LeetCode #125: Valid Palindrome (Solution in Python & Explanation)

In this article, I will be explaining how to solve the Valid Palindrome problem on LeetCode. This is an Easy-level problem.

The Problem

A phrase is a palindrome if, after converting all uppercase letters into lowercase letters and removing all non-alphanumeric characters, it reads the same forward and backward. Alphanumeric characters include letters and numbers.

Given a string s, return true if it is a palindrome, or false otherwise.

Example 1:

Input: s = "A man, a plan, a canal: Panama"
Output: true
Explanation: "amanaplanacanalpanama" is a palindrome.

Example 2:

Input: s = "race a car"
Output: false
Explanation: "raceacar" is not a palindrome.

Example 3:

Input: s = " "
Output: true
Explanation: s is an empty string "" after removing non-alphanumeric characters.
Since an empty string reads the same forward and backward, it is a palindrome.


Alphanumeric characters are the lowercase letters a-z, capital letters A-Z, and integers 0-9.

So, to solve this, we first want to remove all non-alphanumeric characters, including spaces, punctuation, etc., from the input string. Then, we need to check the new string to see if it’s a palindrome. We have two ways we can do this.

Solution #1

For this solution, we create a new string which contains all of the alphanumeric characters from the input string (lowercased). Then, we compare this new string to its reversed version. If they are the same, the function returns true. If not, it returns false.

In order to create the new string, we loop through the characters in the original input string and, if the character is alphanumeric, we append it to the new string.

Finally, we can check if the string (without any special characters or spaces) is a palindrome by using a boolean expression to equate it to its reverse. We can get the reverse of a string by using str[::-1]

def isPalindrome(self, s: str) -> bool:
    newStr = ""
    for char in s:
        if char.isalnum():
            newStr += c.lower()
    return newStr == newStr[::-1]

This solution has a time complexity of O(n) because we have to loop through the string to remove non-alphanumeric characters. It also has a space complexity of O(n) because we are creating another string, which could be at most the size of the input string.

Solution #2

For this solution, we will create pointers to the beginning and end of the array which will make sure that the two sides have the same characters in the same order.

First, create two pointers— one which points to the beginning of the string and the other which points to the end. The two pointers will move inwards, checking each character until the reach the middle or overlap with each other.

As the pointers loop through the characters, they will skip any characters that are not alphanumeric. To do this, we use while loops that will continue to run until the character at the left/right index is alphanumeric. We also add a check to make sure that the left/right index is in bounds by checking left < right / right > left

We must make sure to compare the lowercase versions of the characters to each other. If any character at the left pointer does not match the corresponding character at the right pointer, then we know that it’s not a palindrome and can return False. If the entire loop finishes then that must mean the string is a palindrome and we can return True.

def isPalindrome(self, s: str) -> bool:
    left, right = 0, len(s) - 1 
    while left < right:
        while left < right and not s[left].isalnum():
            left +=1
        while right > left and not s[right].isalnum():
            right -= 1
        if s[left].lower() != s[right].lower():
            return False 
        left, right = left + 1, right - 1
    return True

The time complexity for this solution is O(n) because you are still looping through the entire string. The space complexity is O(1) because you are not making a new string.

LeetCode: Two Sum (Solution in Python & Explanation)

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

The Problem

Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

You can return the answer in any order.

Example 1:

Input: nums = [2,7,11,15], target = 9
Output: [0,1]
Explanation: Because nums[0] + nums[1] == 9, we return [0, 1].

Example 2:

Input: nums = [3,2,4], target = 6
Output: [1,2]

Example 3:

Input: nums = [3,3], target = 6
Output: [0,1]

The Solution

The most straightforward way to solve this problem is to check every combination of two values in the array and see if they can sum up to the target value. To do this, we would iterate through the entire array and for each item we would iterate through the array a second time to check the sums of the current array item with every other item in the array and see if they are the target. The (worst-case) runtime for this solution method is not very efficient: O(n2).

There is a better solution that only requires us to loop through the array once. Basically, if we could loop through the array and, for each array item, check if the number that adds with that value to get the target exists in the array using an operation that doesn’t have O(n) time complexity, then we could make our program much faster. We can do this by using a hash map.

We create a hash map that will store array items as keys and their indexes in the array as values ({array_value: array_index})

We use the enumerate function to get the index of the current item as we loop through the array. For each array item, we compute the difference diff between the target value and the item’s value. Since target - n = diff, that means diff + n = target. If this difference diff is in the hash map, then that means that the value has already been seen in the array. Thus, we’ve found the . If the difference is not in the array, then we add the current array item and its index to the hash map then move to the next array item.

def twoSum(self, nums: List[int], target: int) -> List[int]:
        hm = {} 
        for i, n in enumerate(nums): 
            diff = target - n 
            if diff in hm:
                    return [hm[diff], i]
            hm[n] = i

This solution has a time complexity of O(n) because we only have to loop through the nums array once and the lookup and insert operations into the hash map have a runtime of O(1). Although, we do have to use extra memory to create the hash map. The memory complexity is O(n) because we could potentially add every value to the hash map.