How do you solve Partition Labels on LeetCode?

Partition Labels (LeetCode #763) 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 the last occurrence of characters is crucial for determining valid partitions.

What is the brute force solution for Partition Labels?

The brute force approach for Partition Labels is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking every possible partition of the string and validating if each character appears in at most one part. This is straightforward but inefficient because it explores all combinations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Partition Labels?

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

What are the interview tips for Partition Labels?

Always clarify the problem requirements and constraints before diving into coding. Think about edge cases, such as strings with all unique characters or all identical characters. Practice explaining your thought process as you code; it helps in interviews.

What are common mistakes when solving Partition Labels?

Failing to track the last occurrence of characters, leading to incorrect partition sizes. Not updating the end index correctly during the iteration.

#763

Partition Labels

Medium

Hash TableTwo PointersStringGreedyHash 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 solution uses a greedy approach with a hashmap to track the last occurrence of each character. This allows us to expand our partition until we have included all instances of the characters in the current partition.

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Algorithm

4 steps

1Step 1: Create a hashmap to store the last index of each character in the string.
2Step 2: Initialize variables to track the current partition's end index and the start index.
3Step 3: Iterate through the string, updating the end index to the maximum last index of the characters seen so far.
4Step 4: When the current index matches the end index, record the size of the partition and update the start index.

solution.py13 lines

1# Full working Python code
2from collections import defaultdict
3
4def partition_labels(s):
5    last = {c: i for i, c in enumerate(s)}
6    partitions = []
7    start, end = 0, 0
8    for i, c in enumerate(s):
9        end = max(end, last[c])
10        if i == end:
11            partitions.append(end - start + 1)
12            start = i + 1
13    return partitions

ℹ

Complexity note: This complexity is linear because we make a single pass to record the last indices and another pass to determine the partitions, resulting in O(n) time overall.

1Understanding the last occurrence of characters is crucial for determining valid partitions.
2Greedy approaches can often lead to optimal solutions by making local optimal choices.

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