LeetCode #2213 — HARD

Longest Substring of One Repeating Character

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode
The Problem

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.

The ith query updates the character in s at index queryIndices[i] to the character queryCharacters[i].

Return an array lengths of length k where lengths[i] is the length of the longest substring of s consisting of only one repeating character after the ith query is performed.

Example 1:

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].

Example 2:

Input: s = "abyzz", queryCharacters = "aa", queryIndices = [2,1]
Output: [2,3]
Explanation:
- 1st query updates s = "abazz". The longest substring consisting of one repeating character is "zz" with length 2.
- 2nd query updates s = "aaazz". The longest substring consisting of one repeating character is "aaa" with length 3.
Thus, we return [2,3].

Constraints:

  • 1 <= s.length <= 105
  • s consists of lowercase English letters.
  • k == queryCharacters.length == queryIndices.length
  • 1 <= k <= 105
  • queryCharacters consists of lowercase English letters.
  • 0 <= queryIndices[i] < s.length
Patterns Used
📐Segment Tree→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: 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. The ith query updates the character in s at index queryIndices[i] to the character queryCharacters[i]. Return an array lengths of length k where lengths[i] is the length of the longest substring of s consisting of only one repeating character after the ith query is performed.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Segment Tree

Example 1

"babacc"
"bcb"
[1,3,3]

Example 2

"abyzz"
"aa"
[2,1]

Related Problems

  • Merge Intervals (merge-intervals)
  • Longest Repeating Character Replacement (longest-repeating-character-replacement)
  • Consecutive Characters (consecutive-characters)
  • Create Sorted Array through Instructions (create-sorted-array-through-instructions)
  • Longest Increasing Subsequence II (longest-increasing-subsequence-ii)
Step 02

Core Insight

What unlocks the optimal approach

  • Use a segment tree to perform fast point updates and range queries.
  • We need each segment tree node to store the length of the longest substring of that segment consisting of only 1 repeating character.
  • We will also have each segment tree node store the leftmost and rightmost character of the segment, the max length of a prefix substring consisting of only 1 repeating character, and the max length of a suffix substring consisting of only 1 repeating character.
  • Use this information to properly merge the two segment tree nodes together.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Largest constraint values
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2213: Longest Substring of One Repeating Character
class Node {
    int l, r;
    int lmx, rmx, mx;

    Node(int l, int r) {
        this.l = l;
        this.r = r;
        lmx = rmx = mx = 1;
    }
}

class SegmentTree {
    private char[] s;
    private Node[] tr;

    public SegmentTree(char[] s) {
        int n = s.length;
        this.s = s;
        tr = new Node[n << 2];
        build(1, 1, n);
    }

    public void build(int u, int l, int r) {
        tr[u] = new Node(l, r);
        if (l == r) {
            return;
        }
        int mid = (l + r) >> 1;
        build(u << 1, l, mid);
        build(u << 1 | 1, mid + 1, r);
        pushup(u);
    }

    public void modify(int u, int x, char v) {
        if (tr[u].l == x && tr[u].r == x) {
            s[x - 1] = v;
            return;
        }
        int mid = (tr[u].l + tr[u].r) >> 1;
        if (x <= mid) {
            modify(u << 1, x, v);
        } else {
            modify(u << 1 | 1, x, v);
        }
        pushup(u);
    }

    public int query(int u, int l, int r) {
        if (tr[u].l >= l && tr[u].r <= r) {
            return tr[u].mx;
        }
        int mid = (tr[u].l + tr[u].r) >> 1;
        int ans = 0;
        if (r <= mid) {
            ans = query(u << 1, l, r);
        }
        if (l > mid) {
            ans = Math.max(ans, query(u << 1 | 1, l, r));
        }
        return ans;
    }

    private void pushup(int u) {
        Node root = tr[u];
        Node left = tr[u << 1], right = tr[u << 1 | 1];
        root.mx = Math.max(left.mx, right.mx);
        root.lmx = left.lmx;
        root.rmx = right.rmx;
        int a = left.r - left.l + 1;
        int b = right.r - right.l + 1;
        if (s[left.r - 1] == s[right.l - 1]) {
            if (left.lmx == a) {
                root.lmx += right.lmx;
            }
            if (right.rmx == b) {
                root.rmx += left.rmx;
            }
            root.mx = Math.max(root.mx, left.rmx + right.lmx);
        }
    }
}

class Solution {
    public int[] longestRepeating(String s, String queryCharacters, int[] queryIndices) {
        SegmentTree tree = new SegmentTree(s.toCharArray());
        int k = queryIndices.length;
        int[] ans = new int[k];
        int n = s.length();
        for (int i = 0; i < k; ++i) {
            int x = queryIndices[i] + 1;
            char v = queryCharacters.charAt(i);
            tree.modify(1, x, v);
            ans[i] = tree.query(1, 1, n);
        }
        return ans;
    }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n × log n)
Space
O(n × log n)

Approach Breakdown

BRUTE FORCE
O(n × q) time
O(1) space

For each of q queries, scan the entire range to compute the aggregate (sum, min, max). Each range scan takes up to O(n) for a full-array query. With q queries: O(n × q) total. Point updates are O(1) but queries dominate.

SEGMENT TREE
O(n + q log n) time
O(n) space

Building the tree is O(n). Each query or update traverses O(log n) nodes (tree height). For q queries: O(n + q log n) total. Space is O(4n) ≈ O(n) for the tree array. Lazy propagation adds O(1) per node for deferred updates.

Shortcut: Build O(n), query/update O(log n) each. When you need both range queries AND point updates.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Related Problems
#218The Skyline Problemhard→#220Contains Duplicate IIIhard→#307Range Sum Query - Mutablemedium→#315Count of Smaller Numbers After Selfhard→#327Count of Range Sumhard→