Write an algorithm to determine if a numberFor example, the number
nis happy. A happy number is a number defined by the following process: Starting with any positive integer, replace the number by the sum of the squares of its digits. Repeat the process until the number equals 1 (where it will stay), or it loops endlessly in a cycle which does not include 1. Those numbers for which this process ends in 1 are happy. Return
nis a happy number, and
19is divided into
9^2, then added together as
82, then divided again, etc. After some iterations this reaches
1^2 + 0^2 + 0^2 = 1. On the other hand,
2keeps ever incremententing. The programming aspect is relatively easy to solve in Python when converting the types to a string and then converting the string to a list.
Given the head of a linked list, determine if there is a cycle in the list. A cycle is defined as a chain of arbitrary length that point to the same node. As an added difficulty, the solution should have a space complexity of
The first observation is that the cycle can be of arbitrary length. Thus, saving the visited nodes in a set would have a space complexity of
O(n). Secondly, the numbers cannot be compared because numbers might be duplicated throughout the list. Instead, we need to compare the list objects themselves.
Given an array of
True if there is a duplicated number and
The simple most solution I could find was to use a
set and then see for every number if it is already contained in the set. A map would have worked as well. The time complexity for this is
O(n), assuming that the access to the hash set is
O(1), it takes only one iteration over the whole list.
The task is as follows: Given an integer array
nums, move all 0’s to the end of it while maintaining the relative order of the non-zero elements.
Note that you must do this in-place without making a copy of the array.
The last note is actually the important part. Without this, it would be easy to just use array splitting in Python and add a zero. This would look similar to this:
This obviously overrides
nums and the change is no longer in place. The next best solution that I found was to delete the items at position I from the list and then add a zero to the end. It is important to note, that the index needs to be decremented by one in case a
0 is found to account for double
0. The number of steps needs to always be increased.
Reverse a string and only use one extra variable. The solution is straightforward with two pointers,
p2. THe idea here is to start from both ends and increment
p1 and decrement
p2 until both of them meet in the middle. To switch the values of variables without assigning a temporary variable, the notation of
a, b = b, a can be used. To convert a string into a list of characters, surroung the string with