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
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++ implementation of the approach #include <bits/stdc++.h> 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 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 |
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# 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 // PHP implementation of the approach // Function that returns the second largest // element in the array if exists, else 0 function 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 = $second = -(PHP_INT_MAX - 1);
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 == -(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 ?> |
<script> // JavaScript implementation of the approach // Function that returns the second largest
// element in the array if exists, else 0 function getMaxValue(arr, arr_size)
{
let 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 = Number.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 == Number.MIN_VALUE)
{
return 0;
}
else
{
return second;
}
}
// Driver Code let arr = [4, 5, 1, 8];
let n = arr.length;
document.write(getMaxValue(arr, n));
</script> |
5
Time Complexity: O(n) where n is the size of the array
Auxiliary Space: O(1)