Domande poste durante i colloqui per Sviluppatore Java
Per affrontare un colloquio per la posizione di Sviluppatore java, preparati a rispondere a domande tecniche volte a saggiare la tua conoscenza delle basi del linguaggio Java. È probabile che gli intervistatori pongano anche domande a risposta aperta per valutare le tue capacità in ambito di programmazione e risoluzione dei problemi. Preparati a rispondere a domande non tecniche incentrate su gestione del tempo, comunicazione interpersonale ed etica del lavoro.
Domande tipiche dei colloqui per Sviluppatore java e come rispondere
Ecco le 3 domande più frequenti nei colloqui di lavoro per Sviluppatore Java e consigli su come rispondere:
Domanda 1: Qual è la differenza tra HashTable e HashMap in Java?
Come rispondere: Gli intervistatori si servono di domande tecniche per valutare la tua istruzione formale e le tue conoscenze del linguaggio di programmazione Java. Sfrutta la risposta per dimostrare le tue capacità comunicative. Anche se è già da qualche tempo che svolgi lavori di programmazione, può essere utile ripassare le basi di Java prima di partecipare al colloquio.
Domanda 2: Scrivi un programma per stabilire se 243 è un numero di Armstrong.
Come rispondere: Gli intervistatori utilizzano domande a risposta aperta per testare le tue capacità di programmazione. Questa domanda permette al datore di lavoro di valutare la tua abilità nella risoluzione di problemi utilizzando un algoritmo Java. Esponi il tuo ragionamento per risolvere il problema tramite un editor di codice o una lavagna digitale.
Domanda 3: Come ti organizzi per creare codici di qualità quando devi portare a termine diversi progetti Java contemporaneamente?
Come rispondere: Gli Sviluppatori Java spesso si trovano a gestire diversi progetti contemporaneamente. Gli intervistatori si servono di domande come questa per testare le tue capacità organizzative e di gestione del tempo. I datori di lavoro provano anche a vagliare il tuo impegno sul fronte della qualità. Se hai esperienza come sviluppatore, menziona esempi della tua carriera che dimostrino l'elevata qualità del lavoro svolto e l'efficienza nella gestione dei tempi.
Domande principali poste durante i colloqui
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please let me know if anyone got Job
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36. If you get call then follow this interview experience
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How many of you waiting after 3rd round
Coding test: 1. Given a string, find out if there's repeat characters in it. 2. SQL. Given a Customer table and a Payment table (with Customer ID as primary and foreign key), write a query to output a list of customers who have not paid their bills in the last 3 months.
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Did well on the first Java coding test. Solution using Hashset. Not so well on the SQL query. Able to give a query, but a few errors in it I am sure. Also there was a second questions regarding the SQL query, which we didn't have time to get to. So that was another bad thing. I am pretty sure I did not pass. But good learning experience. Meno
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import java.util.*; public class test { public static void main(String[] args) { String str = "abdc"; char[] arr = str.toCharArray(); HashSet set = new HashSet(); for (char i : arr ) { set.add(i); } if((set.size()) == (arr.length)) System.out.println("unique character"); else System.out.println("repetition"); } } Meno
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No
Write a program to print * * * * * * * * * * * * * * * * * * * * * * * * * * * * * with minimum number of variable use
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class StarPrint { public static void main (String[] args) throws java.lang.Exception { int i = 0, j = 0; for( i = 5; i > 0; i-- ){ for( j = 0; j < 5 - i; j++) System.out.print(" "); for( j = 0; j < i; j++ ) System.out.print("* "); System.out.println();} for( i = 3; i <= 6; i++ ){ for( j = 0; j <= 5 - i; j++) System.out.print(" "); for( j = 1; j < i; j++ ) System.out.print("* "); System.out.println(); } } } Meno
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public class Check { public static void main(String[] args) { int limit = 5; // Specify the number of lines as limit and check output int i=0,j=0; for(i=0;ilimit-i;j--)System.out.print(" "); for(j=limit;j>i;j--)System.out.print("* "); System.out.println(); } j=limit; for(;limit>1;limit--){ for(i=2;ilimit-2;i--)System.out.print("* "); System.out.println(); } } } Meno
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In the previous solution, the LESS THAN and GREATER THAN symbols have nit been parsed. public class Check { public static void main(String[] args) { int limit = 9; int i=0,j=0; for(i=0;i limit-i;j--)System.out.print(" "); for(j=limit;j > i;j--)System.out.print("* "); System.out.println(); } j=limit; for(;limit > 1;limit--){ for(i=2;i limit-2;i--)System.out.print("* "); System.out.println(); } } } Meno
Write a program for odd input the pattern output will as below n=1 *-* n=3 ***** *** *---* *** *****
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import java.util.Scanner; public class Pattern{ public static void main(String[] args) { Scanner sc=new Scanner(System.in); int n=sc.nextInt(); for(int i=n;i>=2;i--){ for(int j=1;j<=2*n-(i-1);j++){ System.out.print(" "); } for(int k=1;k<=2*i-1;k++){ System.out.print("*"); } System.out.println(); } for(int i=1;i<=n-1;i++){ System.out.print(" "); } System.out.print("*"); for(int i=1;i<=n;i++){ System.out.print("-"); } System.out.print("*"); System.out.println(); for(int i=2;i<=n;i++){ for(int j=1;j<=n-i;j++){ System.out.print(" "); } for(int k=1;k<=2*i-1;k++){ System.out.print("*"); } System.out.println(); } } } Meno
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import java.util.Scanner; class New3 { public static void main(String ... args) { Scanner sc =new Scanner(System.in); int n=sc.nextInt(); for(int j=n,p=1;j>=2;j--,p++) { for(int i=1;i<=n+p;i++) System.out.print(" "); for(int k=1;k<=2*j-1;k++) System.out.print("*"); System.out.println(""); } for(int i=1;i<=n-1;i++) System.out.print(" "); System.out.print("*"); for(int i=1;i<=n;i++) System.out.print("-"); System.out.print("*"); System.out.println(""); int q=n-2; for(int j=2;j<=n;j++) { for(int i=1;i<=q;i++) System.out.print(" "); q--; for(int k=1;k<=2*j-1;k++) System.out.print("*"); System.out.println(""); } } } Meno
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import java.util.Scanner; public class webu { public static void main(String[] args) { Scanner sc=new Scanner(System.in); System.out.println("enter the no"); int n=sc.nextInt(); for (int i = n; i >1; i--) { for(int j=1;j1){ for (int l =1 ; l <=2*i-1; l++) { System.out.print("*"); } System.out.println(" "); } } } } Meno
How many credit cards does Amex issue in a year in US?
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Do you remember what type of questions they asked in online coding screen sharing ? I have an interview on next tuesday and i am nervous about typing code in a text editor . Can you give me some sample questions they asked ? Meno
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These were fairly straight forward ones- FizzBuzz, Fibonacci for coding problems, couple of bitwise operations with XOR, regex for US phone number, abt 50% theory questions like interface vs abstract classes and uses of generics and SOLID principles. One question were I messed up a bit was 'Write the api for System.out.println()'. I didn't fully understand the question, then the interviewer clarified. Essentially, he wanted me to write the classes and methods (ignoring the actual logic code) that would allow the user to write System.out.println(). I used inner class to implement it. Good luck. Meno
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Thanks. I was contacted by the Sr.Director about 10 days after the interview and made the offer. Meno
written test on basic java you can crack if you are average concept in java.
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But what is the need of exam no result came 2 months have passed i suggest never go and waste your time. Meno
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you are right . I have never seen any interview experience that got job offer . at every end of year they have walkin drive . Thats look suspicious . At everytime when you go for interview they ask for reference. It might be true that they hire people who have strong refrence Meno
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Yes I got call. May be they call only those people who cracks Aptitude They may have high cut off. Meno
If you add an element at a[1] position in arreylist and access a[2] element,what will happen?
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Let arraylist contains {5,6,4,3,1} If you add 2 at 0th position then arraylist become {2,5,6,4,3,1} So according to question you will get old element of a[1] at a[2]'s position Meno
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We will get runtime exception:ArrayIndexOutOfBoundException
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Let arraylist contains {5,6,4,3,1} If you add 2 at 0th position then arraylist become {2,5,6,4,3,1} So according to question you will get old element of a[1] at a[2]'s position Meno
A company uses a format to exchange messages with us. You need to validate the input. The character encoding ASCII. Valid characters are between 0x20 (space) and 0x7E (~). write validate function to generate valid output or the error message.
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public static void printValidMassage(){ String msg="204a6176617e"; String msg1=convertHextoChars(msg); boolean isValid=(msg1.startsWith(" ") && msg1.endsWith("~")) ? true : false; } private static String convertHextoChars(String hex) { StringBuffer sb = new StringBuffer(); for (int i=0; i Meno
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public static void printValidMassage(){ String msg="204a6176617e"; String msg1=convertHextoChars(msg); boolean isValid=(msg1.startsWith(" ") && msg1.endsWith("~")) ? true : false; System.out.println(isValid); } private static String convertHextoChars(String hex) { StringBuffer sb = new StringBuffer(); for (int i=0; i Meno
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for (int i=0; i
take input from user...3,5,7 etc and draw a pattern if input is 3 then *** *** * * *** * * * if input is 5 then ***** ***** * * * * * * ***** * * * * *
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code for c language #include #include int main() {int n=7,i,j,l,k,m; for(i=1;i<=n;i++) printf("*"); printf(" "); for(i=1;i<=n;i++) printf("*"); printf("\n"); for(j=1;j<=n-2;j++) { for(k=1;k<=n-1;k++) printf(" "); printf("*"); printf(" "); printf("*"); printf("\n"); } for(i=1;i<=n-1;i++) printf(" "); for(i=1;i<=n;i++){ printf("*");} printf("\n"); for(i=1;i<=n;i++) { for(j=1;j<=n;j++) printf(" "); printf("*"); printf("\n"); } return 0; } Meno
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/* if input is 3 then *** *** * * *** * * * if input is 5 then ***** ***** * * * * * * ***** * * * * * */ import java.util.Scanner; public class Pattern3 { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int n=sc.nextInt(); for(int i=1; i<=2; i++){ for(int j=1; j<=n; j++){ System.out.print("*"); }System.out.print(" "); }System.out.println(); for(int i=1; i<=n-2; i++){ for(int j=1; j<=n-1; j++){ System.out.print(" "); }System.out.println("* *"); } for(int i=1; i<=(n+1)/2; i++){ System.out.print(" "); } for(int i=1; i<=n; i++){ System.out.print("*"); }System.out.println(); for(int i=1; i<=n; i++){ for(int j=1; j<=n; j++){ System.out.print(" "); }System.out.println("*"); } sc.close(); } } Meno
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check this import java.util.Scanner; public class MyClass { public static void main(String args[]) { Scanner sc =new Scanner(System.in); int n=sc.nextInt(); for(int i=0;i<2;i++){ for(int j=0;j Meno
The first question was to print pattern that is n=1 *** n=3 * * *** *** ************ *** *** * * n=5 * * *** *** ***** ***** ******************* ***** ***** *** *** * *
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I don't understand the logic of mid line.... import java.util.Scanner; /* 1 *** 3 * * *** *** ************ *** *** * * 5 * * *** *** ***** ***** ******************* ***** ***** *** *** * * */ public class Pattern1 { public static void main(String[] args) { Scanner sc =new Scanner(System.in); int n=sc.nextInt(); if(n>1){ for(int i=1; i1){ for(int i=(n+1)/2; i>=1; i--){ for(int j=1; j<=(((n+1))/2)+1-i; j++){ System.out.print(" "); } int k=0; while(k != (2*i)-1){ System.out.print("*"); k++; }k=0; System.out.print(" "); while(k!=(2*i-1)){ System.out.print("*"); k++; }k=0; System.out.println(); } } sc.close(); } } Meno
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import java.util.Scanner; /* 1 *** 3 * * *** *** ************ *** *** * * 5 * * *** *** ***** ***** ******************* ***** ***** *** *** * * */ public class Pattern1 { public static void main(String[] args) { Scanner sc =new Scanner(System.in); int n=sc.nextInt(); if(n>1){ for(int i=1; i1){ for(int i=(n+1)/2; i>=1; i--){ for(int j=1; j<=(((n+1))/2)+1-i; j++){ System.out.print(" "); } int k=0; while(k != (2*i)-1){ System.out.print("*"); k++; }k=0; System.out.print(" "); while(k!=(2*i-1)){ System.out.print("*"); k++; }k=0; System.out.println(); } } sc.close(); } } Meno
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package pattern; import java.util.Scanner; public class webu1 { public static void main(String[] args) { Scanner sc=new Scanner(System.in); System.out.println("enter the no"); int n=sc.nextInt(); for(int i=1;i=1; i--) { for (int j = 0; j < n-i; j++) { System.out.print(" "); } for(int k=0;k<(2*i-1);k++){ System.out.print("*"); } System.out.print(" "); for(int k=0;k<(2*i-1);k++){ System.out.print("*"); } System.out.println(""); } } } Meno