Related Articles

# Maximum value of |arr[i] – arr[j]| + |i – j|

• Difficulty Level : Hard
• Last Updated : 20 Jul, 2021

Given a array of N positive integers. The task is to find the maximum value of |arr[i] – arr[j]| + |i – j|, where 0 <= i, j <= N – 1 and arr[i], arr[j] belong to the array.

Examples:

Input : N = 4, arr[] = { 1, 2, 3, 1 }
Output : 4
Explanation:
Choose i = 0 and j = 2. This will result in |1-3|+|0-2| = 4 which is the maximum possible value.

Input : N = 3, arr[] = { 1, 1, 1 }
Output : 2

Method 1: The idea is to use brute force i.e iterate in two for loops.

Below is the implementation of this approach:

## C++

 `#include ``using` `namespace` `std;``#define MAX 10` `// Return maximum value of |arr[i] - arr[j]| + |i - j|``int` `findValue(``int` `arr[], ``int` `n)``{``    ``int` `ans = 0;` `    ``// Iterating two for loop, one for``    ``// i and another for j.``    ``for` `(``int` `i = 0; i < n; i++)``        ``for` `(``int` `j = 0; j < n; j++)` `            ``// Evaluating |arr[i] - arr[j]| + |i - j|``            ``// and compare with previous maximum.``            ``ans = max(ans,``                      ``abs``(arr[i] - arr[j]) + ``abs``(i - j));` `    ``return` `ans;``}` `// Driven Program``int` `main()``{``    ``int` `arr[] = { 1, 2, 3, 1 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);` `    ``cout << findValue(arr, n) << endl;` `    ``return` `0;``}`

## Java

 `// java program to find maximum value of``// |arr[i] - arr[j]| + |i - j|``class` `GFG {``    ``static` `final` `int` `MAX = ``10``;` `    ``// Return maximum value of``    ``// |arr[i] - arr[j]| + |i - j|``    ``static` `int` `findValue(``int` `arr[], ``int` `n)``    ``{``        ``int` `ans = ``0``;` `        ``// Iterating two for loop,``        ``// one for i and another for j.``        ``for` `(``int` `i = ``0``; i < n; i++)``            ``for` `(``int` `j = ``0``; j < n; j++)` `                ``// Evaluating |arr[i] - arr[j]|``                ``// + |i - j| and compare with``                ``// previous maximum.``                ``ans = Math.max(ans,``                               ``Math.abs(arr[i] - arr[j])``                               ``+ Math.abs(i - j));` `        ``return` `ans;``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `arr[] = { ``1``, ``2``, ``3``, ``1` `};``        ``int` `n = arr.length;` `        ``System.out.println(findValue(arr, n));``    ``}``}` `// This code is contributed by Anant Agarwal.`

## Python3

 `# Python3 program to find``# maximum value of``# |arr[i] - arr[j]| + |i - j|` `# Return maximum value of``# |arr[i] - arr[j]| + |i - j|``def` `findValue(arr, n):``    ``ans ``=` `0``;``    ` `    ``# Iterating two for loop,``    ``# one for i and another for j.``    ``for` `i ``in` `range``(n):``        ``for` `j ``in` `range``(n):``            ` `            ``# Evaluating |arr[i] -``            ``# arr[j]| + |i - j|``            ``# and compare with``            ``# previous maximum.``            ``ans ``=` `ans ``if` `ans>(``abs``(arr[i] ``-` `arr[j]) ``+``                              ``abs``(i ``-` `j)) ``else` `(``abs``(arr[i] ``-``                                      ``arr[j]) ``+` `abs``(i ``-` `j)) ;``    ``return` `ans;``    ` `# Driver Code``arr ``=` `[``1``, ``2``, ``3``, ``1``];``n ``=` `len``(arr);``print``(findValue(arr, n));` `# This code is contributed by mits.`

## C#

 `// C# program to find maximum value of``// |arr[i] - arr[j]| + |i - j|``using` `System;` `class` `GFG {``    ` `    ``// Return maximum value of``    ``// |arr[i] - arr[j]| + |i - j|``    ``static` `int` `findValue(``int` `[]arr, ``int` `n)``    ``{``        ``int` `ans = 0;``    ` `        ``// Iterating two for loop,``        ``// one for i and another for j.``        ``for` `(``int` `i = 0; i < n; i++)``            ``for` `(``int` `j = 0; j < n; j++)``    ` `                ``// Evaluating |arr[i] - arr[j]|``                ``// + |i - j| and compare with``                ``// previous maximum.``                ``ans = Math.Max(ans,``                    ``Math.Abs(arr[i] - arr[j])``                            ``+ Math.Abs(i - j));``    ` `        ``return` `ans;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `Main ()``    ``{``        ``int` `[]arr = { 1, 2, 3, 1 };``        ``int` `n =arr.Length;``        ` `        ``Console.Write(findValue(arr, n));``    ``}``}` `// This code is contributed by nitin mittal.`

## PHP

 ``

## Javascript

 ``
Output
`4`

Method 2 (tricky):

First of all lets make four equations by removing the absolute value signs (“|”). The following 4 equations will be formed, and we need to find the maximum value of these equations and that will be our answer.

1. arr[i] – arr[j] + i – j = (arr[i] + i) – (arr[j] + j)
2. arr[i] – arr[j] – i + j = (arr[i] – i) – (arr[j] – j)
3. -arr[i] + arr[j] + i – j = -(arr[i] – i) + (arr[j] – j)
4. -arr[i] + arr[j] – i + j = -(arr[i] + i) + (arr[j] + j)

Observe the equations (1) and (4) are identical. Similarly, equations (2) and (3) are identical.
Now the task is to find the maximum value of these equations. So the approach is to form two arrays, first_array[], it will store arr[i] + i, 0 <= i < n, second_array[], it will store arr[i] – i, 0 <= i < n.
Now our task is easy, we just need to find the maximum difference between the two values of these two arrays.
For that, we find the maximum value and minimum value in the first_array and store their difference:
ans1 = (maximum value in first_array – minimum value in first_array)
Similarly, we need to find the maximum value and minimum value in the second_array and store their difference:
ans2 = (maximum value in second_array – minimum value in second_array)
Our answer will be a maximum of ans1 and ans2.

Below is the implementation of the above approach:

## C++

 `// Efficient CPP program to find maximum value``// of |arr[i] - arr[j]| + |i - j|``#include ``using` `namespace` `std;` `// Return maximum |arr[i] - arr[j]| + |i - j|``int` `findValue(``int` `arr[], ``int` `n)``{``    ``int` `a[n], b[n], tmp;` `    ``// Calculating first_array and second_array``    ``for` `(``int` `i = 0; i < n; i++)``    ``{``        ``a[i] = (arr[i] + i);``        ``b[i] = (arr[i] - i);``    ``}` `    ``int` `x = a[0], y = a[0];` `    ``// Finding maximum and minimum value in``    ``// first_array``    ``for` `(``int` `i = 0; i < n; i++)``    ``{``        ``if` `(a[i] > x)``            ``x = a[i];` `        ``if` `(a[i] < y)``            ``y = a[i];``    ``}` `    ``// Storing the difference between maximum and``    ``// minimum value in first_array``    ``int` `ans1 = (x - y);` `    ``x = b[0];``    ``y = b[0];` `    ``// Finding maximum and minimum value in``    ``// second_array``    ``for` `(``int` `i = 0; i < n; i++)``    ``{``        ``if` `(b[i] > x)``            ``x = b[i];` `        ``if` `(b[i] < y)``            ``y = b[i];``    ``}` `    ``// Storing the difference between maximum and``    ``// minimum value in second_array``    ``int` `ans2 = (x - y);` `    ``return` `max(ans1, ans2);``}` `// Driven Code``int` `main()``{``    ``int` `arr[] = { 1, 2, 3, 1 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);` `    ``cout << findValue(arr, n) << endl;` `    ``return` `0;``}`

## Java

 `// Efficient Java program to find maximum``// value of |arr[i] - arr[j]| + |i - j|``import` `java.io.*;``class` `GFG {` `    ``// Return maximum |arr[i] -``    ``// arr[j]| + |i - j|``    ``static` `int` `findValue(``int` `arr[], ``int` `n)``    ``{``        ``int` `a[] = ``new` `int``[n];``        ``int` `b[] = ``new` `int``[n];``        ``int` `tmp;` `        ``// Calculating first_array``        ``// and second_array``        ``for` `(``int` `i = ``0``; i < n; i++)``        ``{``            ``a[i] = (arr[i] + i);``            ``b[i] = (arr[i] - i);``        ``}` `        ``int` `x = a[``0``], y = a[``0``];` `        ``// Finding maximum and``        ``// minimum value in``        ``// first_array``        ``for` `(``int` `i = ``0``; i < n; i++)``        ``{``            ``if` `(a[i] > x)``                ``x = a[i];` `            ``if` `(a[i] < y)``                ``y = a[i];``        ``}` `        ``// Storing the difference``        ``// between maximum and``        ``// minimum value in first_array``        ``int` `ans1 = (x - y);` `        ``x = b[``0``];``        ``y = b[``0``];` `        ``// Finding maximum and``        ``// minimum value in``        ``// second_array``        ``for` `(``int` `i = ``0``; i < n; i++)``        ``{``            ``if` `(b[i] > x)``                ``x = b[i];` `            ``if` `(b[i] < y)``                ``y = b[i];``        ``}` `        ``// Storing the difference``        ``// between maximum and``        ``// minimum value in second_array``        ``int` `ans2 = (x - y);` `        ``return` `Math.max(ans1, ans2);``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `arr[] = { ``1``, ``2``, ``3``, ``1` `};``        ``int` `n = arr.length;``        ``System.out.println(findValue(arr, n));``    ``}``}` `// This code is contributed by anuj_67.`

## Python3

 `# Efficient Python3 program``# to find maximum value``# of |arr[i] - arr[j]| + |i - j|` `# Return maximum |arr[i] -``# arr[j]| + |i - j|`  `def` `findValue(arr, n):``    ``a ``=` `[]``    ``b ``=` `[]` `    ``# Calculating first_array``    ``# and second_array``    ``for` `i ``in` `range``(n):``        ``a.append(arr[i] ``+` `i)``        ``b.append(arr[i] ``-` `i)` `    ``x ``=` `a[``0``]``    ``y ``=` `a[``0``]` `    ``# Finding maximum and``    ``# minimum value in``    ``# first_array``    ``for` `i ``in` `range``(n):``        ``if` `(a[i] > x):``            ``x ``=` `a[i]` `        ``if` `(a[i] < y):``            ``y ``=` `a[i]` `    ``# Storing the difference``    ``# between maximum and``    ``# minimum value in first_array``    ``ans1 ``=` `(x ``-` `y)` `    ``x ``=` `b[``0``]``    ``y ``=` `b[``0``]` `    ``# Finding maximum and``    ``# minimum value in``    ``# second_array``    ``for` `i ``in` `range``(n):``        ``if` `(b[i] > x):``            ``x ``=` `b[i]` `        ``if` `(b[i] < y):``            ``y ``=` `b[i]` `    ``# Storing the difference``    ``# between maximum and``    ``# minimum value in``    ``# second_array``    ``ans2 ``=` `(x ``-` `y)` `    ``return` `max``(ans1, ans2)`  `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ``arr ``=` `[``1``, ``2``, ``3``, ``1``]``    ``n ``=` `len``(arr)` `    ``print``(findValue(arr, n))` `# This code is contributed by mits`

## C#

 `// Efficient Java program to find maximum``// value of |arr[i] - arr[j]| + |i - j|``using` `System;``class` `GFG {` `    ``// Return maximum |arr[i] -``    ``// arr[j]| + |i - j|``    ``static` `int` `findValue(``int``[] arr, ``int` `n)``    ``{``        ``int``[] a = ``new` `int``[n];``        ``int``[] b = ``new` `int``[n];``        ``// int tmp;` `        ``// Calculating first_array``        ``// and second_array``        ``for` `(``int` `i = 0; i < n; i++)``        ``{``            ``a[i] = (arr[i] + i);``            ``b[i] = (arr[i] - i);``        ``}` `        ``int` `x = a[0], y = a[0];` `        ``// Finding maximum and``        ``// minimum value in``        ``// first_array``        ``for` `(``int` `i = 0; i < n; i++)``        ``{``            ``if` `(a[i] > x)``                ``x = a[i];` `            ``if` `(a[i] < y)``                ``y = a[i];``        ``}` `        ``// Storing the difference``        ``// between maximum and``        ``// minimum value in first_array``        ``int` `ans1 = (x - y);` `        ``x = b[0];``        ``y = b[0];` `        ``// Finding maximum and``        ``// minimum value in``        ``// second_array``        ``for` `(``int` `i = 0; i < n; i++)``        ``{``            ``if` `(b[i] > x)``                ``x = b[i];` `            ``if` `(b[i] < y)``                ``y = b[i];``        ``}` `        ``// Storing the difference``        ``// between maximum and``        ``// minimum value in second_array``        ``int` `ans2 = (x - y);` `        ``return` `Math.Max(ans1, ans2);``    ``}` `    ``// Driver Code``    ``public` `static` `void` `Main()``    ``{``        ``int``[] arr = { 1, 2, 3, 1 };``        ``int` `n = arr.Length;``        ``Console.WriteLine(findValue(arr, n));``    ``}``}` `// This code is contributed by anuj_67.`

## PHP

 ` ``\$x``)``        ``\$x` `= ``\$a``[``\$i``];` `        ``if` `(``\$a``[``\$i``] < ``\$y``)``            ``\$y` `= ``\$a``[``\$i``];``    ``}` `    ``// Storing the difference``    ``// between maximum and``    ``// minimum value in first_array``    ``\$ans1` `= (``\$x` `- ``\$y``);` `    ``\$x` `= ``\$b``[0];``    ``\$y` `= ``\$b``[0];` `    ``// Finding maximum and``    ``// minimum value in``    ``// second_array``    ``for` `(``\$i` `= 0; ``\$i` `< ``\$n``; ``\$i``++)``    ``{``        ``if` `(``\$b``[``\$i``] > ``\$x``)``            ``\$x` `= ``\$b``[``\$i``];` `        ``if` `(``\$b``[``\$i``] < ``\$y``)``            ``\$y` `= ``\$b``[``\$i``];``    ``}` `    ``// Storing the difference``    ``// between maximum and``    ``// minimum value in``    ``// second_array``    ``\$ans2` `= (``\$x` `-``\$y``);` `    ``return` `max(``\$ans1``, ``\$ans2``);``}` `    ``// Driver Code``    ``\$arr` `= ``array``(1, 2, 3, 1);``    ``\$n` `= ``count``(``\$arr``);` `    ``echo` `findValue(``\$arr``, ``\$n``);``    ` `// This code is contributed by anuj_67.``?>`

## Javascript

 ``
Output
`4`

Method – 3

```This solution is space optimization on above mentioned (method2) solution.
In Method 2 solution we had used two matrix of size n which laid to O(n) space complexity
but here we only use O(1) space instead of that two n size array```

## C++

 `// Optimized CPP program to find maximum value of``// |arr[i] - arr[j]| + |i - j|``#include ``using` `namespace` `std;` `// Return maximum |arr[i] - arr[j]| + |i - j|``int` `findValue(``int` `arr[], ``int` `n)``{``    ``int` `temp1, temp2;``    ``int` `max1 = INT_MIN, max2 = INT_MIN;``    ``int` `min1 = INT_MAX, min2 = INT_MAX;` `    ``// Calculating max1 , min1 and max2, min2``    ``for` `(``int` `i = 0; i < n; i++) {``        ``temp1 = arr[i] + i;``        ``temp2 = arr[i] - i;``        ``max1 = max(max1, temp1);``        ``min1 = min(min1, temp1);``        ``max2 = max(max2, temp2);``        ``min2 = min(min2, temp2);``    ``}` `    ``// required maximum ans is max of (max1-min1) and``    ``// (max2-min2)``    ``return` `max((max1 - min1), (max2 - min2));``}` `// Driven Code``int` `main()``{``    ``int` `arr[] = { 1, 2, 3, 1 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);` `    ``cout << findValue(arr, n) << endl;` `    ``return` `0;``}` `// code by AJAY MAKVANA`
Output
`4`

Time Complexity : O(N)

Auxiliary Space: O(1)

This article is contributed by Anuj Chauhan. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.