为表达式找出 X 的最小值

给定一个数组arr[] 。任务是找到 X 的值，使得表达式 (A[1] – X)^2 + (A[2] – X)^2 + (A[3] – X)^2 + … (A[n-1] – X)^2 + (A[n] – X)^2 是可能的最小值。
例子 ：

Input : arr[] = {6, 9, 1, 6, 1, 3, 7} 
Output : 5
Input : arr[] = {1, 2, 3, 4, 5} 
Output : 3 
 

方法：
我们可以简化我们需要最小化的表达式。表达式可以写成

(A[1]^2 + A[2]^2 + A[3]^2 + … + A[n]^2) + nX^2 – 2X(A[1] + A[2] + A[3] + … + A[n])

对上述表达式进行微分，我们得到

2nX - 2(A[1] + A[2] + A[3] + … + A[n])

我们可以将项 (A[1] + A[2] + A[3] + … + A[n] ) 表示为 S。我们得到

2nX - 2S 

将 2nX – 2S = 0，我们得到

X = S/N 

下面是上述方法的实现：

C++

// C++ implementation of above approach
 
#include 
using namespace std;
 
// Function to calculate value of X
int valueofX(int ar[], int n)
{
    int sum = 0;
    for (int i = 0; i < n; i++) {
        sum = sum + ar[i];
    }
 
    if (sum % n == 0) {
        return sum / n;
    }
    else {
        int A = sum / n, B = sum / n + 1;
        int ValueA = 0, ValueB = 0;
 
        // Check for both possibilities
        for (int i = 0; i < n; i++) {
            ValueA += (ar[i] - A) * (ar[i] - A);
            ValueB += (ar[i] - B) * (ar[i] - B);
        }
 
        if (ValueA < ValueB) {
            return A;
        }
        else {
            return B;
        }
    }
}
 
// Driver Code
int main()
{
    int n = 7;
    int arr[7] = { 6, 9, 1, 6, 1, 3, 7 };
 
    cout << valueofX(arr, n) << '\n';
 
    return 0;
}

Java

// Java implementation of above approach
class GFG
{
 
// Function to calculate value of X
static int valueofX(int ar[], int n)
{
    int sum = 0;
    for (int i = 0; i < n; i++)
    {
        sum = sum + ar[i];
    }
 
    if (sum % n == 0)
    {
        return sum / n;
    }
    else
    {
        int A = sum / n, B = sum / n + 1;
        int ValueA = 0, ValueB = 0;
 
        // Check for both possibilities
        for (int i = 0; i < n; i++)
        {
            ValueA += (ar[i] - A) * (ar[i] - A);
            ValueB += (ar[i] - B) * (ar[i] - B);
        }
 
        if (ValueA < ValueB)
        {
            return A;
        }
        else
        {
            return B;
        }
    }
}
 
// Driver Code
public static void main(String args[])
{
    int n = 7;
    int arr[] = { 6, 9, 1, 6, 1, 3, 7 };
 
    System.out.println(valueofX(arr, n));
}
}
 
// This code is contributed by PrinciRaj1992

Python3

# Python3 implementation of above approach
 
# Function to calculate value of X
def valueofX(ar, n):
    summ = sum(ar)
 
    if (summ % n == 0):
        return summ // n
    else:
        A = summ // n
        B = summ // n + 1
        ValueA = 0
        ValueB = 0
 
        # Check for both possibilities
        for i in range(n):
            ValueA += (ar[i] - A) * (ar[i] - A)
            ValueB += (ar[i] - B) * (ar[i] - B)
 
        if (ValueA < ValueB):
            return A
        else:
            return B
 
# Driver Code
n = 7
arr = [6, 9, 1, 6, 1, 3, 7]
 
print(valueofX(arr, n))
 
# This code is contributed by Mohit Kumar

C#

// C# implementation of above approach
using System;
                     
class GFG
{
 
// Function to calculate value of X
static int valueofX(int []ar, int n)
{
    int sum = 0;
    for (int i = 0; i < n; i++)
    {
        sum = sum + ar[i];
    }
 
    if (sum % n == 0)
    {
        return sum / n;
    }
    else
    {
        int A = sum / n, B = sum / n + 1;
        int ValueA = 0, ValueB = 0;
 
        // Check for both possibilities
        for (int i = 0; i < n; i++)
        {
            ValueA += (ar[i] - A) * (ar[i] - A);
            ValueB += (ar[i] - B) * (ar[i] - B);
        }
 
        if (ValueA < ValueB)
        {
            return A;
        }
        else
        {
            return B;
        }
    }
}
 
// Driver Code
public static void Main(String []args)
{
    int n = 7;
    int []arr = { 6, 9, 1, 6, 1, 3, 7 };
 
    Console.WriteLine(valueofX(arr, n));
}
}
 
// This code is contributed by 29AjayKumar

Javascript

// JavaScript implementation of above approach
 
// Function to calculate value of X
function valueofX(ar, n)
{
    let sum = 0;
    for (let i = 0; i < n; i++) {
        sum = sum + ar[i];
    }
 
    if (sum % n == 0) {
        return Math.floor(sum / n);
    }
    else {
        let A = Math.floor(sum / n), B = Math.floor(sum / n + 1);
        let ValueA = 0, ValueB = 0;
 
        // Check for both possibilities
        for (let i = 0; i < n; i++) {
            ValueA += (ar[i] - A) * (ar[i] - A);
            ValueB += (ar[i] - B) * (ar[i] - B);
        }
 
        if (ValueA < ValueB) {
            return A;
        }
        else {
            return B;
        }
    }
}
 
// Driver Code
    let n = 7;
    let arr = [ 6, 9, 1, 6, 1, 3, 7 ];
 
    document.write(valueofX(arr, n) + "<br>");
 
// This code is contributed by Surbhi Tyagi.

输出：

5

时间复杂度： O(n)

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