Just Another Programming Puzzle (Finding one non repeating element in list of pairs)
So here is a programming question once again,
Consider a list of unsorted positive numbers in which each number has a pair except one of the number, the problem is to find that number in minimum complexity with O(1) extra space. The complexity in this case should not be more than O(n).
For instance, in the list {5,7,4,9,5,7,4} the answer should be 9. I will post my solution in the comments some time later.
The solution is as follows remember the list is not sorted,
FACT 1: XORing a number with 0 will return the number itself.
FACT 2: XORing a number with itself will return 0.
So when we XOR all the elements with each other in a list of number in which each number is repeated once except one number we will get the non repeated number. For example if List is {5,4,6,4,5} we will get 6 if we execute this: 5 XOR 4 XOR 6 XOR 4 XOR 5.
Here is the C Sharp Source to do the same.
int Foo(int [] nList)
{
int result = 0;
for (int i = 0; i < nList.Length; i++)
result ^= nList[i];
return result;
}
In C Sharp ‘^’ is the operator for XOR, This piece of code takes O(n) time and O(1) extra space.
I think a little bit modification require …
result ^= nList[i]
and what about make it little bit more efficient …
int Foo(int [] nList)
{
int result = nList[0]; // we know the first element
for (int i = 1; i < nList.Length; i++) // running one less than total numbers
result ^= nList[i];
return result;
}
Ohh well Qasim Pasta!! there was a typo, ammended it.. :), well I dont think capturing the first element and running the loop n - 1 times will make any difference, though your comment and intention noted :)…
Thx for poiting out the typo..
ok, I have another solution
Fact: all of list elements has another element equivalent except for one element, which means that list length is an odd number
if you start from the beginning the list and add the numbers to each other, until you reach the middle number, then you start subtracting numbers from the total until you are done, then the final count will be the number with no corresponding peer
int Foo(int[] arr)
{
int count = arr[0];
for(int i = 0; i < arr.Length; i++)
{
if( i <= (size - 1) / 2)
count += arr[i];
else
count -= arr[i];
}
return count;
}
Create Map
Loop thru your array and check
if elememt in aaray exist in map , if not then add
if exists then remove
Finally only one element will left will be your non repeated elemet.
san, the problem definition states that you should solve it in O(1) extra space, using your map, the extra space complexity will be O(n-1)
@ Hisham abd El-Hafez:
Actually your solution works only if the pairs are quasi symmetrically placed and if your pairless number is not the center number then you end up performing the same operation on both the numbers of atlease 1 pair;
eg: 4 4 5 7 5
according to your solution we actuallly have
4 + 4 + 5 - 7 - 5 = 1
Vinayak,
Yeah you are right, my solution will only work if the array is symmetric.
my mistake