LeetCode #3844 — MEDIUM

Longest Almost-Palindromic Substring

Move from brute-force thinking to an efficient approach using core interview patterns strategy.

Solve on LeetCode

The Problem

Problem Statement

You are given a string s consisting of lowercase English letters.

A substring is almost-palindromic if it becomes a palindrome after removing exactly one character from it.

Return an integer denoting the length of the longest almost-palindromic substring in s.

Example 1:

Input: s = "abca"

Output: 4

Explanation:

Choose the substring "abca".

Remove "abca".
The string becomes "aba", which is a palindrome.
Therefore, "abca" is almost-palindromic.

Example 2:

Input: s = "abba"

Output: 4

Explanation:

Choose the substring "abba".

Remove "abba".
The string becomes "aba", which is a palindrome.
Therefore, "abba" is almost-palindromic.

Example 3:

Input: s = "zzabba"

Output: 5

Explanation:

Choose the substring "zzabba".

Remove "zabba".
The string becomes "abba", which is a palindrome.
Therefore, "zabba" is almost-palindromic.

Constraints:

2 <= s.length <= 2500
s consists of only lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: You are given a string s consisting of lowercase English letters. A substring is almost-palindromic if it becomes a palindrome after removing exactly one character from it. Return an integer denoting the length of the longest almost-palindromic substring in s.

Baseline thinking

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

Pattern signal: General problem-solving

Example 1

"abca"

Example 2

"abba"

Example 3

"zzabba"

Step 02

Core Insight

What unlocks the optimal approach

Solve greedily
Fix the center (consider both odd and even centers) and expand outwards
On the first mismatch, try skipping the left character and continue expanding, and also try skipping the right character; take the longer result
Track the maximum length found across all centers

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 #3844: Longest Almost-Palindromic Substring
class Solution {
    private int n;
    private char[] s;

    public int almostPalindromic(String s) {
        n = s.length();
        this.s = s.toCharArray();
        int ans = 0;
        for (int i = 0; i < n; i++) {
            ans = Math.max(ans, f(i, i));
            ans = Math.max(ans, f(i, i + 1));
        }
        return ans;
    }

    private int f(int l, int r) {
        while (l >= 0 && r < n && s[l] == s[r]) {
            l--;
            r++;
        }

        int l1 = l - 1, r1 = r;
        int l2 = l, r2 = r + 1;

        while (l1 >= 0 && r1 < n && s[l1] == s[r1]) {
            l1--;
            r1++;
        }
        while (l2 >= 0 && r2 < n && s[l2] == s[r2]) {
            l2--;
            r2++;
        }

        return Math.min(n, Math.max(r1 - l1 - 1, r2 - l2 - 1));
    }
}

// Accepted solution for LeetCode #3844: Longest Almost-Palindromic Substring
func almostPalindromic(s string) int {
	n := len(s)

	f := func(l, r int) int {
		for l >= 0 && r < n && s[l] == s[r] {
			l--
			r++
		}

		l1, r1 := l-1, r
		l2, r2 := l, r+1

		for l1 >= 0 && r1 < n && s[l1] == s[r1] {
			l1--
			r1++
		}
		for l2 >= 0 && r2 < n && s[l2] == s[r2] {
			l2--
			r2++
		}
		return min(n, max(r1-l1-1, r2-l2-1))
	}

	ans := 0
	for i := 0; i < n; i++ {
		ans = max(ans, f(i, i), f(i, i+1))
	}
	return ans
}

# Accepted solution for LeetCode #3844: Longest Almost-Palindromic Substring
class Solution:
    def almostPalindromic(self, s: str) -> int:
        def f(l: int, r: int) -> int:
            while l >= 0 and r < n and s[l] == s[r]:
                l -= 1
                r += 1
            l1, r1 = l - 1, r
            l2, r2 = l, r + 1
            while l1 >= 0 and r1 < n and s[l1] == s[r1]:
                l1 -= 1
                r1 += 1
            while l2 >= 0 and r2 < n and s[l2] == s[r2]:
                l2 -= 1
                r2 += 1
            return min(n, max(r1 - l1 - 1, r2 - l2 - 1))

        n = len(s)
        ans = 0
        for i in range(n):
            a = f(i, i)
            b = f(i, i + 1)
            ans = max(ans, a, b)
        return ans

// Accepted solution for LeetCode #3844: Longest Almost-Palindromic Substring
// 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 #3844: Longest Almost-Palindromic Substring
// class Solution {
//     private int n;
//     private char[] s;
// 
//     public int almostPalindromic(String s) {
//         n = s.length();
//         this.s = s.toCharArray();
//         int ans = 0;
//         for (int i = 0; i < n; i++) {
//             ans = Math.max(ans, f(i, i));
//             ans = Math.max(ans, f(i, i + 1));
//         }
//         return ans;
//     }
// 
//     private int f(int l, int r) {
//         while (l >= 0 && r < n && s[l] == s[r]) {
//             l--;
//             r++;
//         }
// 
//         int l1 = l - 1, r1 = r;
//         int l2 = l, r2 = r + 1;
// 
//         while (l1 >= 0 && r1 < n && s[l1] == s[r1]) {
//             l1--;
//             r1++;
//         }
//         while (l2 >= 0 && r2 < n && s[l2] == s[r2]) {
//             l2--;
//             r2++;
//         }
// 
//         return Math.min(n, Math.max(r1 - l1 - 1, r2 - l2 - 1));
//     }
// }

// Accepted solution for LeetCode #3844: Longest Almost-Palindromic Substring
function almostPalindromic(s: string): number {
    const n = s.length;

    const f = (l: number, r: number): number => {
        while (l >= 0 && r < n && s[l] === s[r]) {
            l--;
            r++;
        }

        let l1 = l - 1,
            r1 = r;
        let l2 = l,
            r2 = r + 1;

        while (l1 >= 0 && r1 < n && s[l1] === s[r1]) {
            l1--;
            r1++;
        }
        while (l2 >= 0 && r2 < n && s[l2] === s[r2]) {
            l2--;
            r2++;
        }

        return Math.min(n, Math.max(r1 - l1 - 1, r2 - l2 - 1));
    };

    let ans = 0;
    for (let i = 0; i < n; i++) {
        ans = Math.max(ans, f(i, i));
        ans = Math.max(ans, f(i, i + 1));
    }

    return ans;
}

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)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

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.