# 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[0]);     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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