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Minimize value of a given function for any possible value of X

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  • Difficulty Level : Easy
  • Last Updated : 22 Jun, 2021

Given an array A[] consisting of N integers(1-based indexing), the task is to find the minimum value of the function \sum_{i = 1}^{N}(A[i] - (X + i))  for any possible value of X.

Examples:

Input: A[] = {1, 2, 3, 4}  
Output: 0
Explanation:
Consider the value of X as 0, then the value of the given function is (1 – 1 + 2 – 2 + 3 – 3 + 4 – 4) = 0, which is minimum.

Input: A[] = {5, 3, 9}
Output: 5

Approach: The given problem can be solved based on the following observations:

  • Consider a function as (B[i] = A[i] − i), then to minimize the value of \sum_{i = 1}^{N}(B[i] - X), the idea is to choose the value of X as the median of the array B[] such that the sum is minimized.

Follow the steps to solve the problem:

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find minimum value of
// the given function
int minimizeFunction(int A[], int N)
{
    // Stores the value of A[i] - i
    int B[N];
 
    // Traverse the given array A[]
    for (int i = 0; i < N; i++) {
 
        // Update the value of B[i]
        B[i] = A[i] - i - 1;
    }
 
    // Sort the array B[]
    sort(B, B + N);
 
    // Calculate the median of the
    // array B[]
    int median = (B[int(floor((N - 1) / 2.0))]
                  + B[int(ceil((N - 1) / 2.0))])
                 / 2;
 
    // Stores the minimum value of
    // the function
    int minValue = 0;
 
    for (int i = 0; i < N; i++) {
 
        // Update the minValue
        minValue += abs(A[i] - (median + i + 1));
    }
 
    // Return the minimum value
    return minValue;
}
 
// Driver Code
int main()
{
    int A[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9 };
    int N = sizeof(A) / sizeof(A[0]);
    cout << minimizeFunction(A, N);
 
    return 0;
}


Java




/*package whatever //do not write package name here */
import java.io.*;
import java.lang.Math;
import java.util.*;
 
class GFG {
    public static int minimizeFunction(int A[], int N)
    {
       
        // Stores the value of A[i] - i
        int B[] = new int[N];
 
        // Traverse the given array A[]
        for (int i = 0; i < N; i++) {
 
            // Update the value of B[i]
            B[i] = A[i] - i - 1;
        }
 
        // Sort the array B[]
        Arrays.sort(B);
 
        // Calculate the median of the
        // array B[]
        int median = (B[(int)(Math.floor((N - 1) / 2.0))]
                      + B[(int)(Math.ceil((N - 1) / 2.0))])
                     / 2;
 
        // Stores the minimum value of
        // the function
        int minValue = 0;
 
        for (int i = 0; i < N; i++) {
 
            // Update the minValue
            minValue += Math.abs(A[i] - (median + i + 1));
        }
 
        // Return the minimum value
        return minValue;
    }
 
    public static void main(String[] args)
    {
        int A[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9 };
        int N = A.length;
        System.out.println(minimizeFunction(A, N));
    }
}
// This code is contributed by sam_2200.


Python3




# Python3 program for the above approach
from math import floor, ceil
 
# Function to find minimum value of
# the given function
 
 
def minimizeFunction(A, N):
 
    # Stores the value of A[i] - i
    B = [0] * N
 
    # Traverse the given array A[]
    for i in range(N):
 
        # Update the value of B[i]
        B[i] = A[i] - i - 1
 
    # Sort the array B[]
    B = sorted(B)
 
    # Calculate the median of the
    # array B[]
    x, y = int(floor((N - 1) / 2.0)), int(ceil((N - 1) / 2.0))
 
    median = (B[x] + B[y]) / 2
 
    # Stores the minimum value of
    # the function
    minValue = 0
 
    for i in range(N):
 
        # Update the minValue
        minValue += abs(A[i] - (median + i + 1))
 
    # Return the minimum value
    return int(minValue)
 
 
# Driver Code
if __name__ == '__main__':
 
    A = [1, 2, 3, 4, 5, 6, 7, 8, 9]
    N = len(A)
 
    print(minimizeFunction(A, N))
 
# This code is contributed by mohit kumar 29


C#




// C# program for the above approach
using System;
 
class GFG {
    public static int minimizeFunction(int[] A, int N)
    {
       
        // Stores the value of A[i] - i
        int[] B = new int[N];
 
        // Traverse the given array A[]
        for (int i = 0; i < N; i++) {
 
            // Update the value of B[i]
            B[i] = A[i] - i - 1;
        }
 
        // Sort the array B[]
        Array.Sort(B);
 
        // Calculate the median of the
        // array B[]
        int median = (B[(int)(Math.Floor((N - 1) / 2.0))] + B[(int)(Math.Ceiling((N - 1) / 2.0))])
                     / 2;
 
        // Stores the minimum value of
        // the function
        int minValue = 0;
 
        for (int i = 0; i < N; i++) {
 
            // Update the minValue
            minValue += Math.Abs(A[i] - (median + i + 1));
        }
 
        // Return the minimum value
        return minValue;
    }
 
    static void Main()
    {
        int []A = { 1, 2, 3, 4, 5, 6, 7, 8, 9 };
        int N = A.Length;
        Console.WriteLine(minimizeFunction(A, N));
    }
}
// This code is contributed by SoumikMondal.


Javascript




<script>
 
// Javascript program for the above approach
 
function minimizeFunction(A, N){
       
        // Stores the value of A[i] - i
        let B = Array.from({length: N}, (_, i) => 0);
 
        // Traverse the given array A[]
        for (let i = 0; i < N; i++) {
 
            // Update the value of B[i]
            B[i] = A[i] - i - 1;
        }
 
        // Sort the array B[]
        B.sort();
 
        // Calculate the median of the
        // array B[]
        let median = (B[(Math.floor((N - 1) / 2.0))]
                      + B[(Math.ceil((N - 1) / 2.0))])
                     / 2;
 
        // Stores the minimum value of
        // the function
        let minValue = 0;
 
        for (let i = 0; i < N; i++) {
 
            // Update the minValue
            minValue += Math.abs(A[i] - (median + i + 1));
        }
 
        // Return the minimum value
        return minValue;
    }
 
 
// Driver Code
 
    let A = [ 1, 2, 3, 4, 5, 6, 7, 8, 9 ];
    let N = A.length;
    document.write(minimizeFunction(A, N));
 
</script>


Output: 

0

 

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


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