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i tried my best in 2nd round but i got rejected

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I would like to see myself donning one of the senior positions in organization contributing to technology as well as participating in business growth. Less

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This was interview was held on 22nd September,2013. They took about 3 days after the HR round to respond back. Less

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Can,t we do sorting from largest to smallest give arr[1] which is of course second largest Less

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The answer is 6. counter weights of 1, 2, 4, 8, 16 ,32

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Yes, he felt humiliated and he.s right. You don't get to tell somebody that such a certification is useless just cause some company doesn't have a use for it. Thats something you earn with many many hours of study and no one has the right to put you down for it. Less

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package glassdoor; import java.util.HashSet; import java.util.LinkedList; import java.util.List; import java.util.Set; /** * Suppose you have a dictionary that contains valid words. Given an input string with all spaces removed, determine whether the string is composed of valid words or not. You can assume the dictionary is a hashtable that provides O(1) lookup. Some examples: helloworld-> hello world (valid) isitniceinhere-> is it nice in here (valid) zxyy-> invalid */ public class SplitStringIntoWords { private Set dictionary = new HashSet(); public SplitStringIntoWords(Set dictionary) { this.dictionary = dictionary; } /* * method that checks if the string can be split into valid dictionary words */ public boolean isValid(String s) { String clone = new String(s); // a copy of the original string if(null == s || s.isEmpty()) return false; StringBuilder sb = new StringBuilder(""); int x = 0; while(x matchingWords = new LinkedList(); // adding to the list for(String d: dictionary) { if(d.startsWith(sb.toString())) { matchingWords.add(d); } } // if more than one word in dictionary starts with the sub string // find the longest length match that occurs in the given string // remove the selected match from the given string // set the StringBuilder to new and x to -1 // continue the process in the new (string with match removed) string int maxlenght = 0; String match = ""; for(String m : matchingWords) { if(clone.contains((CharSequence) m)) { if(m.length() > maxlenght) { maxlenght = m.length(); match = m; } } } s = s.replace(match, ""); sb = new StringBuilder(""); x = -1; } x++; } if(s.isEmpty() || s=="") { return true; } else { return false; } } public static void main(String[] args) { Set set = new HashSet(); set.add("hello"); set.add("world"); set.add("it"); set.add("is"); set.add("nice"); set.add("in"); set.add("here"); set.add("helloworld"); set.add("dog"); set.add("cat"); set.add("fish"); SplitStringIntoWords ssiw = new SplitStringIntoWords(set); System.out.println(ssiw.isValid("helloworld")); System.out.println(ssiw.isValid("isitniceinhere")); System.out.println(ssiw.isValid("dogcatfish")); System.out.println(ssiw.isValid("dogecatfish")); System.out.println(ssiw.isValid("wqewqe")); } } Less

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Because if its square, the lid can easily slip inside along the diagonal. length of diagonal is greater than side of square. Less

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Really that is what you would answer? What if they stirred the jars around after you distributed the marbles? Personally I think a better answer is to put 1 white marble in the first far and 99 white marbles and 100 black marbles in the second jar. If you choose the jar at random you now have a 74.87% chance of getting a white marble, regardless of the marbles position in the jar. Less