#389

Easy

LeetCode Problem Workspace

Find the Difference

Find the Difference involves identifying the additional letter in one string compared to another, emphasizing hash table and string manipulations.

hash table string bit manipulation sorting

Problem Statement

You are given two strings, s and t. String t is created by shuffling string s and adding one extra letter at a random position. Your task is to return the extra letter in t.

The extra letter is guaranteed to be a single character, and you must determine it efficiently. The problem involves string manipulation and may require utilizing a hash table or bit manipulation techniques to solve.

Examples

Example 1

Input: s = "abcd", t = "abcde"

Output: "e"

'e' is the letter that was added.

Example 2

Input: s = "", t = "y"

Output: "y"

Example details omitted.

Constraints

0 <= s.length <= 1000
t.length == s.length + 1
s and t consist of lowercase English letters.

Solution Approach

Hash Table Approach

By iterating through string t and counting the frequency of each character, you can find the letter that occurs once in t but not in s. This method relies on the hash table to store the frequency of characters and compare the counts between s and t.

Bit Manipulation (XOR)

A more efficient solution uses XOR to find the difference. By XOR-ing all characters of both s and t, the result will be the additional character, as identical characters will cancel each other out. This approach requires O(n) time and O(1) space.

Sorting

Sort both strings and compare them character by character. The first mismatched character will be the extra one in t. Although this solution works, it has a higher time complexity due to sorting, making it less optimal than the other methods.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity for the hash table approach is O(n), while the XOR method also runs in O(n) time but uses O(1) space. Sorting-based solutions, however, run in O(n log n) time, which is less efficient than the linear time methods.

What Interviewers Usually Probe

The candidate chooses a hash table or XOR-based approach for linear time performance.
The candidate uses sorting without considering time complexity trade-offs.
The candidate effectively explains the space-time trade-off between the hash table and XOR solutions.

Common Pitfalls or Variants

Common pitfalls

Using sorting without considering its time complexity can lead to suboptimal solutions.
Not recognizing that XOR can efficiently solve the problem in O(1) space.
Overcomplicating the solution with unnecessary steps when a simple hash table or XOR solution suffices.

Follow-up variants

The problem could involve finding the missing character from a set of strings, rather than two strings.
An extension could involve multiple additional characters being added to t, increasing the complexity.
Consider a scenario where both strings are very large, testing the scalability of different approaches.

FAQ

What is the optimal approach for the Find the Difference problem?

The XOR method is optimal because it runs in linear time and constant space, making it faster and more space-efficient than other methods like sorting or using a hash table.

How does XOR solve the Find the Difference problem?

XOR-ing all the characters in both strings will cancel out the matching characters, leaving the extra character as the result. This technique ensures a time complexity of O(n) and space complexity of O(1).

Can I solve this problem with just a hash table?

Yes, using a hash table is a valid approach. By counting the frequency of each character in both strings, you can identify the extra letter in O(n) time, but this method uses extra space for the hash table.

What are the limitations of sorting as a solution?

Sorting both strings has a time complexity of O(n log n), which is less efficient compared to the linear time solutions using XOR or hash tables.

What other variants could this problem have?

This problem can be extended by adding more characters to string t, testing scalability, or considering larger strings that might require optimization techniques.

terminal

Solution

Solution 1: Counting

We can use a hash table or array $cnt$ to count the occurrence of each character in string $s$, then traverse string $t$. For each character, we subtract its occurrence in $cnt$. If the corresponding count is negative, it means that the occurrence of this character in $t$ is greater than in $s$, so this character is the added character.

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class Solution:
    def findTheDifference(self, s: str, t: str) -> str:
        cnt = Counter(s)
        for c in t:
            cnt[c] -= 1
            if cnt[c] < 0:
                return c

Solution 2: Summation

We can sum the ASCII values of each character in string $t$, then subtract the sum of the ASCII values of each character in string $s$. The final result is the ASCII value of the added character.

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class Solution:
    def findTheDifference(self, s: str, t: str) -> str:
        cnt = Counter(s)
        for c in t:
            cnt[c] -= 1
            if cnt[c] < 0:
                return c

Continue Practicing

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