Find the second largest number in single iteration

Find the second Largest number in single iteration

Program 

int large=0,secod=0;
int i,abc[]={25,60,80,99,9,8,100,5,55,56,59,110,99,111};
for (i=0; i<abc.length-1;i++)
{
if(abc[i] > abc[i+1])
{
if(large < abc[i])
{
large=abc[i];
}
else
{
secod=large;
}
}
else
{
if(large < abc[i+1])
{
large=abc[i+1];
}
else
{
secod=large;
}


}

}
System.out.println(" Large Number is " +large+"Second large=="+secod);

OutPut

 Largr Number is 111Second large==110

Java Program For Printing 0 to 25 Binary Number Like 4=100,5=101 and so on....

Program For Printing 0 to 25 Binary number 

Java Code

                int i=0;

int no=25;

while(i<=no)

{

String BinaryString=Integer.toBinaryString(i);

System.out.println("Number=="+i+"  Binary=="+BinaryString);

i=i+1;

}

Out Put :

Number==0  Binary==0

Number==1  Binary==1

Number==2  Binary==10

Number==3  Binary==11

Number==4  Binary==100

Number==5  Binary==101

Number==6  Binary==110

Number==7  Binary==111

Number==8  Binary==1000

Number==9  Binary==1001

Number==10  Binary==1010

Number==11  Binary==1011

Number==12  Binary==1100

Number==13  Binary==1101

Number==14  Binary==1110

Number==15  Binary==1111

Number==16  Binary==10000

Number==17  Binary==10001

Number==18  Binary==10010

Number==19  Binary==10011

Number==20  Binary==10100

Number==21  Binary==10101

Number==22  Binary==10110

Number==23  Binary==10111

Number==24  Binary==11000

Number==25  Binary==11001