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have you received offer letter?
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Which type of questions are asked in online test
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my dream job is the one which gives me a passion to learn new things to the position in a company Minder
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i tried my best in 2nd round but i got rejected
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Guys i had attend interview in march 2017 ..I cleared all round then HR told me wait mail or call did u tell me how much days I have to wait for result or anyone got offer letter or call..plz tell me Minder
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after how many monthes i could attain syntel again because attained in january 2nd. Minder
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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. Minder
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test
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Because of sharing my knowledge and skills and improving them until I became an expert on the particular task or whatever job assigned to me, I see myself on a higher position within 5 years. Minder
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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. Minder
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Thanks..
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i also gave hr interview on 22sep but havn't received their call till now. Does thi mean i am not selected??? or they will call me for 2nd batch after sometime??? any idea?? Minder
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Can,t we do sorting from largest to smallest give arr[1] which is of course second largest Minder
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It Took Time but Simple logic is needed.
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Would be impressive to solve the general case... find the kth large number: create a minheap of size k, and for each element in the array, if that number is bigger than the top of the minheap, replace the top of the minheap with that number and recreate the minheap. Running time is O(nlgn). If you can have negative numbers, then you can easily modify this code. int findKthLargestNumber(int[] A, int k) { if (k minHeap[2*i+1]) { int temp = minHeap[i]; minHeap[i] = minHeap[2*i+1]; minHeap[ 2*i+1] = temp; i = 2*i+1; } else if (minHeap.length minHeap[2*i+2)]) { int temp = minHeap[i]; minHeap[i] = minHeap[2*i+2]; minHeap[ 2*i+2] = temp; i = 2*i+2; } else { done = true; } } Minder
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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. Minder
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thats good
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Wait, you were “humiliated” because the manager gave you an honest answer that getting your SE is “useless” because we do very little work (some offices do none) which requires an SE stamp? Cry me a river! Minder
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I thought out loud the whole time, almost everything was hints, otherwise it was a tough problem. Minder
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When was your interview? 13th?
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Can you elaborate on String replacement problem? Currently studying for other interviews and was wondering what this was like. Thanks! Minder
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The answer is 6. counter weights of 1, 2, 4, 8, 16 ,32
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Can the answer also be 1, 3, 9, and 27. That means only 4 weights
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Can the answer also be 1, 3, 9, and 27. That means only 4 weights
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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. Minder
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qq is right, that is the optimum combination.
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2 colours, 2 jars. White in one, black in the other
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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")); } } Minder
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the above solution is ok for small dictionary... it is not time or space efficient... typically this problem is solved using dynamic programming... for the purpose of the interview, I am not sure if the above is acceptable Minder
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DP is perfect solution for the above question. Normally, DP will be asked to optimize the solution of recursive one. Minder