How do you solve Longest Substring of One Repeating Character on LeetCode?

Longest Substring of One Repeating Character (LeetCode #2213) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(log n) time complexity and O(n) space complexity. Key insight: Understanding how to efficiently update and query data structures is crucial for performance.

What is the time complexity of Longest Substring of One Repeating Character?

The optimal solution for Longest Substring of One Repeating Character runs in O(log n) time and uses O(n) space. The segment tree allows for efficient updates and queries, reducing the time complexity significantly compared to the brute force approach.

What is the brute force solution for Longest Substring of One Repeating Character?

The brute force approach for Longest Substring of One Repeating Character is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking all possible substrings after each query to find the longest substring of repeating characters. This is straightforward but inefficient, especially for larger strings. The optimal Optimal Solution approach improves this to O(log n).

What algorithm or data structure is used to solve Longest Substring of One Repeating Character?

Longest Substring of One Repeating Character uses the following concepts: Array, String, Segment Tree, Ordered Set. The recommended patterns to study are: Segment Tree, Dynamic Programming.

What are the interview tips for Longest Substring of One Repeating Character?

Always explain your thought process clearly while coding. Consider edge cases and test them during your explanation. Practice building and using data structures like segment trees to become comfortable with their operations.

What are common mistakes when solving Longest Substring of One Repeating Character?

Failing to consider the impact of updates on the overall structure and not updating the segment tree correctly. Not handling edge cases, such as when the query index is at the boundaries of the string.

#2213

Longest Substring of One Repeating Character

Hard

ArrayStringSegment TreeOrdered SetSegment TreeDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(log n)Space O(n)

Using a segment tree allows us to efficiently update the string and query the longest substring of repeating characters in logarithmic time, making it suitable for multiple updates.

⚙️

Algorithm

3 steps

1Step 1: Build a segment tree where each node stores the length of the longest substring of repeating characters for that segment.
2Step 2: For each query, update the character in the string and update the corresponding segment tree nodes.
3Step 3: After each update, query the segment tree to get the longest substring length.

solution.py43 lines

1class SegmentTree:
2    def __init__(self, s):
3        self.n = len(s)
4        self.tree = [0] * (4 * self.n)
5        self.build(s, 0, 0, self.n - 1)
6
7    def build(self, s, node, start, end):
8        if start == end:
9            self.tree[node] = 1
10        else:
11            mid = (start + end) // 2
12            self.build(s, 2 * node + 1, start, mid)
13            self.build(s, 2 * node + 2, mid + 1, end)
14            self.tree[node] = max(self.tree[2 * node + 1], self.tree[2 * node + 2])
15
16    def update(self, idx, value, node, start, end):
17        if start == end:
18            self.tree[node] = value
19        else:
20            mid = (start + end) // 2
21            if start <= idx <= mid:
22                self.update(idx, value, 2 * node + 1, start, mid)
23            else:
24                self.update(idx, value, 2 * node + 2, mid + 1, end)
25            self.tree[node] = max(self.tree[2 * node + 1], self.tree[2 * node + 2])
26
27    def query(self, L, R, node, start, end):
28        if R < start or end < L:
29            return 0
30        if L <= start and end <= R:
31            return self.tree[node]
32        mid = (start + end) // 2
33        left_query = self.query(L, R, 2 * node + 1, start, mid)
34        right_query = self.query(L, R, 2 * node + 2, mid + 1, end)
35        return max(left_query, right_query)
36
37def longestRepeatingSubstring(s, queryCharacters, queryIndices):
38    st = SegmentTree(s)
39    lengths = []
40    for i in range(len(queryCharacters)):
41        st.update(queryIndices[i], queryCharacters[i], 0, 0, len(s) - 1)
42        lengths.append(st.query(0, len(s) - 1, 0, 0, len(s) - 1))
43    return lengths

ℹ

Complexity note: The segment tree allows for efficient updates and queries, reducing the time complexity significantly compared to the brute force approach.

1Understanding how to efficiently update and query data structures is crucial for performance.
2Segment trees can significantly reduce the time complexity of problems involving dynamic updates.

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