# Minimize sum of adjacent difference with removal of one element from array

• Difficulty Level : Easy
• Last Updated : 27 May, 2021

Given an array of positive integers of size greater than 2. The task is to find the minimum value of the sum of consecutive difference modulus of an array, i.e. the value of |A1-A0|+|A2-A1|+|A3-A2|+……+|An-1-An-2|+|An-A(n-1)| after removal of one element from the array, where An represents the nth index of an array element value.

Examples:

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Input: arr[] = [1, 5, 3, 2, 10]
Output:
On removing 10, we get B = {1, 5, 3, 2} i.e. |1-5|+|5-3|+|3-2| = 4+2+1 = 7

Input: arr[] = [6, 12, 7, 8, 10, 15]
Output:
On removing 12, we get B = {6, 12, 7, 8, 10, 15} i.e. |6-7|+|7-8|+|8-10|+|10-15| = 1+1+2+5 = 9

The idea is to traverse the array from start to end, find the element in the array on which we get a maximum difference of consecutive modulus after its removal. Subtract the maximum value obtained from the total value calculated.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of above approach``#include ``using` `namespace` `std;` `// Function to find the element``int` `findMinRemoval(``int` `arr[], ``int` `n)``{``    ``// Value variable for storing the total value``    ``int` `temp, value = 0;` `    ``// Declaring maximum value as zero``    ``int` `maximum = 0;` `    ``// If array contains on element``    ``if` `(n == 1)``        ``return` `0;` `    ``for` `(``int` `i = 0; i < n; i++) {` `        ``// Storing the maximum value in temp variable``        ``if` `(i != 0 && i != n - 1) {``            ``value = value + ``abs``(arr[i] - arr[i + 1]);` `            ``// Adding the adjacent difference modulus``            ``// values of removed element. Removing adjacent``            ``// difference modulus value after removing element``            ``temp = ``abs``(arr[i] - arr[i + 1]) +``                   ``abs``(arr[i] - arr[i - 1]) -``                   ``abs``(arr[i - 1] - arr[i + 1]);``        ``}``        ``else` `if` `(i == 0) {``            ``value = value + ``abs``(arr[i] - arr[i + 1]);``            ``temp = ``abs``(arr[i] - arr[i + 1]);``        ``}``        ``else``            ``temp = ``abs``(arr[i] - arr[i - 1]);` `        ``maximum = max(maximum, temp);``    ``}` `    ``// Returning total value-maximum value``    ``return` `(value - maximum);``}` `// Drivers code``int` `main()``{``    ``int` `arr[] = { 1, 5, 3, 2, 10 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);` `    ``cout << findMinRemoval(arr, n) << ``"\n"``;``    ``return` `0;``}`

## Java

 `// Java implementation of the approach``class` `GFG``{` `// Function to find the element``static` `int` `findMinRemoval(``int` `arr[], ``int` `n)``{``    ``// Value variable for storing the total value``    ``int` `temp, value = ``0``;` `    ``// Declaring maximum value as zero``    ``int` `maximum = ``0``;` `    ``// If array contains on element``    ``if` `(n == ``1``)``        ``return` `0``;` `    ``for` `(``int` `i = ``0``; i < n; i++)``    ``{` `        ``// Storing the maximum value in temp variable``        ``if` `(i != ``0` `&& i != n - ``1``)``        ``{``            ``value = value + Math.abs(arr[i] - arr[i + ``1``]);` `            ``// Adding the adjacent difference modulus``            ``// values of removed element. Removing adjacent``            ``// difference modulus value after removing element``            ``temp = Math.abs(arr[i] - arr[i + ``1``]) +``                ``Math.abs(arr[i] - arr[i - ``1``]) -``                ``Math.abs(arr[i - ``1``] - arr[i + ``1``]);``        ``}``        ``else` `if` `(i == ``0``)``        ``{``            ``value = value + Math.abs(arr[i] - arr[i + ``1``]);``            ``temp = Math.abs(arr[i] - arr[i + ``1``]);``        ``}``        ``else``            ``temp = Math.abs(arr[i] - arr[i - ``1``]);` `        ``maximum = Math.max(maximum, temp);``    ``}` `    ``// Returning total value-maximum value``    ``return` `(value - maximum);``}` `// Drivers code``public` `static` `void` `main(String[] args)``{``    ``int` `arr[] = { ``1``, ``5``, ``3``, ``2``, ``10` `};``    ``int` `n = arr.length;``    ``System.out.print(findMinRemoval(arr, n) + ``"\n"``);``}``}` `// This code contributed by Rajput-Ji`

## Python 3

 `# Python 3 implementation of above approach` `# Function to find the element``def` `findMinRemoval(arr, n):` `    ``# Value variable for storing the``    ``# total value``    ``value ``=` `0` `    ``# Declaring maximum value as zero``    ``maximum ``=` `0` `    ``# If array contains on element``    ``if` `(n ``=``=` `1``):``        ``return` `0` `    ``for` `i ``in` `range``( n):` `        ``# Storing the maximum value in``        ``# temp variable``        ``if` `(i !``=` `0` `and` `i !``=` `n ``-` `1``):``            ``value ``=` `value ``+` `abs``(arr[i] ``-` `arr[i ``+` `1``])` `            ``# Adding the adjacent difference modulus``            ``# values of removed element. Removing``            ``# adjacent difference modulus value after``            ``# removing element``            ``temp ``=` `(``abs``(arr[i] ``-` `arr[i ``+` `1``]) ``+``                    ``abs``(arr[i] ``-` `arr[i ``-` `1``]) ``-``                    ``abs``(arr[i ``-` `1``] ``-` `arr[i ``+` `1``]))``        ` `        ``elif` `(i ``=``=` `0``):``            ``value ``=` `value ``+` `abs``(arr[i] ``-` `arr[i ``+` `1``])``            ``temp ``=` `abs``(arr[i] ``-` `arr[i ``+` `1``])``    ` `        ``else``:``            ``temp ``=` `abs``(arr[i] ``-` `arr[i ``-` `1``])` `        ``maximum ``=` `max``(maximum, temp)` `    ``# Returning total value-maximum value``    ``return` `(value ``-` `maximum)` `# Drivers code``if` `__name__ ``=``=` `"__main__"``:` `    ``arr ``=` `[ ``1``, ``5``, ``3``, ``2``, ``10` `]``    ``n ``=` `len``(arr)` `    ``print``(findMinRemoval(arr, n))` `# This code is contributed by ita_c`

## C#

 `// C# implementation of the approach``using` `System;` `class` `GFG``{` `    ``// Function to find the element``    ``static` `int` `findMinRemoval(``int` `[]arr, ``int` `n)``    ``{``        ``// Value variable for storing the total value``        ``int` `temp, value = 0;``    ` `        ``// Declaring maximum value as zero``        ``int` `maximum = 0;``    ` `        ``// If array contains on element``        ``if` `(n == 1)``            ``return` `0;``    ` `        ``for` `(``int` `i = 0; i < n; i++)``        ``{``    ` `            ``// Storing the maximum value in temp variable``            ``if` `(i != 0 && i != n - 1)``            ``{``                ``value = value + Math.Abs(arr[i] - arr[i + 1]);``    ` `                ``// Adding the adjacent difference modulus``                ``// values of removed element. Removing adjacent``                ``// difference modulus value after removing element``                ``temp = Math.Abs(arr[i] - arr[i + 1]) +``                    ``Math.Abs(arr[i] - arr[i - 1]) -``                    ``Math.Abs(arr[i - 1] - arr[i + 1]);``            ``}``            ``else` `if` `(i == 0)``            ``{``                ``value = value + Math.Abs(arr[i] - arr[i + 1]);``                ``temp = Math.Abs(arr[i] - arr[i + 1]);``            ``}``            ``else``                ``temp = Math.Abs(arr[i] - arr[i - 1]);``    ` `            ``maximum = Math.Max(maximum, temp);``        ``}``    ` `        ``// Returning total value-maximum value``        ``return` `(value - maximum);``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``int` `[]arr = { 1, 5, 3, 2, 10 };``        ``int` `n = arr.Length;``        ``Console.WriteLine(findMinRemoval(arr, n));``    ``}``}` `// This code contributed by Ryuga`

## PHP

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## Javascript

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Output:
`7`

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