给定一个由N 个正整数组成的数组arr[] 。任务是在所有i和j对中找到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对,因此arr[i]应该是数组的第二个最大值,而arr[j]是最大元素,以最大化所需的值。因此,数组上的第二个最大值将是我们的答案。如果不存在任何第二大数,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
?>
Javascript
输出:
5