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Please give more input about this interview
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Please if you see the comment please give more input about this interview
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In these sorts of interviews you really need to drill down and understand what the interviewer is looking for. A good way to simulate a real interview experience is to do a mock with one of the EPAM Systems Senior Test Automation Engineer experts on Prepfully, rated super strongly on TrustPilot... prepfully.com/practice-interviews Less
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public static void SortBinaryArrays(int[] Binary) { int[] sorrtedArray = new int[Binary.length]; int i = 0; for (int a : Binary) { if (a == 0) { sorrtedArray[i] = a; i++; } } for (int a : Binary) { if (a == 1) { sorrtedArray[i] = a; i++; } } for (int a : sorrtedArray) { System.out.print(a + " "); } } Less
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public static int[] sortBinaryArray(int[] a) { int leftptr=0; int rightptr=a.length-1; while(true) { while(a[leftptr]==0) { leftptr++; } while(a[rightptr]==1) { rightptr--; } if(leftptr Less
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int[] a={0,1,0,1,1,1,0,0,0,1}; int index=0; public static int[] sortBinaryArray(int[] a) { for(int i=0;i Less
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//Reverse Words in String public static String reverseWords(String string){ String[] stAr=string.split(" "); String result=""; for(int i=stAr.length-1;i>=0;i--) { result=result+" "+stAr[i]; } return result.subString(1,result.length()); } Less
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public static boolean isPalindrome(String string) { String reverse =""; for(int i=string.length()-1;i>=0;i--) { reverse=reverse+string.charAt(i); } if(string.equalsIgnoreCase(reverse) { return true; } return false; } Less
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//Reverse Words In a String public static String reverseWords(String s) { String res=""; String[] string=s.split(" "); for(int i=0;i=0;i--) { rev=rev+string.charAt(i); } return rev; } Less
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public void missing() { ArrayList arr = new ArrayList(); int a[] = { 1, 3, 4, 5, 6, 7, 10 }; int j = a[0]; for (int i = 0; i < a.length; i++) { if (j == a[i]) { j++; continue; } else { arr.add(j); i--; j++; } } System.out.println("missing numbers are "); for (int r : arr) { System.out.println(" " + r); } } Less
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Does the numbes in an array lies in between a specific range ?
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Does the numbes in an array lies in between a specific range ?
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declare l_orig varchar2(32767):='&input'; l_out varchar2(32767); begin for i in 0..length(l_orig)-1 loop l_out := l_out ||substr(l_orig,length(l_orig)-i,1); end loop; dbms_output.put_line(l_out); end; / Less
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// Java program to Reverse a String by // converting string to characters one // by one import java.lang.*; import java.io.*; import java.util.*; // Class of ReverseString class ReverseString { public static void main(String[] args) { String input = "GeeksForGeeks"; // convert String to character array // by using toCharArray char[] try1 = input.toCharArray(); for (int i = try1.length-1; i>=0; i--) System.out.print(try1[i]); } } Less
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source: GeeksforGeeks
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Work is remotely? or they get paid passage and accommodation of brazil to the United States? Thank you. Less
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Hi, did you pass on this process? What are your impressions of the company? I'm also from Brazil and I'm currently on their recruitment process. Less
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hey! what was the minor project that you had to do? Could you tell us pls?
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public void CountTheRepeatedNumber(int[] arr){ Map map=new Hashmap(); for(Integer num:arr){ if(map.containsKey(num)){ map.put(num,map.get(num)+1); //increase the value if it is repeated } else{ map.put(num,1); // put 1 into value if number is first time present } } Set key=map.KeySet(); //return you all the distinct integer number for(Integer i:key){ if(map.get(i)>=0){ syso(i +" is " +map.get(i) +"Times in an array"); //simply print this so u will get repeated numbers with count and just print it } } Less
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public void unique(){ int[] numbers = {1,2,2,3,4,5,5,6,7}; Arrays.sort(numbers); System.out.println("Sorted : "+numbers.length); for(int i = 1; i < numbers.length; i++) { if(numbers[i] == numbers[i - 1]) { System.out.println("Duplicate: " + numbers[i]); } } } Less
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//Finding the number which appeared twice public static void findDuplicates(int[] a) { Map hashmap = new HashMap(); for(int i=0;i entries : hashmap.entrySet()) { if(entries.getValue()>1) { System.out.print(entries.getKey()+" "); } } } Less
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public class Func35 { public static void main(String[] args) { for(int i=1;i<=100;i++){ if(i%3==0 && i%5==0){ System.out.print("Z"+","); } else if(i%3==0){ System.out.print("X"+","); } else if(i%5==0){ System.out.print("Y"+","); } else { System.out.print(i+","); } } } } Less
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for(int i=1;i<=100;i++) { boolean flag=false; if(i%15==0) { System.out.println(i+"Z"); flag=true; } else if(i%3==0 && flag==false) { System.out.println(i+"X"); flag=true; } else if(i%5==0&&flag==false) { System.out.println(i+" Y"); } } Less
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Used a bool variable added the two division in the loop (with the loop variables) (i%5 and i%3). The first division only sets the bool variable and the second division writes the output according Less
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for(Map.Entry m : map.entrySet()){ system.out.println( m.getKey()+" "+m.getValue()); Less
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map.forEach((key, value ) -> System.out.println(key+"--"+value));
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There's quite an extended back and forth in actual interviews for questions like this, so there's no real replacement for actual practice. The Prepfully EPAM Systems Test Automation Engineer experts have actually worked in this role, so they're able to do a propermock, which really puts you through the paces and lets you assess your readiness level. prepfully.com/practice-interviews Less
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1. Using ChromeOptions class we can run the UTA in the incognito tab ChromeOptions op = new ChromeOptions(); op.addArguments("--incognito"); WebDriver driver = new ChromeDriver(op); Less
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3. How will you execute only the failed testcases after the first run. ? Ans: Run the testng-failed.xml file Less
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4. How to run test cases parallelly? Ans : in the testng.xml file add parallel = "methods" and threadCount= NoOFThreadsNeeded Less