Given an array of n distinct integers. The problem is to find the sum of minimum absolute difference of each array element. For an element x present at index i in the array its minimum absolute difference is calculated as:
Min absolute difference (x) = min(abs(x – arr[j])), where 1 <= j <= n and j != i and abs is the absolute value.
Input Constraint: 2 <= n
Examples:
Input : arr = {4, 1, 5} Output : 5 Sum of absolute differences is |4-5| + |1-4| + |5-4| Input : arr = {5, 10, 1, 4, 8, 7} Output : 9 Input : {12, 10, 15, 22, 21, 20, 1, 8, 9} Output : 18
Naive approach: Using two loops. Pick an element of the array using outer loop and calculate its absolute difference with rest of the array elements using inner loop. Find the minimum absolute value and add it to the sum. Time Complexity O(n2).
// C++ implementation to find the sum // of minimum absolute difference of // each array element #include <bits/stdc++.h> using namespace std;
// function to find the sum of // minimum absolute difference int sumOfMinAbsDifferences( int arr[], int n)
{ int sum = 0;
for ( int i = 0; i < n; i++) {
int diff = INT_MAX;
for ( int j = 0; j < n; j++) {
if (i != j) {
diff = min(diff, abs (arr[i] - arr[j]));
}
}
sum += diff;
}
// required sum
return sum;
} // Driver code int main()
{ int arr[] = { 5, 10, 1, 4, 8, 7 };
int n = 6;
cout << "Sum = " << sumOfMinAbsDifferences(arr, n);
} // This code is contributed by garg28harsh. |
// java implementation to find the sum // of minimum absolute difference of // each array element import java.util.*;
import java.io.*;
public class GFG {
// function to find the sum of
// minimum absolute difference
static int sumOfMinAbsDifferences(
int arr[] , int n)
{
int sum= 0 ;
for ( int i= 0 ;i<n;i++){
int diff=Integer.MAX_VALUE;
for ( int j= 0 ;j<n;j++){
if (i!=j){
diff=Math.min(diff,Math.abs(arr[i]-arr[j]));
}
}
sum+=diff;
}
// required sum
return sum;
}
// Driver code
public static void main(String args[])
{
int arr[] = { 5 , 10 , 1 , 4 , 8 , 7 };
int n = arr.length;
System.out.println( "Sum = "
+ sumOfMinAbsDifferences(arr, n));
}
} |
# Python3 implementation to find the sum # of minimum absolute difference of # each array element import sys
# function to find the sum of # minimum absolute difference def sumOfMinAbsDifferences(arr, n):
sum = 0
for i in range (n):
diff = sys.maxsize
for j in range (n):
if i ! = j:
diff = min (diff, abs (arr[i] - arr[j]))
sum + = diff
# required sum
return sum
# Driver code if __name__ = = "__main__" :
arr = [ 5 , 10 , 1 , 4 , 8 , 7 ]
n = 6
print ( "Sum =" , sumOfMinAbsDifferences(arr, n))
# This code is contributed by akashish__ |
// Include namespace system using System;
public class GFG
{ // function to find the sum of
// minimum absolute difference
public static int sumOfMinAbsDifferences( int [] arr, int n)
{
var sum = 0;
for ( int i = 0; i < n; i++)
{
var diff = int .MaxValue;
for ( int j = 0; j < n; j++)
{
if (i != j)
{
diff = Math.Min(diff,Math.Abs(arr[i] - arr[j]));
}
}
sum += diff;
}
// required sum
return sum;
}
// Driver code
public static void Main(String[] args)
{
int [] arr = {5, 10, 1, 4, 8, 7};
var n = arr.Length;
Console.WriteLine( "Sum = " + GFG.sumOfMinAbsDifferences(arr, n).ToString());
}
} // This code is contributed by aadityaburujwale. |
// Javascript implementation to find the sum // of minimum absolute difference of // each array element // function to find the sum of
// minimum absolute difference
function sumOfMinAbsDifferences(arr, n)
{ let sum = 0;
for (let i = 0; i < n; i++) {
let diff = Number.MAX_VALUE;
for (let j = 0; j < n; j++) {
if (i != j) {
diff = Math.min(diff,
Math.abs(arr[i] - arr[j]));
}
}
sum += diff;
}
// required sum
return sum;
} // Driver code let arr = [ 5, 10, 1, 4, 8, 7 ]; let n = 6; console.log( "Sum = " + sumOfMinAbsDifferences(arr, n));
// This code is contributed by garg28harsh. |
Sum = 9
Time Complexity: O(n2) where n is size of the input array. This is because two nested loops are executing.
Space Complexity: O(1) as no extra space has been used.
Efficient Approach:
The following steps are:
- Sort the array of size n.
- For the 1st element of array its min absolute difference is calculated using the 2nd array element.
- For the last array element its min absolute difference is calculated using the 2nd last array element.
- For the rest of the array elements, 1 <= i <= n-2, minimum absolute difference for an element at index i is calculated as: minAbsDiff = min( abs(arr[i] – arr[i-1]), abs(ar[i] – arr[i+1]) ).
Implementation:
// C++ implementation to find the sum of minimum // absolute difference of each array element #include <bits/stdc++.h> using namespace std;
// function to find the sum of // minimum absolute difference int sumOfMinAbsDifferences( int arr[], int n)
{ // sort the given array
sort(arr, arr+n);
// initialize sum
int sum = 0;
// min absolute difference for
// the 1st array element
sum += abs (arr[0] - arr[1]);
// min absolute difference for
// the last array element
sum += abs (arr[n-1] - arr[n-2]);
// find min absolute difference for rest of the
// array elements and add them to sum
for ( int i=1; i<n-1; i++)
sum += min( abs (arr[i] - arr[i-1]), abs (arr[i] - arr[i+1]));
// required sum
return sum;
} // Driver program to test above int main()
{ int arr[] = {5, 10, 1, 4, 8, 7};
int n = sizeof (arr) / sizeof (arr[0]);
cout << "Sum = "
<< sumOfMinAbsDifferences(arr, n);
} |
// java implementation to find the sum // of minimum absolute difference of // each array element import java.io.*;
import java.util.Arrays;
public class GFG {
// function to find the sum of
// minimum absolute difference
static int sumOfMinAbsDifferences(
int arr[] , int n)
{
// sort the given array
Arrays.sort(arr);
// initialize sum
int sum = 0 ;
// min absolute difference for
// the 1st array element
sum += Math.abs(arr[ 0 ] - arr[ 1 ]);
// min absolute difference for
// the last array element
sum += Math.abs(arr[n- 1 ] - arr[n- 2 ]);
// find min absolute difference for
// rest of the array elements and
// add them to sum
for ( int i = 1 ; i < n - 1 ; i++)
sum +=
Math.min(Math.abs(arr[i] - arr[i- 1 ]),
Math.abs(arr[i] - arr[i+ 1 ]));
// required sum
return sum;
}
// Driver code
public static void main(String args[])
{
int arr[] = { 5 , 10 , 1 , 4 , 8 , 7 };
int n = arr.length;
System.out.println( "Sum = "
+ sumOfMinAbsDifferences(arr, n));
}
} // This code is contributed by Sam007. |
# Python implementation to find the # sum of minimum absolute difference # of each array element # function to find the sum of # minimum absolute difference def sumOfMinAbsDifferences(arr,n):
# sort the given array
arr.sort()
# initialize sum
sum = 0
# min absolute difference for
# the 1st array element
sum + = abs (arr[ 0 ] - arr[ 1 ]);
# min absolute difference for
# the last array element
sum + = abs (arr[n - 1 ] - arr[n - 2 ]);
# find min absolute difference for
# rest of the array elements and
# add them to sum
for i in range ( 1 , n - 1 ):
sum + = min ( abs (arr[i] - arr[i - 1 ]),
abs (arr[i] - arr[i + 1 ]))
# required sum
return sum ;
# Driver code arr = [ 5 , 10 , 1 , 4 , 8 , 7 ]
n = len (arr)
print ( "Sum = " , sumOfMinAbsDifferences(arr, n))
#This code is contributed by Sam007 |
// C# implementation to find the sum // of minimum absolute difference of // each array element using System;
public class GFG {
// function to find the sum of
// minimum absolute difference
static int sumOfMinAbsDifferences(
int []arr , int n)
{
// sort the given array
Array.Sort(arr);
// initialize sum
int sum = 0;
// min absolute difference for
// the 1st array element
sum += Math.Abs(arr[0] - arr[1]);
// min absolute difference for
// the last array element
sum += Math.Abs(arr[n-1] - arr[n-2]);
// find min absolute difference for
// rest of the array elements and
// add them to sum
for ( int i = 1; i < n - 1; i++)
sum +=
Math.Min(Math.Abs(arr[i] - arr[i-1]),
Math.Abs(arr[i] - arr[i+1]));
// required sum
return sum;
}
// Driver code
public static void Main ()
{
int []arr = {5, 10, 1, 4, 8, 7};
int n = arr.Length;
Console.Write( "Sum = "
+ sumOfMinAbsDifferences(arr, n));
}
} // This code is contributed by Sam007. |
<?php // PHP implementation to find // the sum of minimum absolute // difference of each array element // function to find the sum of // minimum absolute difference function sumOfMinAbsDifferences( $arr , $n )
{ // sort the given array
sort( $arr );
sort( $arr , $n );
// initialize sum
$sum = 0;
// min absolute difference for
// the 1st array element
$sum += abs ( $arr [0] - $arr [1]);
// min absolute difference for
// the last array element
$sum += abs ( $arr [ $n - 1] - $arr [ $n - 2]);
// find min absolute difference
// for rest of the array elements
// and add them to sum
for ( $i = 1; $i < $n - 1; $i ++)
$sum += min( abs ( $arr [ $i ] - $arr [ $i - 1]),
abs ( $arr [ $i ] - $arr [ $i + 1]));
// required sum
return $sum ;
} // Driver Code
$arr = array (5, 10, 1, 4, 8, 7);
$n = sizeof( $arr );
echo "Sum = " , sumOfMinAbsDifferences( $arr , $n );
// This code is contributed by nitin mittal. ?> |
<script> // Javascript implementation to find the sum
// of minimum absolute difference of
// each array element
// function to find the sum of
// minimum absolute difference
function sumOfMinAbsDifferences(arr, n)
{
// sort the given array
arr.sort( function (a, b){ return a - b});
// initialize sum
let sum = 0;
// min absolute difference for
// the 1st array element
sum += Math.abs(arr[0] - arr[1]);
// min absolute difference for
// the last array element
sum += Math.abs(arr[n-1] - arr[n-2]);
// find min absolute difference for
// rest of the array elements and
// add them to sum
for (let i = 1; i < n - 1; i++)
sum +=
Math.min(Math.abs(arr[i] - arr[i-1]),
Math.abs(arr[i] - arr[i+1]));
// required sum
return sum;
}
let arr = [5, 10, 1, 4, 8, 7];
let n = arr.length;
document.write( "Sum = " + sumOfMinAbsDifferences(arr, n));
</script> |
Sum = 9
Time Complexity: O(n log n)
Auxiliary Space: O(1)