CareerCup

"you have unsorted array[n] elements. the numbers in the array[n] occurs event times except one number occurs odd time. So, write an algorithm to find that number. Also explain its complexity too. (time and space) both. (On ALGORITHM)"

Sort and XOR the elements the even elements will cancel out.

use the last name of the person..

read abt various techniques to avoid collisions..

for such a question u will have to answer with pros and cons..

3. you have html page directory. in each html page u have written name and phone number. now you have to make normal directory from that information. write an algorithm that convert html directory to normal directory.
how can we do this??

For Q3 i am not getting the question

please can you elaborate the question? what do you mean by directory here??

Q2-Ans : Read each element from array and put it in an Hash-table as key-value pairs(key - number, value- no.of times it occurred). Then read the hash-table values and divide each value by 2. which value gets a remainder 1 that is the number which occurred odd time in an array.

html directory means html files container. Then what is mean by normal directory, what it contains text files or MSoffice files.

Q3 : html directory means html files container. Then what is mean by normal directory, what it contains text files or MSoffice files.

read the array and put the elements in a HashSet.As u keep reading the array, do get for every element and if found ,remove it.Ultimately the only element which is there odd many times will be the item left in hasSet.Reason to use hashSet is the lookup wld be O(1).

"you have unsorted array[n] elements. the numbers in the array[n] occurs event times except one number occurs odd time. So, write an algorithm to find that number. Also explain its complexity too. (time and space) both. (On ALGORITHM)"

You can do it in O(n). The idea is that if you multiply the numbers and divide when it is divisible by the current element, at the end, all even numbers will cancel out, leaving the odd number only:

int number = 1;

For each element i in the array:

if number is divisible by i [i.e. if( number mod (a[i] == 0) ]
then number = number / a[i];
else number = number * a[i];

return number;

public void findodd(int [] ary){
Multimap <int,int>map=new multimap<int,int>();
for(int i=0;i<ary.size();i++){
map.put(a[i],a[i]);
}
for(inti=0;i<ary.size();i++){
Collection<int> list=map.get(a[i]);
if(list.size()%2!=0){
system.out.println("here is the odd one"+list.get(a[i]));
}
}
}