# Find the maximum possible value of a[i] % a[j] over all pairs of i and j

Given an array arr[] of N positive integers. The task is to find the maximum possible value of a[i] % a[j] over all pairs of i and j.

Examples:

Input: arr[] = {4, 5, 1, 8}
Output: 5
If we choose the pair (5, 8), then 5 % 8 gives us 5
which is the maximum possible.

Input: arr[] = {7, 7, 8, 8, 1}
Output: 7

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

Approach: Since we can choose any pair, hence arr[i] should be the second maximum of the array and arr[j] be the maximum element in order to maximize the required value. Hence the second maximum over the array will be our answer. If there does not exist any second largest number, then 0 will be the answer.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function that returns the second largest ` `// element in the array if exists, else 0 ` `int` `getMaxValue(``int` `arr[], ``int` `arr_size) ` `{ ` `    ``int` `i, first, second; ` ` `  `    ``// There must be at least two elements ` `    ``if` `(arr_size < 2) { ` `        ``return` `0; ` `    ``} ` ` `  `    ``// To store the maximum and the second ` `    ``// maximum element from the array ` `    ``first = second = INT_MIN; ` `    ``for` `(i = 0; i < arr_size; i++) { ` ` `  `        ``// If current element is greater than first ` `        ``// then update both first and second ` `        ``if` `(arr[i] > first) { ` `            ``second = first; ` `            ``first = arr[i]; ` `        ``} ` ` `  `        ``// If arr[i] is in between first and ` `        ``// second then update second ` `        ``else` `if` `(arr[i] > second && arr[i] != first) ` `            ``second = arr[i]; ` `    ``} ` ` `  `    ``// No second maximum found ` `    ``if` `(second == INT_MIN) ` `        ``return` `0; ` `    ``else` `        ``return` `second; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 4, 5, 1, 8 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` `    ``cout << getMaxValue(arr, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `class` `GFG ` `{ ` ` `  `    ``// Function that returns the second largest ` `    ``// element in the array if exists, else 0 ` `    ``static` `int` `getMaxValue(``int` `arr[], ``int` `arr_size)  ` `    ``{ ` `        ``int` `i, first, second; ` ` `  `        ``// There must be at least two elements ` `        ``if` `(arr_size < ``2``) ` `        ``{ ` `            ``return` `0``; ` `        ``} ` ` `  `        ``// To store the maximum and the second ` `        ``// maximum element from the array ` `        ``first = second = Integer.MIN_VALUE; ` `        ``for` `(i = ``0``; i < arr_size; i++) ` `        ``{ ` ` `  `            ``// If current element is greater than first ` `            ``// then update both first and second ` `            ``if` `(arr[i] > first) ` `            ``{ ` `                ``second = first; ` `                ``first = arr[i]; ` `            ``}  ` `             `  `            ``// If arr[i] is in between first and ` `            ``// second then update second ` `            ``else` `if` `(arr[i] > second && arr[i] != first)  ` `            ``{ ` `                ``second = arr[i]; ` `            ``} ` `        ``} ` ` `  `        ``// No second maximum found ` `        ``if` `(second == Integer.MIN_VALUE) ` `        ``{ ` `            ``return` `0``; ` `        ``}  ` `        ``else` `        ``{ ` `            ``return` `second; ` `        ``} ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `arr[] = {``4``, ``5``, ``1``, ``8``}; ` `        ``int` `n = arr.length; ` `        ``System.out.println(getMaxValue(arr, n)); ` `    ``} ` `}  ` ` `  `// This code has been contributed by 29AjayKumar `

## Python3

 `import` `sys ` `# Python 3 implementation of the approach ` ` `  `# Function that returns the second largest ` `# element in the array if exists, else 0 ` `def` `getMaxValue(arr,arr_size): ` `     `  `    ``# There must be at least two elements ` `    ``if` `(arr_size < ``2``): ` `        ``return` `0` ` `  `    ``# To store the maximum and the second ` `    ``# maximum element from the array ` `    ``first ``=` `-``sys.maxsize``-``1` `    ``second ``=` `-``sys.maxsize``-``1` `    ``for` `i ``in` `range``(arr_size): ` `         `  `        ``# If current element is greater than first ` `        ``# then update both first and second ` `        ``if` `(arr[i] > first): ` `            ``second ``=` `first ` `            ``first ``=` `arr[i] ` ` `  `        ``# If arr[i] is in between first and ` `        ``# second then update second ` `        ``elif` `(arr[i] > second ``and` `arr[i] !``=` `first): ` `            ``second ``=` `arr[i] ` ` `  `    ``# No second maximum found ` `    ``if` `(second ``=``=` `-``sys.maxsize``-``1``): ` `        ``return` `0` `    ``else``: ` `        ``return` `second ` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``arr ``=` `[``4``, ``5``, ``1``, ``8``] ` `    ``n ``=` `len``(arr) ` `    ``print``(getMaxValue(arr, n)) ` ` `  `# This code is contributed by ` `# Surendra_Gangwar `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` ` `  `    ``// Function that returns the second largest ` `    ``// element in the array if exists, else 0 ` `    ``static` `int` `getMaxValue(``int` `[]arr,  ` `                           ``int` `arr_size)  ` `    ``{ ` `        ``int` `i, first, second; ` ` `  `        ``// There must be at least two elements ` `        ``if` `(arr_size < 2) ` `        ``{ ` `            ``return` `0; ` `        ``} ` ` `  `        ``// To store the maximum and the second ` `        ``// maximum element from the array ` `        ``first = second = ``int``.MinValue; ` `        ``for` `(i = 0; i < arr_size; i++) ` `        ``{ ` ` `  `            ``// If current element is greater than first ` `            ``// then update both first and second ` `            ``if` `(arr[i] > first) ` `            ``{ ` `                ``second = first; ` `                ``first = arr[i]; ` `            ``}  ` `             `  `            ``// If arr[i] is in between first and ` `            ``// second then update second ` `            ``else` `if` `(arr[i] > second &&  ` `                     ``arr[i] != first)  ` `            ``{ ` `                ``second = arr[i]; ` `            ``} ` `        ``} ` ` `  `        ``// No second maximum found ` `        ``if` `(second == ``int``.MinValue) ` `        ``{ ` `            ``return` `0; ` `        ``}  ` `        ``else` `        ``{ ` `            ``return` `second; ` `        ``} ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `[]arr = {4, 5, 1, 8}; ` `        ``int` `n = arr.Length; ` `        ``Console.Write(getMaxValue(arr, n)); ` `    ``} ` `}  ` ` `  `// This code is contributed by Akanksha Rai `

## PHP

 ` ``\$first``)  ` `        ``{  ` `            ``\$second` `= ``\$first``;  ` `            ``\$first` `= ``\$arr``[``\$i``];  ` `        ``}  ` ` `  `        ``// If arr[i] is in between first and  ` `        ``// second then update second  ` `        ``else` `if` `(``\$arr``[``\$i``] > ``\$second` `&&  ` `                 ``\$arr``[``\$i``] != ``\$first``)  ` `            ``\$second` `= ``\$arr``[``\$i``];  ` `    ``}  ` ` `  `    ``// No second maximum found  ` `    ``if` `(``\$second` `== -(PHP_INT_MAX-1))  ` `        ``return` `0;  ` `    ``else` `        ``return` `\$second``;  ` `}  ` ` `  `// Driver code  ` `\$arr` `= ``array``(4, 5, 1, 8);  ` `\$n` `= ``count``(``\$arr``); ` ` `  `echo` `getMaxValue(``\$arr``, ``\$n``);  ` ` `  `// This code is contributed by Ryuga ` `?> `

Output:

```5
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