Given a string
s, find the first non-repeating character in it and return its index. If it does not exist, return
-1. An example is the word
leetcode where the
l only appears once, therefore returning
0 as result.
The obvious, but very slow, solution is to iterate over the string and then for ever character, check the whole string if the character appears again. This has a time complexity of
A better solution is to write all characters in a map and count the number of occurrences. This way, we can iterate again and simply look at the counter in the map. This yields a time complexity of
The task is to determine if a tree is symmetric, meaning mirrored along the
root node. The usual tree structure is given in form of a
The logic itself is quite straightforward look at two nodes and compare the values. If their respective values are the same look at both of their children and compare them. Repeat this process until every node is visited.
There are some special cases here that need to be handled. The left child can be
None while the right child is not and vice versa. If both children are
True and do not proceed because this means we reached a leaf node.
The task is to traverse a binary tree inorder. This means, that starting from the root node, nodes should be visited in the following order: left, self, right. With this knowledge it is easy to come up with a recursive algorithm.
The only special case that needs to be handled are the leaf nodes that don’t have any children. By still visiting the non-existing children and then returning an empty list (
) when a
None is encountered, the whole algorithm fits into two lines.
Runtime: 41 ms, faster than 70.32% of Python3 online submissions for Binary Tree Inorder Traversal.
Memory Usage: 13.8 MB, less than 60.08% of Python3 online submissions for Binary Tree Inorder Traversal.
The task is to implement a square root method without using
pow or any similar function. Any digits after the decimal point do not to be returned. This is effectively the
The first idea is to simply try out every number and multiply it with itself until the square root is found. If the next number is greater than
n, return the previous number. This is however not very efficient as it has a time complexity of
O(sqrt(n)) in every case.
To speed this process up, we can use binary search. The idea is to always double the size of the steps until we reach the target number or overshoot. If we overshoot, take half of the number and start at
1 again, then double, and so on.
The task is to add two binary numbers of various length. There is more than one possible solution. First, the basic solution which doesn’t use too many builtin Python functions works as follows: Reverse the strings containing the numbers. Then Iterate in a loop as long as we didn’t reach the end of both strings. Write the current position to a temporary variable if it’s still in the boundaries of the string. Then add a potential overflow and handle it if the sum of
a + b + overflow > 1. Finally, check for the overflow at the end and add a
1 and reverse the string again before returning it.
This can be solved much simpler with builtin Python functions.