Minimize value of a given function for any possible value of X
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 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 , 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++
#include <bits/stdc++.h>
using namespace std;
int minimizeFunction( int A[], int N)
{
int B[N];
for ( int i = 0; i < N; i++) {
B[i] = A[i] - i - 1;
}
sort(B, B + N);
int median = (B[ int ( floor ((N - 1) / 2.0))]
+ B[ int ( ceil ((N - 1) / 2.0))])
/ 2;
int minValue = 0;
for ( int i = 0; i < N; i++) {
minValue += abs (A[i] - (median + i + 1));
}
return minValue;
}
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
import java.io.*;
import java.lang.Math;
import java.util.*;
class GFG {
public static int minimizeFunction( int A[], int N)
{
int B[] = new int [N];
for ( int i = 0 ; i < N; i++) {
B[i] = A[i] - i - 1 ;
}
Arrays.sort(B);
int median = (B[( int )(Math.floor((N - 1 ) / 2.0 ))]
+ B[( int )(Math.ceil((N - 1 ) / 2.0 ))])
/ 2 ;
int minValue = 0 ;
for ( int i = 0 ; i < N; i++) {
minValue += Math.abs(A[i] - (median + i + 1 ));
}
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));
}
}
|
Python3
from math import floor, ceil
def minimizeFunction(A, N):
B = [ 0 ] * N
for i in range (N):
B[i] = A[i] - i - 1
B = sorted (B)
x, y = int (floor((N - 1 ) / 2.0 )), int (ceil((N - 1 ) / 2.0 ))
median = (B[x] + B[y]) / 2
minValue = 0
for i in range (N):
minValue + = abs (A[i] - (median + i + 1 ))
return int (minValue)
if __name__ = = '__main__' :
A = [ 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 ]
N = len (A)
print (minimizeFunction(A, N))
|
C#
using System;
class GFG {
public static int minimizeFunction( int [] A, int N)
{
int [] B = new int [N];
for ( int i = 0; i < N; i++) {
B[i] = A[i] - i - 1;
}
Array.Sort(B);
int median = (B[( int )(Math.Floor((N - 1) / 2.0))] + B[( int )(Math.Ceiling((N - 1) / 2.0))])
/ 2;
int minValue = 0;
for ( int i = 0; i < N; i++) {
minValue += Math.Abs(A[i] - (median + i + 1));
}
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));
}
}
|
Javascript
<script>
function minimizeFunction(A, N){
let B = Array.from({length: N}, (_, i) => 0);
for (let i = 0; i < N; i++) {
B[i] = A[i] - i - 1;
}
B.sort();
let median = (B[(Math.floor((N - 1) / 2.0))]
+ B[(Math.ceil((N - 1) / 2.0))])
/ 2;
let minValue = 0;
for (let i = 0; i < N; i++) {
minValue += Math.abs(A[i] - (median + i + 1));
}
return minValue;
}
let A = [ 1, 2, 3, 4, 5, 6, 7, 8, 9 ];
let N = A.length;
document.write(minimizeFunction(A, N));
</script>
|
Time Complexity: O(N * log N)
Auxiliary Space: O(N)
Like Article
Suggest improvement
Share your thoughts in the comments
Please Login to comment...