最小化每个元素与 (i+K)%N 处元素的位置乘积之和

给定一个大小为N的整数数组Arr[] ，任务是找到给定函数在应用于数组时的最小值。

S(K) =  1 * Arr[(0 + K)%N] + 2 * Arr[(1 + K)%N] + 3 * Arr[(2 + K)%N] + . . . . . . . + N * Arr[((N-1) + K)%N]
        = ∑[(i+1)∗Arr[(i+K)%N]

where K is a whole number.

例子：

Input: Arr[] = {2, 1, 4, 3}
Output: 20
Explanation: S(0) = 28, S(1) = 26, S(2) = 20, S(3) =26, S(4) = S(0), S(5) = S(1). . . . . 
Thus there are only four possible values of the function and minimum is 20.

Input: Arr[] = {-3, 2, 1}
Output: -5
Explanation: S(0) = 4, S(1) = -5, S(2) = 1, S(3) = S(0), S(4) = S(1), S(5) = S(2) . . . . . 
Thus there are only three possible values of the function and minimum is -5.

Naive Approach：解决该问题的基本方法是针对范围 [0, N-1] 内的所有 K 值计算函数的值。

请按照以下步骤解决问题：

• 可以观察到，如果Arr的长度为N ，那么给定函数只能有N个不同的值。
• 从k = 0 到 N – 1分别找出函数S(k)的所有可能值，并打印其中的最小值。

下面是朴素方法的实现。

C++

// C++ Program to find minimum value of given function
#include 
using namespace std;
 
// Function to find value
// of given function at k
int fnValueAtk(vector Arr, int k)
{
  int sum = 0;
  int N = Arr.size();
  for (int i = 0; i < N; i++) {
    sum += (i + 1) * Arr[(i + k) % N];
  }
  return sum;
}
 
// Function to print minimum value
// of given function
void minValFn(vector Arr)
{
  int N = Arr.size();
  int min_val = INT_MAX;
  for (int k = 0; k < N; k++) {
    int fn_At_k = fnValueAtk(Arr, k);
    min_val = min(min_val, fn_At_k);
  }
  cout << min_val;
}
 
int main()
{
 
  vector Arr = { 2, 1, 4, 3 };
  minValFn(Arr);
  return 0;
}
 
// This code is contributed by hrithikgarg03188.

Java

// Java Program to find minimum value of given function
 
import java.util.*;
 
class GFG {
 
    // Function to print minimum value
    // of given function
    static void minValFn(int[] Arr)
    {
        int N = Arr.length;
        int min_val = Integer.MAX_VALUE;
        for (int k = 0; k < N; k++) {
            int fn_At_k = fnValueAtk(Arr, k);
            min_val = Math.min(min_val, fn_At_k);
        }
        System.out.println(min_val);
    }
 
    // Function to find value
    // of given function at k
    static int fnValueAtk(int[] Arr, int k)
    {
        int sum = 0;
        int N = Arr.length;
        for (int i = 0; i < N; i++) {
            sum += (i + 1) * Arr[(i + k) % N];
        }
        return sum;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int[] Arr = { 2, 1, 4, 3 };
        minValFn(Arr);
    }
}

Python3

# Python3 program to find the minimum value of given function
import sys
 
# function to find the value
# of given function at k
def fnValueAtk(Arr, k):
    sums = 0
    N = len(Arr)
    for i in range(N):
        sums += (i + 1) * Arr[(i + k) % N]
    return sums
 
# function to print minimum value of given function
def minValFn(Arr):
    N = len(Arr)
    min_val = sys.maxsize
    for k in range(N):
        fn_At_k = fnValueAtk(Arr, k)
        min_val = min(min_val, fn_At_k)
 
    print(min_val)
 
# Driver Code
Arr = [2, 1, 4, 3]
minValFn(Arr)
 
# This code is contributed by phasing17

C#

// C# Program to find minimum value of given function
using System;
 
public class GFG{
 
  // Function to print minimum value
  // of given function
  static void minValFn(int[] Arr)
  {
    int N = Arr.Length;
    int min_val = Int32.MaxValue;
    for (int k = 0; k < N; k++) {
      int fn_At_k = fnValueAtk(Arr, k);
      min_val = Math.Min(min_val, fn_At_k);
    }
    Console.WriteLine(min_val);
  }
 
  // Function to find value
  // of given function at k
  static int fnValueAtk(int[] Arr, int k)
  {
    int sum = 0;
    int N = Arr.Length;
    for (int i = 0; i < N; i++) {
      sum += (i + 1) * Arr[(i + k) % N];
    }
    return sum;
  }
 
  // Driver code
  static public void Main (){
 
    int[] Arr = { 2, 1, 4, 3 };
    minValFn(Arr);
  }
}
 
// This code is contributed by hrithikgarg03188.

Javascript

// JavaScript program to find the
// minimum value of the given function
 
// function to find the value of
// given function at k
function fnValueAtk(Arr, k)
{
  var sum = 0;
  var N = Arr.length;
  for (var i = 0; i < N; i++)
  {
    sum += (i + 1) * Arr[(i + k) % N];
  }
  return sum;
}
 
// function to print the minimum value
// of given function
function minValFn(Arr)
{
  var N = Arr.length;
  var min_val = Number.MAX_SAFE_INTEGER;
  for (var k = 0; k < N; k++)
  {
    var fn_At_k = fnValueAtk(Arr, k);
    min_val = Math.min(min_val, fn_At_k);
  }
  document.write(min_val);
}
 
// Driver Code
var Arr = [2, 1, 4, 3];
minValFn(Arr);
 
// This code is contributed by phasing17

输出

20

时间复杂度： O(N ² )。
辅助空间： O(1)

高效方法：解决问题的有效方法基于S(k)和S(k-1) 之间的以下关系：

S(k) = S(k – 1) –  Sum + Arr[(k-1)%N] * N

请参阅下图以获得更好的理解。

插图：

Suppose the array is {2, 1, 4, 3}

Now, to find the value of the function given in the problem:

=  1 * Arr[(0 + k)%N] + 2 * Arr[(1 + k)%N] + 3 * Arr[(2 + k)%N] + . . . . . . . + N * Arr[((N-1) + k)%N]

• Now, for k = 0;
  S(0) = 1* Arr[0] + 2* Arr[1] + 3* Arr[2] + 4* Arr[3]
         = 1*2 + 2*1 + 3*4 + 4*3
         = 28
• similarly, for k = 1;
  S(1) = 1* Arr[1] + 2* Arr[2] + 3* Arr[3] + 4* Arr[0]
         = 1*1 + 2*4 + 3*3 + 4*2
         = 26
• similarly, for k = 2;
  S(2) = 20
• similarly, for k = 3;
  s(3) = 26
• Now, S(4) = S(0), S(5) = S(1). . . . and so on.

Hence the minimum value of function is S(2) = 20;

Now, it can be solved using derived equation:  S(k) = S(k – 1) –  Sum + Arr[(k-1)%N] * N

Firstly, store the value of S(0) (k = 0) in some variable by traversing the array and then use the above mentioned equation to find the values of function for k = 1, 2, 3 and so on. 
Later compare the value of S(k) with S(0) and update it with the minimum one. Follow the below steps:

• Now, the value of function when k = 0 S(0) is 28, and total sum is 10,  then to find the value for k = 1
  using equation,  
  for k = 1: S(1)  = 28 – 10 + Arr[(1 – 1) % 4] * 4
                         = 28 – 10 + 8
                         = 26 (which is less than S(0) hence update the minimum value with S(1))
• For k = 2: S(2) = 26 – 10 + Arr[(2 – 1) % 4]* 4
                         = 26 – 10 + 4
                         = 20 (which is less than S(1) hence update the minimum value with S(2))
• For k = 3: S(3)  = 20 – 10 + Arr[(3 – 1) % 4]* 4
                          = 20 – 10  + 16
                          = 26 (which is not less than S(2) hence, do not update the value with S(3))

Hence, S(2) is the minimum value of function.

按照下面提到的步骤来实现上述观察：

• 声明一个变量'min_val'来存储给定函数的最小值。
• 在数组的单次遍历中找到S(0)并将其存储在'min_val'中。
• 然后使用上面的等式找到k = 1 到 N – 1的函数S(k)的所有可能值。
  • 如果此值小于“min_val”，则更新“ min_val” 。
• min_val的最终值是所需的答案。

下面是上述方法的实现：

C++

// C++ Program to find minimum value
// Of given function
 
#include 
using namespace std;
 
void minValFn(vector Arr)
{
    int N = Arr.size();
    int min_val;
 
    // To store sum of all elements
    int sum = 0;
 
    // Finding S(0)
    int s0 = 0;
    for (int i = 0; i < N; i++) {
        s0 += Arr[i] * (i + 1);
        sum += Arr[i];
    }
 
    min_val = s0;
 
    // To store s(k-1)
    int sk_min_1 = s0;
    for (int k = 1; k < N; k++) {
 
        // Using formula
        int sk = sk_min_1 - sum + Arr[(k - 1) % N] * N;
        min_val = min(sk, min_val);
        sk_min_1 = sk;
    }
    cout << min_val << "\n";
}
 
// Driver code
int main()
{
    vector Arr = { 2, 1, 4, 3 };
    minValFn(Arr);
}
 
// This code is contributed by Taranpreet

Java

// Java Program to find minimum value
// Of given function
 
import java.io.*;
 
public class GFG {
    static void minValFn(int[] Arr)
    {
        int N = Arr.length;
        int min_val;
 
        // To store sum of all elements
        int sum = 0;
 
        // Finding S(0)
        int s0 = 0;
        for (int i = 0; i < N; i++) {
            s0 += Arr[i] * (i + 1);
            sum += Arr[i];
        }
 
        min_val = s0;
 
        // To store s(k-1)
        int sk_min_1 = s0;
        for (int k = 1; k < N; k++) {
 
            // Using formula
            int sk = sk_min_1 - sum
                     + Arr[(k - 1) % N] * N;
            min_val = Math.min(sk, min_val);
            sk_min_1 = sk;
        }
        System.out.println(min_val);
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int[] Arr = { 2, 1, 4, 3 };
        minValFn(Arr);
    }
}

Python3

# Python3 program to find the minimum value of
# the given function
 
# function to find the minimum value
def minValFn(Arr):
    N = len(Arr)
 
    # to store sum of all elements
    sums = 0
 
    # finding S(0)
    s0 = 0
    for i in range(N):
        s0 += Arr[i] * (i + 1)
        sums += Arr[i]
 
    min_val = s0
 
    # To store s(k - 1)
    sk_min_1 = s0
    for k in range(1, N):
        # using formula for minimum value
        sk = sk_min_1 - sums + Arr[(k - 1) % N] * N
        min_val = min(sk, min_val)
        sk_min_1 = sk
 
    print(min_val)
 
# Driver code
Arr = [2, 1, 4, 3]
minValFn(Arr)
 
# This code is contributed by phasing17

C＃

// C# Program to find minimum value
// Of given function
 
using System;
 
public class GFG {
  static void minValFn(int[] Arr)
  {
    int N = Arr.Length;
    int min_val;
 
    // To store sum of all elements
    int sum = 0;
 
    // Finding S(0)
    int s0 = 0;
    for (int i = 0; i < N; i++) {
      s0 += Arr[i] * (i + 1);
      sum += Arr[i];
    }
 
    min_val = s0;
 
    // To store s(k-1)
    int sk_min_1 = s0;
    for (int k = 1; k < N; k++) {
 
      // Using formula
      int sk = sk_min_1 - sum
        + Arr[(k - 1) % N] * N;
      min_val = Math.Min(sk, min_val);
      sk_min_1 = sk;
    }
    Console.Write(min_val);
  }
 
  // Driver code
  public static void Main(string[] args)
  {
    int[] Arr = { 2, 1, 4, 3 };
    minValFn(Arr);
  }
}
 
// This code is contributed by ukasp.

JavaScript

    // JavaScript Program to find minimum value
    // Of given function
 
 
    const minValFn = (Arr) => {
        let N = Arr.length;
        let min_val;
 
        // To store sum of all elements
        let sum = 0;
 
        // Finding S(0)
        let s0 = 0;
        for (let i = 0; i < N; i++) {
            s0 += Arr[i] * (i + 1);
            sum += Arr[i];
        }
 
        min_val = s0;
 
        // To store s(k-1)
        let sk_min_1 = s0;
        for (let k = 1; k < N; k++) {
 
            // Using formula
            let sk = sk_min_1 - sum + Arr[(k - 1) % N] * N;
            min_val = Math.min(sk, min_val);
            sk_min_1 = sk;
        }
        document.write(min_val);
    }
 
    // Driver code
 
    let Arr = [2, 1, 4, 3];
    minValFn(Arr);
 
    // This code is contributed by rakeshsahni
 

输出

20

时间复杂度： O(N)
辅助空间： O(1)