# Maximum absolute difference of value and index sums

Given an unsorted array A of N integers, Return maximum value of f(i, j) for all 1 i, j N.
f(i, j) or absolute difference of two elements of an array A is defined as |A[i] – A[j]| + |i – j|, where |A| denotes
the absolute value of A.

Examples:

```We will calculate the value of f(i, j) for each pair
of (i, j) and return the maximum value thus obtained.

Input : A = {1, 3, -1}
Output : 5
f(1, 1) = f(2, 2) = f(3, 3) = 0
f(1, 2) = f(2, 1) = |1 - 3| + |1 - 2| = 3
f(1, 3) = f(3, 1) = |1 - (-1)| + |1 - 3| = 4
f(2, 3) = f(3, 2) = |3 - (-1)| + |2 - 3| = 5
So, we return 5.

Input : A = {3, -2, 5, -4}
Output : 10
f(1, 1) = f(2, 2) = f(3, 3) = f(4, 4) = 0
f(1, 2) = f(2, 1) = |3 - (-2)| + |1 - 2| = 6
f(1, 3) = f(3, 1) = |3 - 5| + |1 - 3| = 4
f(1, 4) = f(4, 1) = |3 - (-4)| + |1 - 4| = 10
f(2, 3) = f(3, 2) = |(-2) - 5| + |2 - 3| = 8
f(2, 4) = f(4, 2) = |(-2) - (-4)| + |2 - 4| = 4
f(3, 4) = f(4, 3) = |5 - (-4)| + |3 - 4| = 10

So, we return 10
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

A naive brute force approach is to calculate the value f(i, j) by iterating over all such pairs (i, j) and calculating the maximum absolute difference which is implemented below.

## C++

 `// Brute force C++ program to calculate the ` `// maximum absolute difference of an array. ` `#include ` `using` `namespace` `std; ` ` `  `int` `calculateDiff(``int` `i, ``int` `j, ``int` `arr[]) ` `{ ` `    ``// Utility function to calculate ` `    ``// the value of absolute difference ` `    ``// for the pair (i, j). ` `    ``return` `abs``(arr[i] - arr[j]) + ``abs``(i - j); ` `} ` ` `  `// Function to return maximum absolute ` `// difference in brute force. ` `int` `maxDistance(``int` `arr[], ``int` `n) ` `{ ` `    ``// Variable for storing the maximum ` `    ``// absolute distance throughout the ` `    ``// traversal of loops. ` `    ``int` `result = 0; ` ` `  `    ``// Iterate through all pairs. ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``for` `(``int` `j = i; j < n; j++) { ` ` `  `            ``// If the absolute difference of ` `            ``// current pair (i, j) is greater ` `            ``// than the maximum difference ` `            ``// calculated till now, update ` `            ``// the value of result. ` `            ``if` `(calculateDiff(i, j, arr) > result) ` `                ``result = calculateDiff(i, j, arr); ` `        ``} ` `    ``} ` `    ``return` `result; ` `} ` ` `  `// Driver program to test the above function. ` `int` `main() ` `{ ` `    ``int` `arr[] = { -70, -64, -6, -56, 64, ` `                  ``61, -57, 16, 48, -98 }; ` ` `  `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``cout << maxDistance(arr, n) << endl; ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to calculate the maximum ` `// absolute difference of an array. ` `public` `class` `MaximumAbsoluteDifference ` `{ ` `    ``private` `static` `int` `calculateDiff(``int` `i, ``int` `j,  ` `                                     ``int``[] array) ` `    ``{ ` `        ``// Utility function to calculate ` `        ``// the value of absolute difference ` `        ``// for the pair (i, j). ` `        ``return` `Math.abs(array[i] - array[j]) +  ` `                            ``Math.abs(i - j); ` `    ``} ` ` `  `    ``// Function to return maximum absolute ` `    ``// difference in brute force. ` `    ``private` `static` `int` `maxDistance(``int``[] array) ` `    ``{ ` `        ``// Variable for storing the maximum ` `        ``// absolute distance throughout the ` `        ``// traversal of loops. ` `        ``int` `result = ``0``; ` ` `  `        ``// Iterate through all pairs. ` `        ``for` `(``int` `i = ``0``; i < array.length; i++)  ` `        ``{ ` `            ``for` `(``int` `j = i; j < array.length; j++) ` `            ``{ ` ` `  `                ``// If the absolute difference of ` `                ``// current pair (i, j) is greater ` `                ``// than the maximum difference ` `                ``// calculated till now, update ` `                ``// the value of result. ` `                ``result = Math.max(result, calculateDiff(i, j, array)); ` `            ``} ` `        ``} ` `        ``return` `result; ` `    ``} ` ` `  `    ``// Driver program to test above function ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int``[] array = { -``70``, -``64``, -``6``, -``56``, ``64``, ` `                        ``61``, -``57``, ``16``, ``48``, -``98` `}; ` `        ``System.out.println(maxDistance(array)); ` `    ``} ` `} ` ` `  `// This code is contributed by Harikrishnan Rajan `

## Python3

 `# Brute force Python 3 program ` `# to calculate the maximum  ` `# absolute difference of an array. ` ` `  `def` `calculateDiff(i, j, arr): ` ` `  `    ``# Utility function to calculate ` `    ``# the value of absolute difference ` `    ``# for the pair (i, j). ` `    ``return` `abs``(arr[i] ``-` `arr[j]) ``+` `abs``(i ``-` `j) ` ` `  `# Function to return maximum  ` `# absolute difference in  ` `# brute force. ` `def` `maxDistance(arr, n): ` `     `  `    ``# Variable for storing the ` `    ``# maximum absolute distance ` `    ``# throughout the traversal ` `    ``# of loops. ` `    ``result ``=` `0` ` `  `    ``# Iterate through all pairs. ` `    ``for` `i ``in` `range``(``0``,n): ` `        ``for` `j ``in` `range``(i, n): ` ` `  `            ``# If the absolute difference of ` `            ``# current pair (i, j) is greater ` `            ``# than the maximum difference ` `            ``# calculated till now, update ` `            ``# the value of result. ` `            ``if` `(calculateDiff(i, j, arr) > result): ` `                ``result ``=` `calculateDiff(i, j, arr) ` `         `  `    ``return` `result ` ` `  `# Driver program  ` `arr ``=` `[ ``-``70``, ``-``64``, ``-``6``, ``-``56``, ``64``, ` `         ``61``, ``-``57``, ``16``, ``48``, ``-``98` `] ` `n ``=` `len``(arr) ` ` `  `print``(maxDistance(arr, n)) ` ` `  `# This code is contributed by Smitha Dinesh Semwal `

## C#

 `// C# program to calculate the maximum ` `// absolute difference of an array. ` `using` `System; ` ` `  `public` `class` `MaximumAbsoluteDifference ` `{ ` `    ``private` `static` `int` `calculateDiff(``int` `i, ``int` `j,  ` `                                    ``int``[] array) ` `    ``{ ` `        ``// Utility function to calculate ` `        ``// the value of absolute difference ` `        ``// for the pair (i, j). ` `        ``return` `Math.Abs(array[i] - array[j]) +  ` `                            ``Math.Abs(i - j); ` `    ``} ` ` `  `    ``// Function to return maximum absolute ` `    ``// difference in brute force. ` `    ``private` `static` `int` `maxDistance(``int``[] array) ` `    ``{ ` `        ``// Variable for storing the maximum ` `        ``// absolute distance throughout the ` `        ``// traversal of loops. ` `        ``int` `result = 0; ` ` `  `        ``// Iterate through all pairs. ` `        ``for` `(``int` `i = 0; i < array.Length; i++)  ` `        ``{ ` `            ``for` `(``int` `j = i; j < array.Length; j++) ` `            ``{ ` ` `  `                ``// If the absolute difference of ` `                ``// current pair (i, j) is greater ` `                ``// than the maximum difference ` `                ``// calculated till now, update ` `                ``// the value of result. ` `                ``result = Math.Max(result, calculateDiff(i, j, array)); ` `            ``} ` `        ``} ` `        ``return` `result; ` `    ``} ` ` `  `    ``// Driver program ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int``[] array = { -70, -64, -6, -56, 64, ` `                        ``61, -57, 16, 48, -98 }; ` `        ``Console.WriteLine(maxDistance(array)); ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m `

## PHP

 ` ``\$result``) ` `                ``\$result` `= calculateDiff(``\$i``, ``\$j``, ``\$arr``); ` `        ``} ` `    ``} ` `    ``return` `\$result``; ` `} ` ` `  `// Driver Code ` `\$arr` `= ``array``( -70, -64, -6, -56, 64, ` `               ``61, -57, 16, 48, -98 ); ` ` `  `\$n` `= sizeof(``\$arr``); ` ` `  `echo` `maxDistance(``\$arr``, ``\$n``); ` ` `  `// This Code is contributed by mits  ` `?> `

Output:

```167
```

Time complexity: O(n^2)

An efficient solution in O(n) time complexity can be worked out using the properties of absolute values.
f(i, j) = |A[i] – A[j]| + |i – j| can be written in 4 ways (Since we are looking at max value, we don’t even care if the value becomes negative as long as we are also covering the max value in some way).

```Case 1: A[i] > A[j] and i > j
|A[i] - A[j]| = A[i] - A[j]
|i -j| = i - j
hence, f(i, j) = (A[i] + i) - (A[j] + j)

Case 2: A[i] < A[j] and i < j
|A[i] - A[j]| = -(A[i]) + A[j]
|i -j| = -(i) + j
hence, f(i, j) = -(A[i] + i) + (A[j] + j)

Case 3: A[i] > A[j] and i < j
|A[i] - A[j]| = A[i] - A[j]
|i -j| = -(i) + j
hence, f(i, j) = (A[i] - i) - (A[j] - j)

Case 4: A[i] < A[j] and i > j
|A[i] - A[j]| = -(A[i]) + A[j]
|i -j| = i - j
hence, f(i, j) = -(A[i] - i) + (A[j] - j)
```

Note that case 1 and 2 are equivalent and so are case 3 and 4 and hence we can design our algorithm only for two cases as it will cover all the possible cases.

1. Calculate the value of A[i] + i and A[i] – i for every element of the array while traversing through the array.

2. Then for the two equivalent cases, we find the maximum possible value. For that, we have to store minimum and maximum values of expressions A[i] + i and A[i] – i for all i.

3. Hence the required maximum absolute difference is maximum of two values i.e. max((A[i] + i) – (A[j] + j)) and max((A[i] – i) – (A[j] – j)). These values can be found easily in linear time.
a. For max((A[i] + i) – (A[j] + j)) Maintain two variables max1 and min1 which will store maximum and minimum values of A[i] + i respectively. max((A[i] + i) – (A[j] + j)) = max1 – min1
b. For max((A[i] – i) – (A[j] – j)). Maintain two variables max2 and min2 which will store maximum and minimum values of A[i] – i respectively. max((A[i] – i) – (A[j] – j)) = max2 – min2

Implementation using the above fast algorithm is given below.

## C++

 `// C++ program to calculate the maximum ` `// absolute difference of an array. ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return maximum absolue ` `// difference in linear time. ` `int` `maxDistance(``int` `arr[], ``int` `n) ` `{ ` `    ``// max and min variables as described ` `    ``// in algorithm. ` `    ``int` `max1 = INT_MIN, min1 = INT_MAX; ` `    ``int` `max2 = INT_MIN, min2 = INT_MAX; ` ` `  `    ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `        ``// Updating max and min variables ` `        ``// as described in algorithm. ` `        ``max1 = max(max1, arr[i] + i); ` `        ``min1 = min(min1, arr[i] + i); ` `        ``max2 = max(max2, arr[i] - i); ` `        ``min2 = min(min2, arr[i] - i); ` `    ``} ` ` `  `    ``// Calculating maximum absolute difference. ` `    ``return` `max(max1 - min1, max2 - min2); ` `} ` ` `  `// Driver program to test the above function. ` `int` `main() ` `{ ` `    ``int` `arr[] = { -70, -64, -6, -56, 64, ` `                  ``61, -57, 16, 48, -98 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` `    ``cout << maxDistance(arr, n) << endl; ` `    ``return` `0; ` `} `

## Java

 `// Java program to calculate the maximum ` `// absolute difference of an array. ` `public` `class` `MaximumAbsoluteDifference ` `{ ` `    ``// Function to return maximum absolue ` `    ``// difference in linear time. ` `    ``private` `static` `int` `maxDistance(``int``[] array) ` `    ``{ ` `        ``// max and min variables as described ` `        ``// in algorithm. ` `        ``int` `max1 = Integer.MIN_VALUE; ` `        ``int` `min1 = Integer.MAX_VALUE; ` `        ``int` `max2 = Integer.MIN_VALUE; ` `        ``int` `min2 = Integer.MAX_VALUE; ` ` `  `        ``for` `(``int` `i = ``0``; i < array.length; i++) ` `        ``{ ` ` `  `            ``// Updating max and min variables ` `            ``// as described in algorithm. ` `            ``max1 = Math.max(max1, array[i] + i); ` `            ``min1 = Math.min(min1, array[i] + i); ` `            ``max2 = Math.max(max2, array[i] - i); ` `            ``min2 = Math.min(min2, array[i] - i); ` `        ``} ` ` `  `        ``// Calculating maximum absolute difference. ` `        ``return` `Math.max(max1 - min1, max2 - min2); ` `    ``} ` ` `  `    ``// Driver program to test above function ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int``[] array = { -``70``, -``64``, -``6``, -``56``, ``64``, ` `                         ``61``, -``57``, ``16``, ``48``, -``98` `}; ` `        ``System.out.println(maxDistance(array)); ` `    ``} ` `} ` ` `  `// This code is contributed by Harikrishnan Rajan `

## Python3

 `# Python program to ` `# calculate the maximum ` `# absolute difference ` `# of an array. ` ` `  `# Function to return ` `# maximum absolue ` `# difference in linear time. ` `def` `maxDistance(array): ` `     `  `    ``# max and min variables as described ` `    ``# in algorithm. ` `    ``max1 ``=` `-``2147483648` `    ``min1 ``=` `+``2147483647` `    ``max2 ``=` `-``2147483648` `    ``min2 ``=` `+``2147483647` `  `  `    ``for` `i ``in` `range``(``len``(array)): ` ` `  `  `  `        ``# Updating max and min variables ` `        ``# as described in algorithm. ` `        ``max1 ``=` `max``(max1, array[i] ``+` `i) ` `        ``min1 ``=` `min``(min1, array[i] ``+` `i) ` `        ``max2 ``=` `max``(max2, array[i] ``-` `i) ` `        ``min2 ``=` `min``(min2, array[i] ``-` `i) ` `     `  `  `  `    ``# Calculating maximum absolute difference. ` `    ``return` `max``(max1 ``-` `min1, max2 ``-` `min2) ` ` `  `  `  `# Driver program to ` `# test above function ` ` `  `array ``=` `[ ``-``70``, ``-``64``, ``-``6``, ``-``56``, ``64``, ` `           ``61``, ``-``57``, ``16``, ``48``, ``-``98` `] ` ` `  `print``(maxDistance(array)) ` ` `  `# This code is contributed ` `# by Anant Agarwal. `

## C#

 `// C# program to calculate the maximum ` `// absolute difference of an array. ` `using` `System; ` ` `  `public` `class` `MaximumAbsoluteDifference ` `{ ` `    ``// Function to return maximum absolue ` `    ``// difference in linear time. ` `    ``private` `static` `int` `maxDistance(``int``[] array) ` `    ``{ ` `        ``// max and min variables as described ` `        ``// in algorithm. ` `        ``int` `max1 = ``int``.MinValue ; ` `        ``int` `min1 = ``int``.MaxValue ; ` `        ``int` `max2 = ``int``.MinValue ; ` `        ``int` `min2 =``int``.MaxValue ; ` ` `  `        ``for` `(``int` `i = 0; i < array.Length; i++) ` `        ``{ ` ` `  `            ``// Updating max and min variables ` `            ``// as described in algorithm. ` `            ``max1 = Math.Max(max1, array[i] + i); ` `            ``min1 = Math.Min(min1, array[i] + i); ` `            ``max2 = Math.Max(max2, array[i] - i); ` `            ``min2 = Math.Min(min2, array[i] - i); ` `        ``} ` ` `  `        ``// Calculating maximum absolute difference. ` `        ``return` `Math.Max(max1 - min1, max2 - min2); ` `    ``} ` ` `  `    ``// Driver program  ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int``[] array = { -70, -64, -6, -56, 64, ` `                        ``61, -57, 16, 48, -98 }; ` `        ``Console.WriteLine(maxDistance(array)); ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m `

## PHP

 ` `

Output:

```167
```

Time Complexity: O(n)

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