Leetcode #2213: Longest Substring of One Repeating Character
In this guide, we solve Leetcode #2213 Longest Substring of One Repeating Character in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given a 0-indexed string s. You are also given a 0-indexed string queryCharacters of length k and a 0-indexed array of integer indices queryIndices of length k, both of which are used to describe k queries.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Segment Tree, Array, String, Ordered Set
Intuition
We need to scan characters while tracking positions or counts.
A simple state machine keeps the logic precise.
Approach
Iterate through the string once and update the state for each character.
Use a map or array if you need fast lookups.
Steps:
- Iterate through characters.
- Maintain necessary state.
- Build or validate the output.
Example
Input: s = "babacc", queryCharacters = "bcb", queryIndices = [1,3,3]
Output: [3,3,4]
Explanation:
- 1st query updates s = "bbbacc". The longest substring consisting of one repeating character is "bbb" with length 3.
- 2nd query updates s = "bbbccc".
The longest substring consisting of one repeating character can be "bbb" or "ccc" with length 3.
- 3rd query updates s = "bbbbcc". The longest substring consisting of one repeating character is "bbbb" with length 4.
Thus, we return [3,3,4].
Python Solution
def max(a: int, b: int) -> int:
return a if a > b else b
class Node:
__slots__ = "l", "r", "lmx", "rmx", "mx"
def __init__(self, l: int, r: int):
self.l = l
self.r = r
self.lmx = self.rmx = self.mx = 1
class SegmentTree:
__slots__ = "s", "tr"
def __init__(self, s: str):
self.s = list(s)
n = len(s)
self.tr: List[Node | None] = [None] * (n * 4)
self.build(1, 1, n)
def build(self, u: int, l: int, r: int):
self.tr[u] = Node(l, r)
if l == r:
return
mid = (l + r) // 2
self.build(u << 1, l, mid)
self.build(u << 1 | 1, mid + 1, r)
self.pushup(u)
def query(self, u: int, l: int, r: int) -> int:
if self.tr[u].l >= l and self.tr[u].r <= r:
return self.tr[u].mx
mid = (self.tr[u].l + self.tr[u].r) // 2
ans = 0
if r <= mid:
ans = self.query(u << 1, l, r)
if l > mid:
ans = max(ans, self.query(u << 1 | 1, l, r))
return ans
def modify(self, u: int, x: int, v: str):
if self.tr[u].l == self.tr[u].r:
self.s[x - 1] = v
return
mid = (self.tr[u].l + self.tr[u].r) // 2
if x <= mid:
self.modify(u << 1, x, v)
else:
self.modify(u << 1 | 1, x, v)
self.pushup(u)
def pushup(self, u: int):
root, left, right = self.tr[u], self.tr[u << 1], self.tr[u << 1 | 1]
root.lmx = left.lmx
root.rmx = right.rmx
root.mx = max(left.mx, right.mx)
a, b = left.r - left.l + 1, right.r - right.l + 1
if self.s[left.r - 1] == self.s[right.l - 1]:
if left.lmx == a:
root.lmx += right.lmx
if right.rmx == b:
root.rmx += left.rmx
root.mx = max(root.mx, left.rmx + right.lmx)
class Solution:
def longestRepeating(
self, s: str, queryCharacters: str, queryIndices: List[int]
) -> List[int]:
tree = SegmentTree(s)
ans = []
for x, v in zip(queryIndices, queryCharacters):
tree.modify(1, x + 1, v)
ans.append(tree.query(1, 1, len(s)))
return ans
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.