Find the minimum value of X for an expression

Given an array arr[]. The task is t find the value of X such that the result of the expression (A[1] – X)^2 + (A[2] – X)^2 + (A[3] – X)^2 + … (A[n-1] – X)^2 + (A[n] – X)^2 is minimum possible.

Examples :

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



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

Approach:

We can simplify the expression that we need to minimize. The expression can be written as

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

On differentiating the above expression, we get

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

We can denote the term (A[1] + A[2] + A[3] + … + A[n] ) as S. We get

 2nX - 2S 

Putting 2nX – 2S = 0, we get

 X = S/N 

Below is the implementation of the above approach:

C++

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// C++ implementation of above approach
  
#include <bits/stdc++.h>
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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output:

5

Time Complexity: O(n)



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