@Arpit
i think finding max & second max does n't reuire to mach time as u
have told n + log(n) -2 step
check out my solution if i am wrong plz xplain me n + log(n) -2 step
required to solve this problem
public class array
{
public void getMax( double ar[] )
{
double max1 = ar[0]; // Assume the first
double max2 = ar[0]; // element in the array
int ZERO = 0; // Variable to store inside it the index of the
max
value to set it to zero.
for( int i = 0; i < ar.length; i++ )
{
if( ar[i] >= max1)
{
max1 = ar[i];
ZERO = i;
}
}
ar[ZERO] = 0; // Set the index contains the 1st max to ZERO.
//remove elemmn from reset length of array i..e. shrink
array
for( int j = 0; j < ar.length; j++ )
{
if( ar[j] >= max2 )
{
max2 = ar[j];
ZERO = j;
}
}
System.out.println("The 1st maximum element in the array is: " +
max1 + ", the 2nd is: " + max2);
}
public static void main(String[] args)
{
// Creating an object from the class Array to be able to use its
methods.
array array = new array();
// Creating an array of type double.
double a[] = {2.2, 3.4, 5.5, 5.5, 6.6, 5.6};
array.getMax( a ); // Calling the method that'll find the 1st
max,
2nd max, and 3rd max.
}
}
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