What is the brute force solution for Compare Strings by Frequency of the Smallest Character?

The brute force approach for Compare Strings by Frequency of the Smallest Character is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves calculating the frequency of the smallest character for each query and comparing it with every word in the list. This is straightforward but inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n log n + m log m).

What algorithm or data structure is used to solve Compare Strings by Frequency of the Smallest Character?

Compare Strings by Frequency of the Smallest Character uses the following concepts: Array, Hash Table, String, Binary Search, Sorting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Compare Strings by Frequency of the Smallest Character?

Always clarify the problem requirements and constraints before jumping into coding. Think about edge cases, such as strings with all identical characters. Practice explaining your thought process clearly while coding, as communication is key in interviews.

What are common mistakes when solving Compare Strings by Frequency of the Smallest Character?

Not calculating the frequency correctly, especially when dealing with multiple characters. Failing to sort the frequencies before performing binary search.

#1170

Compare Strings by Frequency of the Smallest Character

Medium

ArrayHash TableStringBinary SearchSortingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n log n + m log m)Space O(n)

By precomputing the frequencies of the smallest characters in the words and sorting them, we can efficiently answer each query using binary search. This reduces the number of comparisons needed.

⚙️

Algorithm

3 steps

1Step 1: Calculate the frequency of the smallest character for each word and store these frequencies in a list.
2Step 2: Sort the list of frequencies.
3Step 3: For each query, calculate its frequency and use binary search to find how many frequencies in the sorted list are greater than this frequency.

solution.py5 lines

1def f(s): return s.count(min(s))
2
3def numSmallerByFrequency(queries, words):
4    freq = sorted(f(w) for w in words)
5    return [len(freq) - bisect.bisect_right(freq, f(q)) for q in queries]

ℹ

Complexity note: The sorting step takes O(n log n) for the words and O(m log m) for the queries, where n is the number of words and m is the number of queries. The space complexity accounts for storing the frequencies.

1Understanding how to compute character frequencies is crucial for this problem.
2Sorting and binary search can significantly reduce the time complexity of the solution.

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