head of a linked list, determine if the linked list is a palindrome and return
True if it is,
False otherwise. For added difficulty, find an algorithm that runs in
O(n) and uses
Given two arrays,
nums2, return the intersection of those arrays in any order.
The easiest way I was able to come up with was to sort the arrays first and then look at the beginning of the arrays. If both are equal, add the element to the output and pop the first element from each array. Otherwise, pop the smaller element of both arrays. Repeat this until one of the arrays is empty at which point there can be no more intersecting elements.
Given a number
n, check if this number is a power of three, that is that it can be represented by
A straightforward solution looks as follows. Multiply by
3 until we reach
n. If we’re greater than
n*n matrix, find the
kth smallest element. Each row in the matrix is sorted in ascending order and independent of the other rows. Do not use
O(n*n) space when solving this. One example matrix looks like this:
If we could use all the space we wanted, we could just perform an n-way merge sort on every row at once. This would work because the rows itself are sorted already. Afterwards, we would need to pick the kth element from the list and would be done.
head of a list with pointers to
next and a
random element, copy the list to a new list without any random pointers pointing to the old list.