求出所有i和j对中a [i]％a [j]的最大可能值

给定N个正整数的数组arr [] 。任务是在所有i和j对上找到a [i]％a [j]的最大可能值。

例子：

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

方法：由于我们可以选择任何一对，因此arr [i]应该是数组的第二个最大值，而arr [j]应该是最大元素，以便最大化所需的值。因此，数组上的第二个最大值将是我们的答案。如果不存在第二大数字，则答案为0 。

下面是上述方法的实现：

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
?>

输出：

5