We use cookies to ensure you have the best browsing experience on our website. Please read our cookie policy for more information about how we use cookies.
For me, it's just an alternative way. There's not only one way to think for the same problem whatever the solution is good or bad. I shared my solution because I hadn't seen anyone think the same way and I thought it was good to learn from people's comments on my solution.
I like your solution, elegant and efficient. It took my two days for mine and even now has timeout issues when testing. I have good envy now. Well done!
This solution will not work when rotation is greater than array size, meaning d > n. Java returns negative result if you perform % operation on a negative number. And that negative number will throw an
java.lang.ArrayIndexOutOfBoundsException
Eg. n = 4, d = 7
Starting with i=0,
a[(i+n-d)%n] will give
a[(0 + 4 - 7)%4] => a[(0 + (-3))%4] => a[(-3)%4] => a[-3]
Left Rotation
You are viewing a single comment's thread. Return to all comments →
For me, it's just an alternative way. There's not only one way to think for the same problem whatever the solution is good or bad. I shared my solution because I hadn't seen anyone think the same way and I thought it was good to learn from people's comments on my solution.
A good solution by you!
I like your solution, elegant and efficient. It took my two days for mine and even now has timeout issues when testing. I have good envy now. Well done!
I really like your solution as you have reduce the time complexity to O(n^2) while is very efficeint compare to my program which has a T(n) = O(n^3)
How the complexity is quadratic? It is O(n), right?
This solution will not work when rotation is greater than array size, meaning d > n. Java returns negative result if you perform % operation on a negative number. And that negative number will throw an
Eg. n = 4, d = 7
Starting with i=0, a[(i+n-d)%n] will give a[(0 + 4 - 7)%4] => a[(0 + (-3))%4] => a[(-3)%4] => a[-3]
Range of a is from 0 to n. Hence the error
for this case you can check for d>n and d = d%n then d will be in range of 0 to n and the above solution will work perfectly :)
this is the simplest and most efficient solution. I was struggling Time out error with mine.