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📜  将给定序列转换为几何级数的最小操作数。套装2

📅  最后修改于: 2021-05-06 17:35:08             🧑  作者: Mango

给定一个由N个整数组成的数组arr [] ，可以一次对任意一个元素执行以下三个操作：

  • 将一个添加到元素。
  • 从元素中减去一个。
  • 保持元素不变。

任务是找到将其转换为几何级数所需的最低成本，并找到公比。
注意：每次加减法费用为1个单位。

例子：

方法：想法是在可能的通用比率范围内进行迭代，并检查所需的最小操作数。请按照以下步骤解决问题：

  • 对数组进行排序。
  • 使用公式Ceil(arr [maximum_element] /(N – 1))查找可能的公共比率范围。
  • 计算所有可能的通用比率所需的操作次数。
  • 查找任何常见比率所需的最小操作。

下面是上述方法的实现：

C++
// C++ program for above approach
 
#include 
using namespace std;
 
// Function to find minimum cost
void minCost(int arr[], int n)
{
    if (n == 1) {
        cout << 0 << endl;
        return;
    }
 
    // Sort the array
    sort(arr, arr + n);
 
    // Maximum possible common ratios
    float raised = 1 / float(n - 1);
    float temp = pow(arr[n - 1], raised);
    int r = round(temp) + 1;
 
    int i, j, min_cost = INT_MAX;
    int common_ratio = 1;
 
    // Iterate over all possible common ratios
    for (j = 1; j <= r; j++) {
        int curr_cost = 0, prod = 1;
 
        // Calculate operations required
        // for the current common ratio
        for (i = 0; i < n; i++) {
 
            curr_cost += abs(arr[i] - prod);
            prod *= j;
            if (curr_cost >= min_cost)
                break;
        }
 
        // Calculate minimum cost
        if (i == n) {
            min_cost = min(min_cost,
                           curr_cost);
            common_ratio = j;
        }
    }
 
    cout << min_cost << ' ';
    cout << common_ratio << ' ';
}
 
// Driver Code
int main()
{
    // Given N
    int N = 6;
 
    // Given arr[]
    int arr[] = { 1, 11, 4, 27, 15, 33 };
 
    // Function Calling
    minCost(arr, N);
 
    return 0;
}

Java
// Java program for
// the above approach
import java.util.*;
class GFG{
 
// Function to find minimum cost
static void minCost(int arr[],
                    int n)
{
  if (n == 1)
  {
    System.out.print(0 + "\n");
    return;
  }
 
  // Sort the array
  Arrays.sort(arr);
 
  // Maximum possible common ratios
  float raised = 1 / (float)(n - 1);
  float temp = (float)Math.pow(arr[n - 1],
                               raised);
  int r = (int)(temp) + 1;
 
  int i, j, min_cost = Integer.MAX_VALUE;
  int common_ratio = 1;
 
  // Iterate over all possible
  // common ratios
  for (j = 1; j <= r; j++)
  {
    int curr_cost = 0, prod = 1;
 
    // Calculate operations required
    // for the current common ratio
    for (i = 0; i < n; i++)
    {
      curr_cost += Math.abs(arr[i] -
                            prod);
      prod *= j;
      if (curr_cost >= min_cost)
        break;
    }
 
    // Calculate minimum cost
    if (i == n)
    {
      min_cost = Math.min(min_cost,
                          curr_cost);
      common_ratio = j;
    }
  }
 
  System.out.print(min_cost + " ");
  System.out.print(common_ratio + " ");
}
 
// Driver Code
public static void main(String[] args)
{
  // Given N
  int N = 6;
 
  // Given arr[]
  int arr[] = {1, 11, 4,
               27, 15, 33};
 
  // Function Calling
  minCost(arr, N);
}
}
 
// This code is contributed by shikhasingrajput

Python3
# Python3 program for above approach
import sys
 
# Function to find minimum cost
def minCost(arr, n):
     
    if (n == 1):
        print(0)
        return
 
    # Sort the array
    arr = sorted(arr)
 
    # Maximum possible common ratios
    raised = 1 / (n - 1)
    temp = pow(arr[n - 1], raised)
    r = round(temp) + 1
 
    min_cost = sys.maxsize
    common_ratio = 1
 
    # Iterate over all possible
    # common ratios
    for j in range(1, r + 1):
        curr_cost = 0
        prod = 1
 
        # Calculate operations required
        # for the current common ratio
        i = 0
        while i < n:
            curr_cost += abs(arr[i] - prod)
            prod *= j
             
            if (curr_cost >= min_cost):
                break
             
            i += 1
 
        # Calculate minimum cost
        if (i == n):
            min_cost = min(min_cost,
                          curr_cost)
            common_ratio = j
 
    print(min_cost, common_ratio)
 
# Driver Code
if __name__ == '__main__':
     
    # Given N
    N = 6
 
    # Given arr[]
    arr = [ 1, 11, 4, 27, 15, 33 ]
 
    # Function calling
    minCost(arr, N)
 
# This code is contributed by mohit kumar 29

C#
// C# program for
// the above approach
using System;
 
class GFG{
 
// Function to find minimum cost
static void minCost(int []arr,
                    int n)
{
  if (n == 1)
  {
    Console.Write(0 + "\n");
    return;
  }
 
  // Sort the array
  Array.Sort(arr);
 
  // Maximum possible common ratios
  float raised = 1 / (float)(n - 1);
  float temp = (float)Math.Pow(arr[n - 1],
                               raised);
  int r = (int)(temp) + 1;
 
  int i, j, min_cost = int.MaxValue;
  int common_ratio = 1;
 
  // Iterate over all possible
  // common ratios
  for (j = 1; j <= r; j++)
  {
    int curr_cost = 0, prod = 1;
 
    // Calculate operations required
    // for the current common ratio
    for (i = 0; i < n; i++)
    {
      curr_cost += Math.Abs(arr[i] -
                            prod);
      prod *= j;
       
      if (curr_cost >= min_cost)
        break;
    }
 
    // Calculate minimum cost
    if (i == n)
    {
      min_cost = Math.Min(min_cost,
                          curr_cost);
      common_ratio = j;
    }
  }
 
  Console.Write(min_cost + " ");
  Console.Write(common_ratio + " ");
}
 
// Driver Code
public static void Main(String[] args)
{
   
  // Given N
  int N = 6;
 
  // Given []arr
  int []arr = { 1, 11, 4,
                27, 15, 33 };
 
  // Function Calling
  minCost(arr, N);
}
}
 
// This code is contributed by gauravrajput1

输出： 
28 2

时间复杂度： O(N * K)，其中K是最大可能的公共比率。
辅助空间： O(1)