LeetCode 389: Find the Difference Solution in Python – A Step-by-Step Guide

What Is LeetCode 389: Find the Difference?

LeetCode 389: Find the Difference provides two strings, s and t, where t is a shuffled version of s with exactly one additional character. Your task is to return the character that was added to t. It’s a problem that tests your ability to compare strings and isolate differences efficiently!

Problem Statement

• Input:
• s: Original string.
• t: Shuffled string with one extra character.
• Output: Character - The added character in t.
• Rules:
• t is s shuffled with one character appended.
• Both strings contain lowercase English letters.

Constraints

• 0 <= s.length <= 1000
• t.length == s.length + 1
• s and t contain only lowercase letters.

Examples

• Input: s = "abcd", t = "abcde"
• Output: 'e'
• 'e' is the extra character in t.
• Input: s = "", t = "y"
• Output: 'y'
• Empty s, 'y' added to t.
• Input: s = "a", t = "aa"
• Output: 'a'
• Second 'a' is the extra character.

These show the difference hunt—let’s solve it!

Understanding the Problem: Spotting the Extra Character

To tackle LeetCode 389 in Python, we need to: 1. Identify the single character in t that’s not part of the original s. 2. Account for t being a shuffled version of s plus one extra character. 3. Find an efficient method given the string lengths.

A naive approach might compare each character, but that’s inefficient. We’ll use:

• Best Solution: XOR—O(n) time, O(1) space—fast and clever.
• Alternative Solution: Hash map—O(n) time, O(n) space—intuitive but less space-efficient.

Let’s spot it with the best solution!

Best Solution: XOR

Why This Is the Best Solution

The XOR approach runs in O(n) time (n is length of t) with O(1) space by leveraging the XOR operation’s property: XORing a number with itself cancels it out (0), and XORing with 0 leaves it unchanged. It’s the most efficient—finding the extra character in one pass without extra storage beyond a single variable!

How It Works

Think of it like a magic trick:

• XOR Property: a ^ a = 0, a ^ 0 = a, XOR is associative.
• Process:
• XOR all characters in s and t.
• Paired characters in s and t cancel out (e.g., 'a' ^ 'a' = 0).
• The extra character in t remains (e.g., 0 ^ 'e' = 'e').
• Result: The leftover value is the added character.

It’s like canceling out twins to find the singleton!

Step-by-Step Example

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

• Init: result = 0.
• XOR s:
• 0 ^ 'a' = 'a'.
• 'a' ^ 'b' = (97 ^ 98 = 3).
• 3 ^ 'c' = (3 ^ 99 = 96).
• 96 ^ 'd' = (96 ^ 100 = 4).
• XOR t:
• 4 ^ 'a' = (4 ^ 97 = 101).
• 101 ^ 'b' = (101 ^ 98 = 7).
• 7 ^ 'c' = (7 ^ 99 = 100).
• 100 ^ 'd' = (100 ^ 100 = 0).
• 0 ^ 'e' = (0 ^ 101 = 101 = 'e').
• Result: 'e'.

For s = "", t = "y":

• XOR s: result = 0 (empty).
• XOR t: 0 ^ 'y' = (0 ^ 121 = 121 = 'y').
• Result: 'y'.

For s = "a", t = "aa":

• XOR s: 0 ^ 'a' = 97.
• XOR t:
• 97 ^ 'a' = (97 ^ 97 = 0).
• 0 ^ 'a' = (0 ^ 97 = 97 = 'a').
• Result: 'a'.

Code with Explanation

Here’s the Python code using XOR:

Copy codeclass Solution:
    def findTheDifference(self, s: str, t: str) -> str:
        # Initialize result with 0
        result = 0

        # XOR all characters in s
        for char in s:
            result ^= ord(char)

        # XOR all characters in t
        for char in t:
            result ^= ord(char)

        # Convert back to character
        return chr(result)

Explanation

• Line 4: result: Start with 0 for XOR.
• Lines 6-7: XOR each char in s (ord converts to ASCII)—O(n).
• Lines 9-10: XOR each char in t—O(m), where m = n + 1.
• Line 12: Convert final XOR result back to char—O(1).
• Time: O(n)—single pass over both strings (n is length of s).
• Space: O(1)—single integer variable.

This is like a bitwise magician—fast and lean!

Alternative Solution: Hash Map

Why an Alternative Approach?

The hash map solution also runs in O(n) time but uses O(n) space by counting character frequencies in both strings, then finding the character with a count difference. It’s more intuitive—great for understanding frequency-based comparison, though it uses more space than XOR!

How It Works

• Hash Map: Count frequencies in s and t.
• Compare: Find char in t with count one more than in s.
• Result: Return that char.

Step-by-Step Example

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

• Count s: {'a': 1, 'b': 1, 'c': 1, 'd': 1}.
• Count t: {'a': 1, 'b': 1, 'c': 1, 'd': 1, 'e': 1}.
• Compare:
• 'e': t=1, s=0, t-s=1, return 'e'.
• Result: 'e'.

For s = "", t = "y":

• Count s: {}.
• Count t: {'y': 1}.
• Compare: 'y': t=1, s=0, return 'y'.
• Result: 'y'.

Code with Explanation

Copy codeclass Solution:
    def findTheDifference(self, s: str, t: str) -> str:
        # Hash maps for frequencies
        s_count = {}
        t_count = {}

        # Count frequencies in s
        for char in s:
            s_count[char] = s_count.get(char, 0) + 1

        # Count frequencies in t
        for char in t:
            t_count[char] = t_count.get(char, 0) + 1

        # Find the extra character
        for char in t_count:
            if char not in s_count or t_count[char] > s_count[char]:
                return char

        return ""

Explanation

• Lines 4-5: s_count, t_count: Dicts for frequencies.
• Lines 7-8: Count chars in s—O(n).
• Lines 10-11: Count chars in t—O(n+1).
• Lines 13-16: Find char with higher count in t—O(n).
• Line 18: Fallback (shouldn’t occur)—O(1).
• Time: O(n)—linear passes.
• Space: O(n)—two hash maps.

It’s a frequency spotter—clear but bulkier!

Comparing the Two Solutions

• XOR (Best):
• Time: O(n)—linear.
• Space: O(1)—minimal.
• Pros: Fast, space-efficient, clever.
• Cons: Less intuitive.
• Hash Map (Alternative):
• Time: O(n)—linear.
• Space: O(n)—maps.
• Pros: Easy to understand.
• Cons: More space, two structures.

XOR wins for speed and space!

Additional Examples and Edge Cases

• s="a", t="ab": Both return 'b'.
• s="abc", t="cabd": Both return 'd'.
• Empty s: Both handle correctly.

XOR is optimal, hash map works too.

Complexity Breakdown

• XOR:
• Time: O(n).
• Space: O(1).
• Hash Map:
• Time: O(n).
• Space: O(n).

XOR excels for efficiency.

Key Takeaways

• XOR: Magic bit trick—fast and lean!
• Hash Map: Count and compare—clear but heavy!
• Difference: Efficiency meets intuition.
• Python Tip: Bit ops rock—see [Python Basics](/python/basics).

Final Thoughts: Find That Extra Char

LeetCode 389: Find the Difference in Python is a string difference challenge. XOR is the efficiency champ, while hash map offers a straightforward alternative. Want more? Try LeetCode 387: First Unique Character or LeetCode 136: Single Number. Ready to spot? Visit Solve LeetCode 389 on LeetCode and uncover that char today!