write java code to print second max number in the array
Answers were Sorted based on User's Feedback
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);
}
}
| Is This Answer Correct ? | 28 Yes | 8 No |
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));
}
}
| Is This Answer Correct ? | 16 Yes | 5 No |
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.
| Is This Answer Correct ? | 5 Yes | 0 No |
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.
| Is This Answer Correct ? | 4 Yes | 0 No |
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);
}
}
| Is This Answer Correct ? | 7 Yes | 5 No |
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
| Is This Answer Correct ? | 2 Yes | 0 No |
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);
}
}
| Is This Answer Correct ? | 0 Yes | 0 No |
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));
}
}
| Is This Answer Correct ? | 0 Yes | 0 No |
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);
}
}
| Is This Answer Correct ? | 2 Yes | 3 No |
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