LeetCode #242: Valid Anagrams (Solution in Python & Explanation)

The Problem

Given two strings s and t, return true if t is an anagram of s, and false otherwise.

An Anagram is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.

Example 1:

Input: s = "anagram", t = "nagaram"
Output: true

Example 2:

Input: s = "rat", t = "car"
Output: false

Solution

We know that the two strings are anagrams if:

  1. They have the same length
  2. They have the same characters and the same quantity of each character

So, if we can verify that the strings pass these two criteria, we can verify that they are valid anagrams.

We can check if they are the same length by simply using len:

len(s) == len(t)   # boolean expression

But how can we check if they have the same characters and the same quantity of each character? We can use hash maps. For each string, we can create a hash map that includes every unique character in the string and its frequency in the string.

For example, this would be the hash map of the string s, which has a value of "anagram":

s = "anagram"
countS = {"a": 3, "n": 1, "g": 1, "m": 1}

Once, we have created the hash maps, we can check if they are equivalent by using a boolean expression:

countS == countT

So, let’s put it all together. These are all the steps we have to take:

  1. We check if the strings are not the same length. If they’re not, we can immediately return False without doing any other code.
  2. We create empty hash maps for the two strings s and t.
  3. We loop through each character in s and t and add them to the hash maps. If the character does not exist in the hash map, a new key for the character is added and the value is initialized to 0 (then we add 1). If a key for the character does exist in the hash map, we add 1 to the value.
  4. We check if the two hash maps are equivalent. If they are, we return True. If they are not, we return False.

This is the final code.

   def isAnagram(self, s: str, t: str) -> bool:
        # Step 1: check if same length
        if len(s) != len(t):
            return False
        
        # Step 2: create hash maps
        countS = {}
        countT = {}
        
        # Step 3: loop through characters to add to hash map
        for i in range(len(s)):
            countS[s[i]] = 1 + countS.get(s[i], 0)
            countT[t[i]] = 1 + countT.get(t[i], 0)
        
        # Step 4: check if hash maps are equivalent  
        if (countS == countT): 
            return True
        else: 
            return False

This solution has a time complexity of O(s + t) or simply O(n) because we have to iterate through each of the strings. The space complexity is also O(s + t) because we are building hash maps that are the sizes of arrays s and t.

There is another solution that is much more simple, but less efficient. We can sort both of the strings and see if the sorted versions are equivalent. This works because, if they are anagrams, then they will have the same sorted value.

def isAnagram(self, s: str, t: str) -> bool:
    return sorted(s) == sorted(t)

A good sorting algorithm can sort an array of characters in O(n * log(n)) time. The space complexity could be either O(1) or O(n)

LeetCode: Contains Duplicates (Solution in Python & Explanation)

In this article, I will be explaining how to solve the Contain Duplicates problem on LeetCode. This is an Easy-level question for Arrays.

The Problem

Given an integer array nums, return true if any value appears at least twice in the array, and return false if every element is distinct.

Example 1:

Input: nums = [1,2,3,1]
Output: true

Example 2:

Input: nums = [1,2,3,4]
Output: false

Example 3:

Input: nums = [1,1,1,3,3,4,3,2,4,2]
Output: true

Solution

This problem requires that we loop through the array. We may not have to loop through the entire array depending on when / if we find the first duplicate.

As we loop through the array, we check each array item to see if we’ve seen it before. How can we check if we’ve seen a value in the array before? The naive method would be to do a second loop through the array for each. The program will have a complexity of O(n2) because you are looping through each array element and then, within that loop, you are looping through each array element again.

def containsDuplicate(self, nums: List[int]) -> bool:
        dups = set()
        for n in nums:
            for s in nums:
                if n == s:
                   return True
            dups.add(n)
        return False

This solution is slow and we can solve this problem in a more efficient way.

As we loop through the array, we can keep track of the values that we’ve seen thus far. Then, as we continue looping through and checking new values, we can check if they are already in the set of values that we’ve seen before. How can we keep track of the values that we’ve seen? We can use a hash set. This solution takes O(n) time because we are only looping through the array once instead of twice.

def containsDuplicate(self, nums: List[int]) -> bool:
        dups = set()
        for n in nums:
            if n in dups:
                return True
            dups.add(n)
        return False

There is another solution that is even more simple in terms of readability. It is only one line. Basically, we convert the list of numbers into a hash set and compare the length of the set to the length of the original list. If the lengths are different then that must mean that there are duplicate items in the original list because the duplicates are taken out of the set.

This solution uses the no duplicates property of hash sets to solve the problem in an extremely simple way. The time complexity of this solution is also O(n) because that is the complexity of converting a list to a set.

def containsDuplicate(self, nums: List[int]) -> bool:
        return len(set(nums))!=len(nums)

Data Structures: Arrays

In this article, we will be talking about the most popular data structure: arrays. You will learn what an array is, how it’s stored in your computer’s memory, the complexity of array operations. This will help you to be successful when using arrays in code interviews.

What is an array?

An array, also known as a list, is a collection of items stored in contiguous space in memory. The items of an array are organized sequentially in memory.

This image gives a visual of how arrays are stored in memory. Each array element has a value, memory address, and an index. The index is the integer (0-n) that corresponds to the element’s order position in the array. The first element has an index of 0 and the last element in the array has an of n-1 where n is the length of the array.

Array Operations

OperationComplexity
lookupO(1)
pushO(1)
insertO(n)
deleteO(n)
searchO(n)

The lookup operation takes O(1) time, or constant time because, since each array item is allocated to an equal-size memory space and each memory space is indexed by contiguous integers, your computer is able to immediately jump to the correct memory address using the array index.

The insert operation takes O(n) time, or linear time, because if the item is not at the very end of the list, the other items in the last after the index of the insertion have to be shifted down in memory to make space for the new item. This requires looping through each of the items to copy and move the contents of the memory address to a different memory address.

The push operation also has a complexity of O(1) because since you are adding the item at the last array index, you don’t have to loop through each array element to shift them. This operation has different names in different languages. For example, in Python it’s called append. There is also the pop operation in which the last array item is removed from the array. This also takes constant time.

The delete operation takes O(n) time for the same reason as the insertion operation. When an item is deleted from the array, all of the items after that time must be shifted forward.

The search operation takes O(n) time because you must look at each item in the array in order to see if it is the item you are looking for. This can be quick if it’s one of the first items in the array, but if it’s at the end of the array it will take longer.

Pros & Cons of Arrays

Pros of arrays:

  • Fast lookups
  • Fast push/pop
  • Ordered (sorting)

Cons of arrays:

  • Slow inserts
  • Slow deletes
  • Fixed size (if using static arrays)

Static vs Dynamic Array

Static arrays are a fixed in size. You specify the number of items your array will hold ahead of time.

Dynamic arrays allow us to copy and rebuild an array at a new location with more space in memory.

How to Divide an Array in Half in JavaScript

In this tutorial, I will show you how to divide a JavaScript array into two equal parts (or almost equal if the array has an odd length).

Use the Array instance slice() method:

const test_array = ["a", "b", "c", "d", "e", "f"];

const half = Math.ceil(list.length / 2); 

const firstHalf = list.slice(0, half)
const secondHalf = list.slice(-half)

If the list contains an even number of items, the result is split with exactly half the items:

>>> firstHalf
["a", "b", "c"]
>>> secondHalf
["d", "e", "f"]

The Math.ceil() function always rounds a number up to the next largest integer. We add this in case the array length is odd and list.length / 2 is not a whole number.

If the array contains an odd number of elements, the result is split as close to half as possible.

const odd_array = [1, 2, 3, 4, 5]
const half = Math.ceil(list.length / 2); 
const firstHalf = list.slice(0, half)
const secondHalf = list.slice(-half)

Since there are 5 elements, the half variable will be Math.ceil(5 / 2) = 3. So, the array will split on the third element.

>>> firstHalf
[1, 2, 3]
>>> secondHalf
[4, 5]