Answer / rajaram

public static void main(String[] args){
int maxNumber = 0;
int secondMaxNumber = 0;
if(args.length == 0){
System.err.println("Number array is empty");
return;
}
for(int i=0; i < args.length; i++){
int currNumber = Integer.parseInt(args[i]);
if(maxNumber < currNumber){
secondMaxNumber = maxNumber;
maxNumber = currNumber;
}else if(secondMaxNumber < currNumber){
secondMaxNumber = currNumber;
}
}
System.err.println("Max. number is "+maxNumber);
System.err.println("Second Max. is "+secondMaxNumber);
}
}

Answer / himesh mistry

import java.util.ArrayList;
import java.util.Collections;
import java.util.HashSet;
import java.util.Set;


public class SecondMax {

/**
* @param args
*/
public static void main(String[] args) {
int[] numbers = {9,4,8,0,0,5,9,1,4,2};
Set arrSet = new HashSet();
for (int i=0;i<numbers.length;i++) {
arrSet.add(numbers[i]);
}
ArrayList s = new ArrayList(arrSet);
Collections.sort(s);

System.out.println("Second element : " +
s.get(s.size()-2));
}
}

Answer / geetha

Himesh Mistry -best solution

Answer / sunny

I think, the except for Himesh's program, all other ones
doesnt handle duplicate values ...
I mean if the numbers are :- 9 9 8 8 7 7 7 1 3 2
the answer should be 8 ideally.
But I doubt whether thats really happening in any other
programs, except for Himesh's.
Do let me know, if I am wrong. Thanks.

Answer / rajesh s

This is I tried.
public class SecondLargestNumber {
/**
* @param args
*/
public static void main(String[] args) {
int[] num = new int[]{-101,-105,-2,-7,-22,-
104,-8,-10,-100,-102,-102};
int big=-1;
int secbig=-1;
if(num.length == 1) {
big = num[0];
secbig = num[0];
} else {
if(num[0] > num[1]) {
big = num[0];
secbig = num[1];
} else {
big = num[1];
secbig = num[0];
}
}
if(num.length > 2){
for(int i=2;i<num.length;++i){
if(num[i] > secbig && num
[i] <big) {
secbig = num[i];
}
if(num[i]>=big){
secbig = big;
big = num[i];
}
}
}
System.out.println(big);
System.out.println(secbig);
}
}

Please check is it working and let me know. For removing
duplicates, we will have one more method to remove the
duplicate elements.

Answer / raja ram

public class SecondMaximumNumber{
public static void main(String[] args){
int maxNumber = 0;
int secondMaxNumber = 0;
if(args.length == 0){
System.err.println("Number array is empty");
return;
}
for(int i=0; i < args.length; i++){
int currNumber = Integer.parseInt(args[i]);
if(maxNumber < currNumber){
secondMaxNumber = maxNumber;
maxNumber = currNumber;
}else if(secondMaxNumber < currNumber){
secondMaxNumber = currNumber;
}
}
System.err.println("Max. number is "+maxNumber);
System.err.println("Second Max. number is
"+secondMaxNumber);
}
}

Answer / hitman

hey using collections solving this type of problem becomes
easy but whenever u go for interview most of the time they
ask you to write program without using collection API thts
why i think rajaram's code is correct

Answer / ibrahim

import java.util.ArrayList;
public class maxSec {

public static void main(String[] args) {
int maxFi=0;
int maxSe=0;
ArrayList<Integer> arl=new ArrayList<Integer>();

arl.add(4); arl.add(8); arl.add(2);
arl.add(6); arl.add(7); arl.add(8);

for(int i=0; i<arl.size(); i++){
if (maxFi< arl.get(i))
maxFi= arl.get(i);
}

for(int i=0; i<arl.size(); i++){
if (maxSe <arl.get(i)){
if (arl.get(i)<maxFi ){

maxSe=arl.get(i);
}
}
}
System.out.println(maxFi);
System.out.println(maxSe);

}

}

Answer / laddu

import java.util.Arrays;
import java.util.Collections;
import java.util.*;
public class Main1 {
public static void main(String[] args) {
Integer[] numbers = { 8, 2, 7, 1, 4, 9, 5};
int min = (int) Collections.min(Arrays.asList(numbers));
int max = (int) Collections.max(Arrays.asList(numbers));

ArrayList s = new ArrayList(Arrays.asList(numbers));
Collections.sort(s);

System.out.println("Second element : " + s.get(s.size()-2));
}
}

Answer / sujeev

Ranjaram is correct.I modified and testes it.
public class test123 {
public static void main(String[] args){
int maxNumber = 0;
int secondMaxNumber = 0;

int[] anArray;
anArray =new int [10];
anArray[0] = 100;
anArray[1] = 200;
anArray[2] = 300;
anArray[3] = 400;
anArray[4] = 500;

if(anArray.length == 0){
System.err.println("Number array is empty");
return;
}
for(int i=0; i < anArray.length; i++){
int currNumber = anArray[i];
if(maxNumber < currNumber){
secondMaxNumber = maxNumber;
maxNumber = currNumber;
}else if(secondMaxNumber < currNumber){
secondMaxNumber = currNumber;
}
}
System.err.println("Max. number is "+maxNumber);
System.err.println("Second Max.
is "+secondMaxNumber);
}
}

What is casting?

3 Answers  

why would you use a synchronized block vs. Synchronized method? : Java thread

0 Answers  

What is meant by Static query and Dynamic query?

2 Answers  

What is a class object?

0 Answers  

what is business objects?

2 Answers  

Write a java program to find the route that connects between Red and Green Cells. General Rules for traversal 1. You can traverse from one cell to another vertically, horizontally or diagonally. 2. You cannot traverse through Black cells. 3. There should be only one Red and Green cell and at least one of each should be present. Otherwise the array is invalid. 4. You cannot revisit a cell that you have already traversed. 5. The maze need not be in the same as given in the above example

0 Answers  

Difference between concurrent hashmap and hashtable and collections

0 Answers  

Explain wait() method of object class ?

0 Answers  

Explain abstract class in java?

0 Answers  

What is merge sort in java?

0 Answers  

How to create a base64 decoder in java8?

0 Answers  

Explain the selection sort algorithm?

0 Answers   Hexaware,