First enter all the length of the sticks in an array, sort it in O(nlogn) and then traverse the array from the right end, checking for multiple instances of any stick. This also covers the special case of a square.
A different approach,which has O(N*logN) time complexity but uses Only One Array is as follows:
After storing the array containing the lengths of sticks, sort it.
We must note that the same lengths will automatically come together after sorting.
For example,
the array 5 2 8 3 1 2 4 3 5
becomes 1 2 2 3 3 4 5 5 8
Then if we traverse the array from backwards picking up the first two occurrences of a pair of same numbers(for e.g (5,5) would be the first occurrence and (3,3) would be the second in this case.Don’t forget to break the loop here. )
the answer would be 5*3=15.If we are unable to find two occurrences we print -1.
The JAVA solution of the problem using this method is:
Can anyone help with the code. I have included the square case too. But I don’t know why my solution is not getting accepted. CodeChef: Practical coding for everyone
CodeChef: Practical coding for everyone I cant seem to figure out whats wrong with my solution. It passes most of the test cases I can come up with.
Please help out. Thanks in advance
PS: I looked through some solutions… They use the size of count array as 10000 whereas the mentioned size is 10^3 = 1000. Why is that so?
I tried both ways by the way. Still wrong answer.