LeetCode #336 — HARD

Palindrome Pairs

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 array of unique strings words.

A palindrome pair is a pair of integers (i, j) such that:

0 <= i, j < words.length,
i != j, and
words[i] + words[j] (the concatenation of the two strings) is a palindrome.

Return an array of all the palindrome pairs of words.

You must write an algorithm with O(sum of words[i].length) runtime complexity.

Example 1:

Input: words = ["abcd","dcba","lls","s","sssll"]
Output: [[0,1],[1,0],[3,2],[2,4]]
Explanation: The palindromes are ["abcddcba","dcbaabcd","slls","llssssll"]

Example 2:

Input: words = ["bat","tab","cat"]
Output: [[0,1],[1,0]]
Explanation: The palindromes are ["battab","tabbat"]

Example 3:

Input: words = ["a",""]
Output: [[0,1],[1,0]]
Explanation: The palindromes are ["a","a"]

Constraints:

1 <= words.length <= 5000
0 <= words[i].length <= 300
words[i] consists of lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed array of unique strings words. A palindrome pair is a pair of integers (i, j) such that: 0 <= i, j < words.length, i != j, and words[i] + words[j] (the concatenation of the two strings) is a palindrome. Return an array of all the palindrome pairs of words. You must write an algorithm with O(sum of words[i].length) runtime complexity.

Baseline thinking

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

Pattern signal: Array · Hash Map · Trie

Example 1

["abcd","dcba","lls","s","sssll"]

Example 2

["bat","tab","cat"]

Example 3

["a",""]

Core Insight

What unlocks the optimal approach

Checking every two pairs will exceed the time limit. It will be O(n^2 * k). We need a faster way.
If we hash every string in the array, how can we check if two pairs form a palindrome after the concatenation?
We can check every string in words and consider it as words[j] (i.e., the suffix of the target palindrome). We can check if there is a hash of string that can be the prefix to make it a palindrome.

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

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 #336: Palindrome Pairs
class Solution {
    private static final int BASE = 131;
    private static final long[] MUL = new long[310];
    private static final int MOD = (int) 1e9 + 7;
    static {
        MUL[0] = 1;
        for (int i = 1; i < MUL.length; ++i) {
            MUL[i] = (MUL[i - 1] * BASE) % MOD;
        }
    }
    public List<List<Integer>> palindromePairs(String[] words) {
        int n = words.length;
        long[] prefix = new long[n];
        long[] suffix = new long[n];
        for (int i = 0; i < n; ++i) {
            String word = words[i];
            int m = word.length();
            for (int j = 0; j < m; ++j) {
                int t = word.charAt(j) - 'a' + 1;
                int s = word.charAt(m - j - 1) - 'a' + 1;
                prefix[i] = (prefix[i] * BASE) % MOD + t;
                suffix[i] = (suffix[i] * BASE) % MOD + s;
            }
        }
        List<List<Integer>> ans = new ArrayList<>();
        for (int i = 0; i < n; ++i) {
            for (int j = i + 1; j < n; ++j) {
                if (check(i, j, words[j].length(), words[i].length(), prefix, suffix)) {
                    ans.add(Arrays.asList(i, j));
                }
                if (check(j, i, words[i].length(), words[j].length(), prefix, suffix)) {
                    ans.add(Arrays.asList(j, i));
                }
            }
        }
        return ans;
    }

    private boolean check(int i, int j, int n, int m, long[] prefix, long[] suffix) {
        long t = ((prefix[i] * MUL[n]) % MOD + prefix[j]) % MOD;
        long s = ((suffix[j] * MUL[m]) % MOD + suffix[i]) % MOD;
        return t == s;
    }
}

// Accepted solution for LeetCode #336: Palindrome Pairs
func palindromePairs(words []string) [][]int {
	base := 131
	mod := int(1e9) + 7
	mul := make([]int, 310)
	mul[0] = 1
	for i := 1; i < len(mul); i++ {
		mul[i] = (mul[i-1] * base) % mod
	}
	n := len(words)
	prefix := make([]int, n)
	suffix := make([]int, n)
	for i, word := range words {
		m := len(word)
		for j, c := range word {
			t := int(c-'a') + 1
			s := int(word[m-j-1]-'a') + 1
			prefix[i] = (prefix[i]*base)%mod + t
			suffix[i] = (suffix[i]*base)%mod + s
		}
	}
	check := func(i, j, n, m int) bool {
		t := ((prefix[i]*mul[n])%mod + prefix[j]) % mod
		s := ((suffix[j]*mul[m])%mod + suffix[i]) % mod
		return t == s
	}
	var ans [][]int
	for i := 0; i < n; i++ {
		for j := i + 1; j < n; j++ {
			if check(i, j, len(words[j]), len(words[i])) {
				ans = append(ans, []int{i, j})
			}
			if check(j, i, len(words[i]), len(words[j])) {
				ans = append(ans, []int{j, i})
			}
		}
	}
	return ans
}

# Accepted solution for LeetCode #336: Palindrome Pairs
class Solution:
    def palindromePairs(self, words: List[str]) -> List[List[int]]:
        d = {w: i for i, w in enumerate(words)}
        ans = []
        for i, w in enumerate(words):
            for j in range(len(w) + 1):
                a, b = w[:j], w[j:]
                ra, rb = a[::-1], b[::-1]
                if ra in d and d[ra] != i and b == rb:
                    ans.append([i, d[ra]])
                if j and rb in d and d[rb] != i and a == ra:
                    ans.append([d[rb], i])
        return ans

// Accepted solution for LeetCode #336: Palindrome Pairs
struct Solution;

use std::collections::HashMap;
use std::collections::HashSet;

impl Solution {
    fn palindrome_pairs(words: Vec<String>) -> Vec<Vec<i32>> {
        let mut id: HashMap<String, usize> = HashMap::new();
        let n = words.len();
        let mut res = HashSet::new();
        for i in 0..n {
            id.insert(words[i].to_string(), i);
        }
        for i in 0..n {
            let k = words[i].len();
            for mid in 0..=k {
                let left: String = words[i][0..mid].to_string();
                let right: String = words[i][mid..].to_string();
                if Self::is_palindrome(&left) {
                    let right_r: String = right.chars().rev().collect();
                    if let Some(&j) = id.get(&right_r) {
                        if i != j {
                            res.insert(vec![j as i32, i as i32]);
                        }
                    }
                }
                if Self::is_palindrome(&right) {
                    let left_r: String = left.chars().rev().collect();
                    if let Some(&j) = id.get(&left_r) {
                        if i != j {
                            res.insert(vec![i as i32, j as i32]);
                        }
                    }
                }
            }
        }
        res.into_iter().collect()
    }

    fn is_palindrome(s: &str) -> bool {
        !s.chars().zip(s.chars().rev()).any(|(a, b)| a != b)
    }
}

#[test]
fn test() {
    let words = vec_string!["abcd", "dcba", "lls", "s", "sssll"];
    let mut res = vec![[0, 1], [1, 0], [3, 2], [2, 4]];
    let mut ans = Solution::palindrome_pairs(words);
    res.sort_unstable();
    ans.sort_unstable();
    assert_eq!(ans, res);
    let words = vec_string!["bat", "tab", "cat"];
    let mut res = vec![[0, 1], [1, 0]];
    let mut ans = Solution::palindrome_pairs(words);
    res.sort_unstable();
    ans.sort_unstable();
    assert_eq!(ans, res);
}

// Accepted solution for LeetCode #336: Palindrome Pairs
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #336: Palindrome Pairs
// class Solution {
//     private static final int BASE = 131;
//     private static final long[] MUL = new long[310];
//     private static final int MOD = (int) 1e9 + 7;
//     static {
//         MUL[0] = 1;
//         for (int i = 1; i < MUL.length; ++i) {
//             MUL[i] = (MUL[i - 1] * BASE) % MOD;
//         }
//     }
//     public List<List<Integer>> palindromePairs(String[] words) {
//         int n = words.length;
//         long[] prefix = new long[n];
//         long[] suffix = new long[n];
//         for (int i = 0; i < n; ++i) {
//             String word = words[i];
//             int m = word.length();
//             for (int j = 0; j < m; ++j) {
//                 int t = word.charAt(j) - 'a' + 1;
//                 int s = word.charAt(m - j - 1) - 'a' + 1;
//                 prefix[i] = (prefix[i] * BASE) % MOD + t;
//                 suffix[i] = (suffix[i] * BASE) % MOD + s;
//             }
//         }
//         List<List<Integer>> ans = new ArrayList<>();
//         for (int i = 0; i < n; ++i) {
//             for (int j = i + 1; j < n; ++j) {
//                 if (check(i, j, words[j].length(), words[i].length(), prefix, suffix)) {
//                     ans.add(Arrays.asList(i, j));
//                 }
//                 if (check(j, i, words[i].length(), words[j].length(), prefix, suffix)) {
//                     ans.add(Arrays.asList(j, i));
//                 }
//             }
//         }
//         return ans;
//     }
// 
//     private boolean check(int i, int j, int n, int m, long[] prefix, long[] suffix) {
//         long t = ((prefix[i] * MUL[n]) % MOD + prefix[j]) % MOD;
//         long s = ((suffix[j] * MUL[m]) % MOD + suffix[i]) % MOD;
//         return t == s;
//     }
// }

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(L)

Space

O(N × L)

Approach Breakdown

HASH SET

O(N × L) time

O(N × L) space

Store all N words in a hash set. Each insert/lookup hashes the entire word of length L, giving O(L) per operation. Prefix queries require checking every stored word against the prefix — O(N × L) per prefix search. Space is O(N × L) for storing all characters.

TRIE

O(L) time

O(N × L) space

Each operation (insert, search, prefix) takes O(L) time where L is the word length — one node visited per character. Total space is bounded by the sum of all stored word lengths. Tries win over hash sets when you need prefix matching: O(L) prefix search vs. checking every stored word.

Shortcut: One node per character → O(L) per operation. Prefix queries are what make tries worthwhile.

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.

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.