Find the Difference

You can find the letter that was added to string t by using the concept of character frequency. The idea is to count the frequency of characters in both strings s and t and then find the character that has one more occurrence in t than in s.

Here’s the code to do that:

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from collections import Counter

class Solution:
    def findTheDifference(self, s: str, t: str) -> str:
        # Count the frequency of characters in both s and t
        count_s = Counter(s)
        count_t = Counter(t)

        # Find the character that has one more occurrence in t than in s
        for char, count in count_t.items():
            if count_s[char] != count:
                return char

Alternatively, you could simply use the Counter’s subtraction method to subtract the count of characters in s from the count in t. The remaining item(s) in the result will be the added character.

Here’s a more concise version using this approach:

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from collections import Counter

class Solution:
    def findTheDifference(self, s: str, t: str) -> str:
        # Subtract the frequency of characters in s from the frequency in t
        difference = Counter(t) - Counter(s)

        # The remaining character is the one that was added
        return list(difference.keys())[0]

Both of these solutions adhere to the given constraints, with a time complexity of O(n), where n is the length of the strings s and t.

Identifying Problem Isomorphism

“Find the Difference” can be mapped to the following problems:

  1. “Single Number” (#136) on LeetCode
  2. “Single Number II” (#137) on LeetCode
  3. “Single Number III” (#260) on LeetCode

Reasoning:

All these problems have to deal with finding unique elements in the given input - an array in the case of “Single Number” problems and strings for the “Find the Difference” problem.

  1. “Single Number” is the simplest problem, asking to find the single number that appears once in an array where every other number appears twice. This problem can be considered an easier version of the “Find the Difference” problem as you are looking for the element that doesn’t follow the general rule.

  2. “Single Number II” takes this concept a step further by asking for the single number that appears once in an array where every other number appears three times. This is a more complex problem than “Single Number” as the array contains a higher level of repetitions.

  3. “Single Number III” asks for two numbers that appear once in an array where every other number appears twice. This problem is the most complex among the listed problems as it requires finding more than one unique element in the array.

While these problems are not exactly the same, they all involve identifying elements that break a pattern of repetition. The problem-solving skills developed in solving one of these problems can be helpful in tackling the others.

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class Solution:
    def findTheDifference(self, s: str, t: str) -> str:
        for i in t:
            if s.count(i) != t.count(i):
                return i

Problem Classification

  1. Strings Manipulation: The problem involves working with strings, where string t is a shuffled version of string s with one extra letter.

  2. Character Comparison: We need to find the extra character that was added to string t, which requires comparison of characters between two strings.

  3. Randomness: The problem introduces randomness in shuffling string s and adding an extra character to create string t.

  4. Searching: The problem essentially requires searching for the additional character in string t.

  5. Counting: The problem can be solved by counting the occurrences of characters in both strings and finding the one that is different.

This problem can be classified as a string manipulation and search problem with elements of counting.

Language Agnostic Coding Drills

  1. String Manipulation: Understanding how strings are manipulated is crucial for solving this problem. In particular, you need to understand how to traverse a string and access individual characters. This is a fundamental aspect of string manipulation in most languages.

  2. Character Counting: Another important concept is understanding how to count the occurrences of each character in a string. In many languages, this can be achieved by using a for loop to traverse the string, and using a dictionary or an array to keep track of the count of each character.

  3. Dictionary/Hash Map Manipulation: This problem involves using a dictionary or hash map to store the counts of characters. You need to understand how to insert items into a dictionary, retrieve items, and check if an item exists in a dictionary.

  4. Character Comparison: The problem involves comparing the counts of characters between two strings. This requires understanding of how to compare values in a language-agnostic way.

  5. Looping: This problem involves looping over the strings and their individual characters, which requires understanding of how to implement and control loops in your language of choice.

The step by step problem-solving approach for this problem would be:

  1. Start by traversing the second string t.

  2. For each character in t, count the occurrences of this character in both s and t.

  3. Compare the counts. If the counts are not the same, this means that the character is the extra character added in t.

  4. Return the extra character.

Targeted Drills in Python

  1. String Manipulation: Write a function that receives a string as input and prints each character separately.
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def print_chars(s):
    for c in s:
        print(c)

print_chars("hello")
  1. Character Counting: Write a function that counts the occurrences of each character in a string. Return a dictionary with characters as keys and counts as values.
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def count_chars(s):
    count_dict = {}
    for c in s:
        if c in count_dict:
            count_dict[c] += 1
        else:
            count_dict[c] = 1
    return count_dict

print(count_chars("hello"))
  1. Dictionary/Hash Map Manipulation: Write a function that receives a dictionary and a key, and returns the value for the given key if it exists, or None otherwise.
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def get_value(dictionary, key):
    return dictionary.get(key)

print(get_value({'a': 1, 'b': 2}, 'a'))
  1. Character Comparison: Write a function that receives two characters and prints “Same” if they are the same or “Different” otherwise.
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def compare_chars(c1, c2):
    if c1 == c2:
        print("Same")
    else:
        print("Different")

compare_chars('a', 'a')
compare_chars('a', 'b')
  1. Looping: Write a function that receives a list of integers and prints each number in the list.
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def print_numbers(numbers):
    for num in numbers:
        print(num)

print_numbers([1, 2, 3, 4, 5])

After you’ve become comfortable with these drills, you can try combining these concepts to solve the problem. Remember that the problem is asking you to compare the counts of characters between two strings, and find the character that occurs more times in the second string. This means you’ll need to use string manipulation, character counting, dictionary manipulation, character comparison, and looping in your solution.