How do you solve Number of Good Ways to Split a String on LeetCode?

Number of Good Ways to Split a String (LeetCode #1525) 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 hash maps allows efficient counting of distinct characters.

What is the brute force solution for Number of Good Ways to Split a String?

The brute force approach for Number of Good Ways to Split a String is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible split of the string and counting the distinct characters in both parts. This is straightforward but inefficient for larger strings. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Number of Good Ways to Split a String?

Number of Good Ways to Split a String uses the following concepts: Hash Table, String, Dynamic Programming, Bit Manipulation. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Number of Good Ways to Split a String?

Explain your thought process clearly while coding. Consider edge cases and discuss them with the interviewer. Practice writing clean and efficient code to save time.

What are common mistakes when solving Number of Good Ways to Split a String?

Not considering the case where the string length is less than 2. Forgetting to update the right hash map correctly.

#1525

Number of Good Ways to Split a String

Medium

Hash TableStringDynamic ProgrammingBit ManipulationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

The optimal approach uses two hash maps to keep track of the distinct character counts for the left and right parts of the string as we iterate through it. This allows us to efficiently compare counts without needing to recreate substrings.

⚙️

Algorithm

5 steps

1Step 1: Initialize two hash maps to count characters in the left and right parts of the string.
2Step 2: Populate the right hash map with all characters from the string.
3Step 3: Iterate through the string, updating the left hash map and removing characters from the right hash map as you go.
4Step 4: After each character addition, compare the sizes of the two hash maps. If they are equal, increment the good split count.
5Step 5: Return the total count of good splits.

solution.py14 lines

1def numSplits(s):
2    left_count, right_count = {}, {}
3    for char in s:
4        right_count[char] = right_count.get(char, 0) + 1
5    good_splits = 0
6    for i in range(len(s) - 1):
7        char = s[i]
8        left_count[char] = left_count.get(char, 0) + 1
9        right_count[char] -= 1
10        if right_count[char] == 0:
11            del right_count[char]
12        if len(left_count) == len(right_count):
13            good_splits += 1
14    return good_splits

ℹ

Complexity note: The time complexity is O(n) because we make a single pass through the string to populate the right hash map and another pass to check splits. The space complexity is O(n) due to the storage of character counts in the hash maps.

1Using hash maps allows efficient counting of distinct characters.
2Iterating through the string while updating counts helps avoid redundant calculations.

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