How do you solve Find the Difference on LeetCode?

Find the Difference (LeetCode #389) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a frequency map can significantly reduce the time complexity.

What is the brute force solution for Find the Difference?

The brute force approach for Find the Difference is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking each character in string t to see if it exists in string s. If a character from t is not found in s, that character is the one that was added. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find the Difference?

Find the Difference uses the following concepts: Hash Table, String, Bit Manipulation, Sorting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Find the Difference?

Always clarify the problem statement and constraints before jumping into coding. Think about the time and space complexity of your solution and be ready to explain it. Practice dry-running your code with examples to catch potential errors.

What are common mistakes when solving Find the Difference?

Not considering edge cases like empty strings. Overlooking the need to check for character counts in the optimal solution.

#389

Find the Difference

Easy

Hash TableStringBit ManipulationSortingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a HashMap (or dictionary) to count the occurrences of each character in string s, then decrements the count for each character found in t. The character that remains with a count of 1 is the added character.

⚙️

Algorithm

3 steps

1Step 1: Create a frequency map for characters in string s.
2Step 2: Iterate through each character in string t, decrementing the count in the frequency map.
3Step 3: The character with a count of 1 in the frequency map is the added character.

solution.py9 lines

1from collections import Counter
2s_counter = Counter(s)
3for char in t:
4    if char in s_counter:
5        s_counter[char] -= 1
6        if s_counter[char] == 0:
7            del s_counter[char]
8    else:
9        print(char)

ℹ

Complexity note: The time complexity is O(n) because we traverse both strings once. The space complexity is O(n) due to the storage of character counts in the HashMap.

1Using a frequency map can significantly reduce the time complexity.
2Understanding how to manipulate character counts is crucial for string problems.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.