Skip to content
Related Articles

Related Articles

Improve Article
Minimize value of a given function for any possible value of X
  • Last Updated : 16 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.
Output: 
0

 

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with industry experts, please refer Geeks Classes Live 




My Personal Notes arrow_drop_up
Recommended Articles
Page :