How do you solve Optimal Partition of String on LeetCode?

Optimal Partition of String (LeetCode #2405) 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: The problem can be solved efficiently by using a greedy approach that focuses on character uniqueness.

What is the brute force solution for Optimal Partition of String?

The brute force approach for Optimal Partition of String is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible partitions of the string and checking each partition to see if all characters in each substring are unique. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Optimal Partition of String?

Optimal Partition of String uses the following concepts: Hash Table, String, Greedy. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Optimal Partition of String?

Always explain your thought process while coding; it helps interviewers understand your approach. Consider edge cases and test your code with them during the interview. Practice explaining your solution clearly and concisely, as communication is key in interviews.

What are common mistakes when solving Optimal Partition of String?

Failing to clear the set when starting a new substring, which leads to incorrect counts. Not considering edge cases, such as strings with all identical characters.

#2405

Optimal Partition of String

Medium

Hash TableStringGreedyHash 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 greedy approach where we iterate through the string and keep track of characters we've seen in the current substring. When we encounter a duplicate character, we start a new substring.

⚙️

Algorithm

5 steps

1Step 1: Initialize a set to keep track of characters in the current substring.
2Step 2: Initialize a counter for the number of substrings.
3Step 3: Iterate through each character in the string. If the character is already in the set, increment the substring counter and clear the set.
4Step 4: Add the character to the set.
5Step 5: Return the total count of substrings.

solution.py12 lines

1# Full working Python code
2
3def min_partitions(s):
4    seen = set()
5    count = 1
6    for char in s:
7        if char in seen:
8            count += 1
9            seen.clear()
10        seen.add(char)
11    return count
12

ℹ

Complexity note: The time complexity is O(n) because we only make a single pass through the string, and the space complexity is O(n) due to the set used to track characters.

1The problem can be solved efficiently by using a greedy approach that focuses on character uniqueness.
2Tracking characters with a set allows for quick checks and updates, optimizing performance.

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