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

Given a array of N positive integers. The task is to find 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
Choose i = 0 and j = 4

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

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

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); ` ` `  `    ``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

 ` `

Output:

```4
```

Method 2 (tricky):
First of all lets make four equations by removing the absolute value signs (“|”). Following 4 equations will be formed, and we need to find the maximum value of these equation 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 equation (1) and (4) are identical, similarly equation (2) and (3) are identical.

Now the task is to find the maximum value of these equation. 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 two values these two arrays.

For that, we find maximum value and minimum value in first_array and store their difference:
ans1 = (maximum value in first_array – minimum value in first_array)
Similarly, we need to find maximum value and minimum value in second_array and store their difference:
ans2 = (maximum value in second_array – minimum value in second_array)
Our answer will be maximum of ans1 and ans2.

Below is the implementation of 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, y = a; ` ` `  `    ``// 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; ` `    ``y = b; ` ` `  `    ``// 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 Program ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, 2, 3, 1 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``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, y = a; ` ` `  `    ``// 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; ` `    ``y = b; ` ` `  `    ``// 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``; ` `    ``\$y` `= ``\$b``; ` ` `  `    ``// 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. ` `?> `

Output:

```4
```

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