LeetCode #2981 — MEDIUM

Find Longest Special Substring That Occurs Thrice I

Move from brute-force thinking to an efficient approach using hash map strategy.

The Problem

Problem Statement

You are given a string s that consists of lowercase English letters.

A string is called special if it is made up of only a single character. For example, the string "abc" is not special, whereas the strings "ddd", "zz", and "f" are special.

Return the length of the longest special substring of s which occurs at least thrice, or -1 if no special substring occurs at least thrice.

A substring is a contiguous non-empty sequence of characters within a string.

Example 1:

Input: s = "aaaa"
Output: 2
Explanation: The longest special substring which occurs thrice is "aa": substrings "aaaa", "aaaa", and "aaaa".
It can be shown that the maximum length achievable is 2.

Example 2:

Input: s = "abcdef"
Output: -1
Explanation: There exists no special substring which occurs at least thrice. Hence return -1.

Example 3:

Input: s = "abcaba"
Output: 1
Explanation: The longest special substring which occurs thrice is "a": substrings "abcaba", "abcaba", and "abcaba".
It can be shown that the maximum length achievable is 1.

Constraints:

3 <= s.length <= 50
s consists of only lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: You are given a string s that consists of lowercase English letters. A string is called special if it is made up of only a single character. For example, the string "abc" is not special, whereas the strings "ddd", "zz", and "f" are special. Return the length of the longest special substring of s which occurs at least thrice, or -1 if no special substring occurs at least thrice. A substring is a contiguous non-empty sequence of characters within a string.

Baseline thinking

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

Pattern signal: Hash Map · Binary Search · Sliding Window

Example 1

"aaaa"

Example 2

"abcdef"

Example 3

"abcaba"

Core Insight

What unlocks the optimal approach

The constraints are small.
Brute force checking all substrings.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
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

Upper-end input sizes

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 #2981: Find Longest Special Substring That Occurs Thrice I
class Solution {
    private String s;
    private int n;

    public int maximumLength(String s) {
        this.s = s;
        n = s.length();
        int l = 0, r = n;
        while (l < r) {
            int mid = (l + r + 1) >> 1;
            if (check(mid)) {
                l = mid;
            } else {
                r = mid - 1;
            }
        }
        return l == 0 ? -1 : l;
    }

    private boolean check(int x) {
        int[] cnt = new int[26];
        for (int i = 0; i < n;) {
            int j = i + 1;
            while (j < n && s.charAt(j) == s.charAt(i)) {
                j++;
            }
            int k = s.charAt(i) - 'a';
            cnt[k] += Math.max(0, j - i - x + 1);
            if (cnt[k] >= 3) {
                return true;
            }
            i = j;
        }
        return false;
    }
}

// Accepted solution for LeetCode #2981: Find Longest Special Substring That Occurs Thrice I
func maximumLength(s string) int {
	n := len(s)
	l, r := 0, n
	check := func(x int) bool {
		cnt := [26]int{}
		for i := 0; i < n; {
			j := i + 1
			for j < n && s[j] == s[i] {
				j++
			}
			k := s[i] - 'a'
			cnt[k] += max(0, j-i-x+1)
			if cnt[k] >= 3 {
				return true
			}
			i = j
		}
		return false
	}
	for l < r {
		mid := (l + r + 1) >> 1
		if check(mid) {
			l = mid
		} else {
			r = mid - 1
		}
	}
	if l == 0 {
		return -1
	}
	return l
}

# Accepted solution for LeetCode #2981: Find Longest Special Substring That Occurs Thrice I
class Solution:
    def maximumLength(self, s: str) -> int:
        def check(x: int) -> bool:
            cnt = defaultdict(int)
            i = 0
            while i < n:
                j = i + 1
                while j < n and s[j] == s[i]:
                    j += 1
                cnt[s[i]] += max(0, j - i - x + 1)
                i = j
            return max(cnt.values()) >= 3

        n = len(s)
        l, r = 0, n
        while l < r:
            mid = (l + r + 1) >> 1
            if check(mid):
                l = mid
            else:
                r = mid - 1
        return -1 if l == 0 else l

// Accepted solution for LeetCode #2981: Find Longest Special Substring That Occurs Thrice I
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2981: Find Longest Special Substring That Occurs Thrice I
// class Solution {
//     private String s;
//     private int n;
// 
//     public int maximumLength(String s) {
//         this.s = s;
//         n = s.length();
//         int l = 0, r = n;
//         while (l < r) {
//             int mid = (l + r + 1) >> 1;
//             if (check(mid)) {
//                 l = mid;
//             } else {
//                 r = mid - 1;
//             }
//         }
//         return l == 0 ? -1 : l;
//     }
// 
//     private boolean check(int x) {
//         int[] cnt = new int[26];
//         for (int i = 0; i < n;) {
//             int j = i + 1;
//             while (j < n && s.charAt(j) == s.charAt(i)) {
//                 j++;
//             }
//             int k = s.charAt(i) - 'a';
//             cnt[k] += Math.max(0, j - i - x + 1);
//             if (cnt[k] >= 3) {
//                 return true;
//             }
//             i = j;
//         }
//         return false;
//     }
// }

// Accepted solution for LeetCode #2981: Find Longest Special Substring That Occurs Thrice I
function maximumLength(s: string): number {
    const n = s.length;
    let [l, r] = [0, n];
    const check = (x: number): boolean => {
        const cnt: number[] = Array(26).fill(0);
        for (let i = 0; i < n; ) {
            let j = i + 1;
            while (j < n && s[j] === s[i]) {
                j++;
            }
            const k = s[i].charCodeAt(0) - 'a'.charCodeAt(0);
            cnt[k] += Math.max(0, j - i - x + 1);
            if (cnt[k] >= 3) {
                return true;
            }
            i = j;
        }
        return false;
    };
    while (l < r) {
        const mid = (l + r + 1) >> 1;
        if (check(mid)) {
            l = mid;
        } else {
            r = mid - 1;
        }
    }
    return l === 0 ? -1 : l;
}

Step 06

Interactive Study Demo

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

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O((n + |\Sigma|)

Space

O(|\Sigma|)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

Shrinking the window only once

Wrong move: Using `if` instead of `while` leaves the window invalid for multiple iterations.

Usually fails on: Over-limit windows stay invalid and produce wrong lengths/counts.

Fix: Shrink in a `while` loop until the invariant is valid again.