# Maximize value of (arr[i] – i) – (arr[j] – j) in an array

Given an array, arr[] find the maximum value of (arr[i] – i) – (arr[j] – j) where i is not equal to j. Where i and j vary from 0 to n-1 and n is size of input array arr[].

Examples:

```Input : arr[] = {9, 15, 4, 12, 13}
Output : 12
We get the maximum value for i = 1 and j = 2
(15 - 1) - (4 - 2) = 12

Input : arr[] = {-1, -2, -3, 4, 10}
Output : 6
We get the maximum value for i = 4 and j = 2
(10 - 4) - (-3 - 2) = 11
```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

One important observation is, value of (arr[i] – i) – (arr[j] – j) can never be negative. We can alway swap i and j to convert a negative value into positive. So the condition i not equal to j is bogus and doesn’t require explicit check.

Method 1 (Naive : O(n2))

The idea is to run two loops to consider all possible pairs and keep track of maximum value of expression (arr[i]-i)-(arr[j]-j). Below is the implementation of this idea.

## C++

 `// C++ program to find maximum value (arr[i]-i) ` `// - (arr[j]-j) in an array. ` `#include ` `using` `namespace` `std; ` ` `  `// Returns maximum value of (arr[i]-i) - (arr[j]-j) ` `int` `findMaxDiff(``int` `arr[], ``int` `n) ` `{ ` `    ``if` `(n < 2) ` `    ``{ ` `        ``cout << ``"Invalid "``; ` `        ``return` `0; ` `    ``} ` ` `  `    ``// Use two nested loops to find the result ` `    ``int` `res = INT_MIN; ` `    ``for` `(``int` `i=0; i

## Java

 `// Java program to find maximum value (arr[i]-i) ` `// - (arr[j]-j) in an array. ` `import` `java.util.*; ` ` `  `class` `GFG { ` `     `  `// Returns maximum value of ` `// (arr[i]-i) - (arr[j]-j) ` `static` `int` `findMaxDiff(``int` `arr[], ``int` `n)  ` `{ ` `    ``if` `(n < ``2``) { ` `    ``System.out.print(``"Invalid "``); ` `    ``return` `0``; ` `    ``} ` ` `  `    ``// Use two nested loops to find the result ` `    ``int` `res = Integer.MIN_VALUE; ` `    ``for` `(``int` `i = ``0``; i < n; i++) ` `    ``for` `(``int` `j = ``0``; j < n; j++) ` `        ``if` `(res < (arr[i] - arr[j] - i + j)) ` `        ``res = (arr[i] - arr[j] - i + j); ` ` `  `    ``return` `res; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args)  ` `{ ` `    ``int` `arr[] = {``9``, ``15``, ``4``, ``12``, ``13``}; ` `    ``int` `n = arr.length; ` `    ``System.out.print(findMaxDiff(arr, n)); ` `} ` `} ` ` `  `// This code is contributed by Anant Agarwal. `

## Python3

 `# Python program to find ` `# maximum value (arr[i]-i) ` `# - (arr[j]-j) in an array. ` ` `  `# Returns maximum value of ` `# (arr[i]-i) - (arr[j]-j) ` `def` `findMaxDiff(arr,n): ` ` `  `    ``if` `(n < ``2``): ` `     `  `        ``print``(``"Invalid "``) ` `        ``return` `0` `     `  `  `  `    ``# Use two nested loops ` `    ``# to find the result ` `    ``res ``=` `-``2147483648` `    ``for` `i ``in` `range``(n): ` `        ``for` `j ``in` `range``(n): ` `            ``if` `( res < (arr[i]``-``arr[j]``-``i``+``j) ): ` `                ``res ``=` `(arr[i]``-``arr[j]``-``i``+``j) ` `  `  `    ``return` `res ` ` `  `# Driver code ` ` `  `arr``=` `[``9``, ``15``, ``4``, ``12``, ``13``] ` `n ``=` `len``(arr) ` ` `  `print``(findMaxDiff(arr, n)) ` ` `  `# This code is contributed ` `# by Anant Agarwal. `

## C#

 `// C# program to find maximum  ` `// value (arr[i]-i)- (arr[j]-j)  ` `// in an array. ` `using` `System; ` `class` `GFG { ` `     `  `// Returns maximum value of ` `// (arr[i]-i) - (arr[j]-j) ` `static` `int` `findMaxDiff(``int` `[]arr, ``int` `n)  ` `{ ` `    ``if` `(n < 2) { ` `    ``Console.WriteLine(``"Invalid "``); ` `    ``return` `0; ` `    ``} ` ` `  `    ``// Use two nested loops to ` `    ``// find the result ` `    ``int` `res = ``int``.MinValue; ` `    ``for` `(``int` `i = 0; i < n; i++) ` `    ``for` `(``int` `j = 0; j < n; j++) ` `        ``if` `(res < (arr[i] - arr[j] - i + j)) ` `            ``res = (arr[i] - arr[j] - i + j); ` ` `  `    ``return` `res; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main()  ` `{ ` `    ``int` `[]arr = {9, 15, 4, 12, 13}; ` `    ``int` `n = arr.Length; ` `    ``Console.WriteLine(findMaxDiff(arr, n)); ` `} ` `} ` ` `  `// This code is contributed by anjuj_67. `

## PHP

 ` `

Output:

```12
```

Method 2 (Tricky : O(n))
1) Find maximum value of arr[i] – i in whole array.
2) Find minimum value of arr[i] – i in whole array.
3) Return difference of above two values.

## C++

 `// C++ program to find maximum value (arr[i]-i) ` `// - (arr[j]-j) in an array. ` `#include ` `using` `namespace` `std; ` ` `  `// Returns maximum value of (arr[i]-i) - (arr[j]-j) ` `int` `findMaxDiff(``int` `a[], ``int` `n) ` `{ ` `    ``if` `(n < 2) ` `    ``{ ` `        ``cout << ``"Invalid "``; ` `        ``return` `0; ` `    ``} ` ` `  `    ``// Find maximum of value of arr[i] - i for all ` `    ``// possible values of i. Let this value be max_val. ` `    ``// Find minimum of value of arr[i] - i for all ` `    ``// possible values of i. Let this value be min_val. ` `    ``// The difference max_val - min_v ` `    ``int` `min_val = INT_MAX, max_val =  ` `                           ``INT_MIN; ` `    ``for` `(``int` `i=0; i max_val) ` `            ``max_val = a[i] - i; ` `        ``if` `((a[i]-i) < min_val) ` `            ``min_val = a[i]-i; ` `    ``} ` ` `  `    ``return` `(max_val - min_val); ` `} ` ` `  `// Driver program ` `int` `main() ` `{ ` `    ``int` `arr[] = {9, 15, 4, 12, 13}; ` `    ``int` `n = ``sizeof``(arr)/``sizeof``(arr[0]); ` `    ``cout << findMaxDiff(arr, n); ` `    ``return` `0; ` `} `

## Java

 `// Java program to find maximum value (arr[i]-i) ` `// - (arr[j]-j) in an array. ` `import` `java.io.*; ` ` `  `class` `GFG { ` ` `  `    ``// Returns maximum value of (arr[i]-i) -  ` `    ``// (arr[j]-j) ` `    ``static` `int` `findMaxDiff(``int` `arr[], ``int` `n)     ` `    ``{ ` `        ``if` `(n < ``2``) ` `        ``{ ` `            ``System.out.println(``"Invalid "``); ` `            ``return` `0``; ` `        ``} ` ` `  `        ``// Find maximum of value of arr[i] - i  ` `        ``// for all possible values of i. Let  ` `        ``// this value be max_val. Find minimum ` `        ``// of value of arr[i] - i for all ` `        ``// possible values of i. Let this value ` `        ``// be min_val. The difference max_val - ` `        ``// min_v ` `        ``int` `min_val = Integer.MAX_VALUE, ` `            ``max_val = Integer.MIN_VALUE; ` `         `  `        ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``{ ` `            ``if` `((arr[i]-i) > max_val) ` `                ``max_val = arr[i] - i; ` `                 `  `            ``if` `((arr[i]-i) < min_val) ` `                ``min_val = arr[i]-i; ` `        ``} ` ` `  `        ``return` `(max_val - min_val); ` `    ``} ` ` `  `    ``// Driver program ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `arr[] = {``9``, ``15``, ``4``, ``12``, ``13``}; ` `        ``int` `n = arr.length; ` `        ``System.out.println(findMaxDiff(arr, n)); ` `    ``} ` `} ` `// This code is contributed by Prerna Saini `

## Python3

 `# Python 3 program to find maximum value ` `# (arr[i]-i) - (arr[j]-j) in an array. ` `import` `sys  ` ` `  `# Returns maximum value of  ` `# (arr[i]-i) - (arr[j]-j) ` `def` `findMaxDiff(a, n): ` `    ``if` `(n < ``2``): ` `        ``print``(``"Invalid "``) ` `        ``return` `0` ` `  `    ``# Find maximum of value of arr[i] - i ` `    ``# for all possible values of i. Let  ` `    ``# this value be max_val. Find minimum ` `    ``# of value of arr[i] - i for all possible  ` `    ``# values of i. Let this value be min_val. ` `    ``# The difference max_val - min_v ` `    ``min_val ``=` `sys.maxsize ` `    ``max_val ``=` `-``sys.maxsize ``-` `1` `    ``for` `i ``in` `range``(n): ` `        ``if` `((a[i] ``-` `i) > max_val): ` `            ``max_val ``=` `a[i] ``-` `i ` `        ``if` `((a[i] ``-` `i) < min_val): ` `            ``min_val ``=` `a[i] ``-` `i ` ` `  `    ``return` `(max_val ``-` `min_val) ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``arr ``=` `[``9``, ``15``, ``4``, ``12``, ``13``] ` `    ``n ``=` `len``(arr) ` `    ``print``(findMaxDiff(arr, n)) ` ` `  `# This code is contributed by Rajput-Ji `

## C#

 `// C# program to find maximum value (arr[i]-i) ` `// - (arr[j]-j) in an array. ` `using` `System; ` `  `  `class` `GFG { ` `  `  `    ``// Returns maximum value of (arr[i]-i) -  ` `    ``// (arr[j]-j) ` `    ``static` `int` `findMaxDiff(``int` `[]arr, ``int` `n)     ` `    ``{ ` `        ``if` `(n < 2) ` `        ``{ ` `            ``Console.Write(``"Invalid "``); ` `            ``return` `0; ` `        ``} ` `  `  `        ``// Find maximum of value of arr[i] - i  ` `        ``// for all possible values of i. Let  ` `        ``// this value be max_val. Find minimum ` `        ``// of value of arr[i] - i for all ` `        ``// possible values of i. Let this value ` `        ``// be min_val. The difference max_val - ` `        ``// min_v ` `        ``int` `min_val = ``int``.MaxValue, ` `            ``max_val = ``int``.MinValue; ` `          `  `        ``for` `(``int` `i = 0; i < n; i++) ` `        ``{ ` `            ``if` `((arr[i] - i) > max_val) ` `                ``max_val = arr[i] - i; ` `                  `  `            ``if` `((arr[i] - i) < min_val) ` `                ``min_val = arr[i] - i; ` `        ``} ` `  `  `        ``return` `(max_val - min_val); ` `    ``} ` `  `  `    ``// Driver program ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `[]arr = {9, 15, 4, 12, 13}; ` `        ``int` `n = arr.Length; ` `        ``Console.Write(findMaxDiff(arr, n)); ` `    ``} ` `} ` ` `  `// This code is contributed by nitin mittal. `

## PHP

 ` ``\$max_val``) ` `            ``\$max_val` `= ``\$a``[``\$i``] - ``\$i``; ` `        ``if` `((``\$a``[``\$i``] - ``\$i``) < ``\$min_val``) ` `            ``\$min_val` `= ``\$a``[``\$i``] - ``\$i``; ` `    ``} ` ` `  `    ``return` `(``\$max_val` `- ``\$min_val``); ` `} ` ` `  `// Driver Code ` `\$arr` `= ``array``(9, 15, 4, 12, 13); ` `\$n` `= sizeof(``\$arr``); ` `echo` `findMaxDiff(``\$arr``, ``\$n``); ` ` `  `// This code is contributed by Sachin. ` `?> `

Output:

```12
```

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