Given an array A[] consisting of N integers(1-based indexing), the task is to find the minimum value of the function
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:
- Initialize an array, say B[] that stores the value of (A[i] – i) for every possible value of the array A[].
- Traverse the given array A[] and for each index i, update the value of B[i] as (A[i] – i).
- Sort the array B[] in ascending order and find the value X as the median of the array B[].
- After completing the above steps, find the value of the given function
for the calculated value of X.
Below is the implementation of the above approach:
// 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;
} |
/*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 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# 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. |
<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)