How do you solve Select K Disjoint Special Substrings on LeetCode?

Select K Disjoint Special Substrings (LeetCode #3458) 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: Each special substring must have unique characters not found elsewhere.

What is the time complexity of Select K Disjoint Special Substrings?

The optimal solution for Select K Disjoint Special Substrings runs in O(n) time and uses O(n) space. We only traverse the string a few times, leading to O(n) time complexity.

What is the brute force solution for Select K Disjoint Special Substrings?

The brute force approach for Select K Disjoint Special Substrings is Brute Force, with O(n²) time complexity and O(1) space complexity. We can generate all possible substrings and check if they are special. This involves checking overlaps and character occurrences outside the substring. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Select K Disjoint Special Substrings?

Select K Disjoint Special Substrings uses the following concepts: Hash Table, String, Dynamic Programming, Greedy, Sorting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Select K Disjoint Special Substrings?

Clarify the definition of special substrings. Discuss edge cases, such as strings with all unique characters. Use examples to illustrate your thought process.

What are common mistakes when solving Select K Disjoint Special Substrings?

Not considering the case where characters repeat in the substring. Confusing overlapping intervals with disjoint ones.

#3458

Select K Disjoint Special Substrings

Medium

Hash TableStringDynamic ProgrammingGreedySortingHash 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)

Identify character boundaries to create special substrings efficiently. Use a greedy approach to select the maximum number of disjoint special substrings.

⚙️

Algorithm

3 steps

1Step 1: Track first and last occurrences of each character in the string.
2Step 2: Create intervals for each character based on its first and last occurrence.
3Step 3: Sort intervals and use a greedy approach to count disjoint intervals, ensuring no overlaps.

solution.py14 lines

1def canSelectKDisjointSpecialSubstrings(s, k):
2    first, last = {}, {}
3    for i, c in enumerate(s):
4        if c not in first:
5            first[c] = i
6        last[c] = i
7    intervals = [(first[c], last[c]) for c in first if first[c] != last[c]]
8    intervals.sort(key=lambda x: x[1])
9    count, end = 0, -1
10    for start, finish in intervals:
11        if start > end:
12            count += 1
13            end = finish
14    return count >= k

ℹ

Complexity note: We only traverse the string a few times, leading to O(n) time complexity.

1Each special substring must have unique characters not found elsewhere.
2Disjoint intervals can be efficiently counted using sorting.

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