How do you solve Odd String Difference on LeetCode?

Odd String Difference (LeetCode #2451) 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: Understanding how to calculate the difference array is crucial.

What is the time complexity of Odd String Difference?

The optimal solution for Odd String Difference runs in O(n) time and uses O(n) space. This complexity is due to a single pass through the words to build the hash map (O(n)) and another pass to find the unique string (O(n)).

What is the brute force solution for Odd String Difference?

The brute force approach for Odd String Difference is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves calculating the difference integer array for each string and then comparing these arrays to find the one that is different. This method is straightforward but inefficient as it checks all pairs. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Odd String Difference?

Odd String Difference uses the following concepts: Array, Hash Table, String. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Odd String Difference?

Explain your thought process clearly while coding. Consider edge cases and constraints upfront. Practice writing clean and efficient code.

What are common mistakes when solving Odd String Difference?

Not correctly calculating the difference array. Overlooking the need to check for uniqueness in the difference arrays.

#2451

Odd String Difference

Easy

ArrayHash TableStringHash 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 approach uses a hash map to count the occurrences of each difference array. This allows us to find the odd string in a single pass through the list, making it much more efficient.

⚙️

Algorithm

4 steps

1Step 1: Create a hash map to store the frequency of each difference array.
2Step 2: For each string, calculate its difference array and update the hash map.
3Step 3: Iterate through the strings again to find the one with a unique difference array in the hash map.
4Step 4: Return the corresponding string.

solution.py14 lines

1# Full working Python code
2
3def oddString(words):
4    def get_diff_array(s):
5        return [ord(s[i+1]) - ord(s[i]) for i in range(len(s)-1)]
6
7    diff_count = {}
8    for word in words:
9        diff = tuple(get_diff_array(word))
10        diff_count[diff] = diff_count.get(diff, 0) + 1
11
12    for word in words:
13        if diff_count[tuple(get_diff_array(word))] == 1:
14            return word

ℹ

Complexity note: This complexity is due to a single pass through the words to build the hash map (O(n)) and another pass to find the unique string (O(n)).

1Understanding how to calculate the difference array is crucial.
2Using a hash map can significantly reduce time complexity.

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