操作数。在单个操作中，可以将K添加到矩阵的任何元素或从中减去。
示例
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算法
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
示例
#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;
}当您编译并执行上述程序时。它产生以下输出
输出结果
Minimum required operations = 27