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++
// C++ implementation of above approach#include <bits/stdc++.h>using namespace std;// Function to calculate value of Xint 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 Codeint 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 approachclass GFG {// Function to calculate value of Xstatic 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 Codepublic 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 Xdef 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 Coden = 7arr = [6, 9, 1, 6, 1, 3, 7]print(valueofX(arr, n))# This code is contributed by Mohit Kumar |
C#
// C# implementation of above approachusing System; class GFG {// Function to calculate value of Xstatic 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 Codepublic 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
<script>// 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.</script> |
Output:
5
Time Complexity: O(n)
Auxiliary Space: O(1)
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