# Minimize value of x that minimizes value of |a1−x|^c+|a2−x|^c+···+|an−x|^c for value of c as 1 and 2

Given an array arr[] of N elements, the task is to find the value of x that minimizes the value of expression for c = 1.

|a1−x|c+|a2−x|c+···+|an−x|c  = |a1−x|+|a2−x|+···+|an−x|

Examples:

Input: arr[] = { 1, 2, 9, 2, 6 }
Output: 2
Explanation: The best solution is to select x = 2 which produces the sum  |1−2| + |2−2| + |9−2| + |2−2| + |6−2| = 12 , which is the minimum possible sum, for all other values, the sum so obtained will be greater than 2

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

Approach: In the general case, the best choice for x is the median of the given numbers, The median is an optimal choice, because if x is smaller than the median, the sum becomes smaller by increasing x, and if x is larger than the median, the sum becomes smaller by decreasing x. Hence, the optimal solution is that x is the median.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach ``#include ``using` `namespace` `std; `` ` `// Function to print the possible ``// values of x that minimizes the sum ``void` `findX(``int` `arr[], ``int` `n) ``{ ``    ``// Sort the numbers ``    ``sort(arr, arr + n); `` ` `    ``// Stores the median ``    ``int` `x; `` ` `    ``// Only one median if n is odd ``    ``if` `(n % 2 != 0) { ``        ``x = arr[n / 2]; ``    ``} `` ` `    ``// Two medians if n is even ``    ``// and every value between them ``    ``// is optimal print any of them ``    ``else` `{ ``        ``int` `a = arr[n / 2 - 1]; ``        ``int` `b = arr[n / 2]; ``        ``x = a; ``    ``} `` ` `    ``int` `sum = 0; `` ` `    ``// Find minimum sum ``    ``for` `(``int` `i = 0; i < n; i++) { ``        ``sum += ``abs``(arr[i] - x); ``    ``} `` ` `    ``cout << sum; ``} `` ` `// Driver Code ``int` `main() ``{ ``    ``int` `arr1[] = { 1, 2, 9, 2, 6 }; ``    ``int` `n1 = ``sizeof``(arr1) / ``sizeof``(arr1[0]); `` ` `    ``findX(arr1, n1); ``    ``return` `0; ``}`

## Java

 `// Java code for the above approach ``import` `java.util.*; `` ` `class` `GFG ``{ ``   ` `  ``// Function to print the possible ``// values of x that minimizes the sum ``static` `void` `findX(``int` `arr[], ``int` `n) ``{ ``   ` `    ``// Sort the numbers ``    ``Arrays.sort(arr); `` ` `    ``// Stores the median ``    ``int` `x; `` ` `    ``// Only one median if n is odd ``    ``if` `(n % ``2` `!= ``0``) { ``        ``x = arr[(``int``)Math.floor(n / ``2``)]; ``    ``} `` ` `    ``// Two medians if n is even ``    ``// and every value between them ``    ``// is optimal print any of them ``    ``else` `{ ``        ``int` `a = arr[n / ``2` `- ``1``]; ``        ``int` `b = arr[n / ``2``]; ``        ``x = a; ``    ``} `` ` `    ``int` `sum = ``0``; `` ` `    ``// Find minimum sum ``    ``for` `(``int` `i = ``0``; i < n; i++) { ``        ``sum += Math.abs(arr[i] - x); ``    ``} `` ` `   ``System.out.println( sum); ``} `` ` `    ``public` `static` `void` `main (String[] args) { ``         ` `    ``int` `arr1[] = { ``1``, ``2``, ``9``, ``2``, ``6` `}; ``    ``int` `n1 = arr1.length; `` ` `    ``findX(arr1, n1); ``    ``} ``} `` ` `// This code is contributed by Potta Lokesh`

## Python3

 `# Python program for the above approach `` ` `# Function to print the possible ``# values of x that minimizes the sum ``def` `findX(arr, n): ``   ` `  ``# Sort the numbers ``  ``arr.sort(); `` ` `  ``# Stores the median ``  ``x ``=` `None``; `` ` `  ``# Only one median if n is odd ``  ``if` `(n ``%` `2` `!``=` `0``): ``    ``x ``=` `arr[n ``/``/` `2``]; ``   ` `  ``# Two medians if n is even ``  ``# and every value between them ``  ``# is optimal print any of them ``  ``else``: ``    ``a ``=` `arr[(n ``/``/` `2``) ``-` `1``]; ``    ``b ``=` `arr[n ``/``/` `2``]; ``    ``x ``=` `a; ``  ``sum` `=` `0``; `` ` `  ``# Find minimum sum ``  ``for` `i ``in` `range``(n): ``    ``sum` `+``=` `abs``(arr[i] ``-` `x); `` ` ` ` `  ``print``(``sum``); `` ` `# Driver Code ``arr1 ``=` `[``1``, ``2``, ``9``, ``2``, ``6``]; ``n1 ``=` `len``(arr1) `` ` `findX(arr1, n1); `` ` `# This code is contributed by gfgking.`

## C#

 `// C# code for the above approach ``using` `System; `` ` `class` `GFG { `` ` `    ``// Function to print the possible ``    ``// values of x that minimizes the sum ``    ``static` `void` `findX(``int``[] arr, ``int` `n) ``    ``{ `` ` `        ``// Sort the numbers ``        ``Array.Sort(arr); `` ` `        ``// Stores the median ``        ``int` `x; `` ` `        ``// Only one median if n is odd ``        ``if` `(n % 2 != 0) { ``            ``x = arr[(``int``)Math.Floor((``float``)(n / 2))]; ``        ``} `` ` `        ``// Two medians if n is even ``        ``// and every value between them ``        ``// is optimal print any of them ``        ``else` `{ ``            ``int` `a = arr[n / 2 - 1]; `` ` `            ``x = a; ``        ``} `` ` `        ``int` `sum = 0; `` ` `        ``// Find minimum sum ``        ``for` `(``int` `i = 0; i < n; i++) { ``            ``sum += Math.Abs(arr[i] - x); ``        ``} `` ` `        ``Console.WriteLine(sum); ``    ``} `` ` `    ``public` `static` `void` `Main(``string``[] args) ``    ``{ `` ` `        ``int``[] arr1 = { 1, 2, 9, 2, 6 }; ``        ``int` `n1 = arr1.Length; `` ` `        ``findX(arr1, n1); ``    ``} ``} `` ` `// This code is contributed by ukasp.`

## Javascript

 ``

Output
`12`

Time Complexity: O(N*log N)
Auxiliary Space: O(1)

Given an array arr[] of N elements, the task is to find the value of x that minimizes the value of expression for c = 2.

|a1−x|c+|a2−x|c+···+|an−x|c  = (a1−x)2+(a2−x)2+···+(an−x)2.

Examples :

Input: arr[] = { 1, 2, 9, 2, 6 }
Output: 4
Explanation:  The best solution is to select x = 4 which produces the sum  (1−4)^2 + (2−4)^2 + (9−4)^2 + (2−4)^2 + (6−4)^2 = 46, which is the minimum possible sum.

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

Approach: In the general case, the best choice for x is the average of the numbers. This result can be derived by expanding the sum as follows:

nx2−2x(a1+a2+···+an) + (a12+a22+···+an2

The last part does not depend on x. The remaining parts form a function nx2 − 2xs where s=a1+a2+···+an. Applying derivative to this equation w.r.t x and equating the result to zero gives us x = s / n, which is the value that minimizes the sum.

Below is the implementation of the above approach:

## C++

 `// C++ implementation for the above approach ``#include ``using` `namespace` `std; `` ` `// Function to find the value of x ``// that minimizes the sum ``void` `findX(``int` `arr[], ``int` `n) ``{ ``    ``// Store the sum ``    ``double` `sum = 0; ``    ``for` `(``int` `i = 0; i < n; i++) { ``        ``sum += arr[i]; ``    ``} `` ` `    ``// Store the average of numbers ``    ``double` `x = sum / n; `` ` `    ``double` `minSum = 0; `` ` `    ``// Find minimum sum ``    ``for` `(``int` `i = 0; i < n; i++) { ``        ``minSum += ``pow``((arr[i] - x), 2); ``    ``} `` ` `    ``cout << minSum; ``} `` ` `// Driver Code ``int` `main() ``{ ``    ``int` `arr[] = { 1, 2, 9, 2, 6 }; ``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]); `` ` `    ``findX(arr, n); `` ` `    ``return` `0; ``}`

## Java

 `// Java implementation for the above approach ``import` `java.util.*; ``public` `class` `GFG ``{ ``// Function to find the value of x ``// that minimizes the sum ``static` `void` `findX(``int` `[]arr, ``int` `n) ``{ ``    ``// Store the sum ``    ``int` `sum = ``0``; ``    ``for` `(``int` `i = ``0``; i < n; i++) { ``        ``sum += arr[i]; ``    ``} `` ` `    ``// Store the average of numbers ``    ``int` `x = sum / n; `` ` `    ``int` `minSum = ``0``; `` ` `    ``// Find minimum sum ``    ``for` `(``int` `i = ``0``; i < n; i++) { ``        ``minSum += Math.pow((arr[i] - x), ``2``); ``    ``} `` ` `    ``System.out.print(minSum); ``} `` ` `// Driver Code ``public` `static` `void` `main(String args[]) ``{ ``    ``int` `[]arr = { ``1``, ``2``, ``9``, ``2``, ``6` `}; ``    ``int` `n = arr.length; `` ` `    ``findX(arr, n); ``} ``} ``// This code is contributed by Samim Hossain Mondal.`

## Python3

 `# Python implementation for the above approach `` ` `# Function to find the value of x ``# that minimizes the sum ``def` `findX(arr, n): ``   ` `    ``# Store the sum ``    ``sum` `=` `0``; ``    ``for` `i ``in` `range``(n): ``        ``sum` `+``=` `arr[i]; ``     ` `    ``# Store the average of numbers ``    ``x ``=` `sum` `/``/` `n; `` ` `    ``minSum ``=` `0``; `` ` `    ``# Find minimum sum ``    ``for` `i ``in` `range``(n): ``        ``minSum ``+``=` `pow``((arr[i] ``-` `x), ``2``); ``    ``print``(minSum); `` ` `# Driver Code ``if` `__name__ ``=``=` `'__main__'``: ``    ``arr ``=` `[ ``1``, ``2``, ``9``, ``2``, ``6` `]; ``    ``n ``=` `len``(arr); `` ` `    ``findX(arr, n); `` ` `# This code is contributed by shikhasingrajput`

## C#

 `// C# implementation for the above approach ``using` `System; ``class` `GFG ``{ ``// Function to find the value of x ``// that minimizes the sum ``static` `void` `findX(``int` `[]arr, ``int` `n) ``{ ``    ``// Store the sum ``    ``int` `sum = 0; ``    ``for` `(``int` `i = 0; i < n; i++) { ``        ``sum += arr[i]; ``    ``} `` ` `    ``// Store the average of numbers ``    ``int` `x = sum / n; `` ` `    ``int` `minSum = 0; `` ` `    ``// Find minimum sum ``    ``for` `(``int` `i = 0; i < n; i++) { ``        ``minSum += (``int``)Math.Pow((arr[i] - x), 2); ``    ``} `` ` `    ``Console.Write(minSum); ``} `` ` `// Driver Code ``public` `static` `void` `Main() ``{ ``    ``int` `[]arr = { 1, 2, 9, 2, 6 }; ``    ``int` `n = arr.Length; `` ` `    ``findX(arr, n); ``} ``} ``// This code is contributed by Samim Hossain Mondal.`

## Javascript

 ``

Output
`46`

Time Complexity: O(N)
Auxiliary Space: O(1)

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