This repo contains my solutions to the tasks from dailycodingproblem.com.
- How to solve a hard programming interview questions
- How to formulaically solve tree interview questions
| Number | Title | Difficulty | Questioner | Description |
|---|---|---|---|---|
| 11 | autocomplete | Medium | Implement an autocomplete system. That is, given a query string s and a set of all possible query strings, return all strings in the set that have s as a prefix. For example, given the query string de and the set of strings [dog, deer, deal], return [deer, deal]. Hint: Try preprocessing the dictionary into a more efficient data structure to speed up queries. |
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| 9 | largest sum of non adjacent numbers | Hard | Airbnb | Given a list of integers, write a function that returns the largest sum of non-adjacent numbers. Numbers can be 0 or negative. For example, [2, 4, 6, 2, 5] should return 13, since we pick 2, 6, and 5. [5, 1, 1, 5] should return 10, since we pick 5 and 5. Follow-up: Can you do this in O(N) time and constant space? |
| 8 | unival tree | Easy | A unival tree (which stands for "universal value") is a tree where all nodes under it have the same value. Given the root to a binary tree, count the number of unival subtrees. |
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| 7 | number of interpretations | Medium | Given the mapping a = 1, b = 2, ... z = 26, and an encoded message, count the number of ways it can be decoded. For example, the message '111' would give 3, since it could be decoded as 'aaa', 'ka', and 'ak'. You can assume that the messages are decodable. For example, '001' is not allowed. |
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| 6 | xor linked list | Hard | An XOR linked list is a more memory efficient doubly linked list. Instead of each node holding next and prev fields, it holds a field named both, which is an XOR of the next node and the previous node. Implement an XOR linked list; it has an add(element) which adds the element to the end, and a get(index) which returns the node at index. If using a language that has no pointers (such as Python), you can assume you have access to get_pointer and dereference_pointer functions that converts between nodes and memory addresses. |
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| 5 | cons car cdr | Medium | Jane Street | cons(a, b) constructs a pair, and car(pair) and cdr(pair) returns the first and last element of that pair. For example, car(cons(3, 4)) returns 3, and cdr(cons(3, 4)) returns 4. Given this implementation of cons: def cons(a, b): def pair(f): return f(a, b) return pair Implement car and cdr. |
| 4 | lowest positive integer not in array | Hard | Stripe | Given an array of integers, find the first missing positive integer in linear time and constant space. In other words, find the lowest positive integer that does not exist in the array. The array can contain duplicates and negative numbers as well. For example, the input [3, 4, -1, 1] should give 2. The input [1, 2, 0] should give 3. You can modify the input array in-place. |
| 3 | binary tree serialization and deserialization | Medium | Given the root to a binary tree, implement serialize(root), which serializes the tree into a string, and deserialize(s), which deserializes the string back into the tree. For example, given the following Node class class Node: def init(self, val, left=None, right=None): self.val = val self.left = left self.right = right The following test should pass: node = Node('root', Node('left', Node('left.left')), Node('right')) assert deserialize(serialize(node)).left.left.val == 'left.left' |
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| 2 | product of other array elements | Hard | Uber | Given an array of integers, return a new array such that each element at index i of the new array is the product of all the numbers in the original array except the one at i. For example, if our input was [1, 2, 3, 4, 5], the expected output would be [120, 60, 40, 30, 24]. If our input was [3, 2, 1], the expected output would be [2, 3, 6]. Follow-up: what if you can't use division? |
| 1 | is sum in array | Easy | Given a list of numbers and a number k, return whether any two numbers from the list add up to k. For example, given [10, 15, 3, 7] and k of 17, return true since 10 + 7 is 17. Bonus: Can you do this in one pass? |