Palindrome Pairs
Time O(n * k^2) · Space O(n * k) · Official statement on LeetCode
Solutions
// Time: O(n * k^2), n is the number of the words, k is the max length of the words.
// Space: O(n * k)
class Solution {
public:
vector<vector<int>> palindromePairs(vector<string>& words) {
vector<vector<int>> res;
unordered_map<int, unordered_map<string, int>> lookup;
for (int i = 0; i < words.size(); ++i) {
lookup[size(words[i])][words[i]] = i;
}
for (int i = 0; i < words.size(); ++i) {
for (int j = 0; j <= words[i].length(); ++j) {
if (lookup.count(j) && is_palindrome(words[i], j, words[i].length() - 1)) {
string suffix = words[i].substr(0, j);
reverse(suffix.begin(), suffix.end());
auto& bucket = lookup[size(suffix)];
if (bucket.count(suffix) && i != bucket[suffix]) {
res.push_back({i, bucket[suffix]});
}
}
if (j > 0 && lookup.count(size(words[i]) - j) && is_palindrome(words[i], 0, j - 1)) {
string prefix = words[i].substr(j);
reverse(prefix.begin(), prefix.end());
auto& bucket = lookup[size(prefix)];
if (bucket.count(prefix) && bucket[prefix] != i) {
res.push_back({bucket[prefix], i});
}
}
}
}
return res;
}
private:
bool is_palindrome(string& s, int start, int end) {
while (start < end) {
if (s[start++] != s[end--]) {
return false;
}
}
return true;
}
};
// Time: O(n * k^2), n is the number of the words, k is the max length of the words.
// Space: O(n * k^2)
// Manacher solution.
class Solution2_TLE {
public:
vector<vector<int>> palindromePairs(vector<string>& words) {
unordered_multimap<string, int> prefix, suffix;
for (int i = 0; i < words.size(); ++i) { // O(n)
vector<int> P;
manacher(words[i], &P);
for (int j = 0; j < P.size(); ++j) { // O(k)
if (j - P[j] == 1) {
prefix.emplace(words[i].substr((j + P[j]) / 2), i); // O(k)
}
if (j + P[j] == P.size() - 2) {
suffix.emplace(words[i].substr(0, (j - P[j]) / 2), i);
}
}
}
vector<vector<int>> res;
for (int i = 0; i < words.size(); ++i) { // O(n)
string reversed_word(words[i].rbegin(), words[i].rend()); // O(k)
auto its = prefix.equal_range(reversed_word);
for (auto it = its.first; it != its.second; ++it) {
if (it->second != i) {
res.push_back({i, it->second});
}
}
its = suffix.equal_range(reversed_word);
for (auto it = its.first; it != its.second; ++it) {
if (words[i].size() != words[it->second].size()) {
res.push_back({it->second, i});
}
}
}
return res;
}
void manacher(const string& s, vector<int> *P) {
string T = preProcess(s);
const int n = T.length();
P->resize(n);
int C = 0, R = 0;
for (int i = 1; i < n - 1; ++i) {
int i_mirror = 2 * C - i;
(*P)[i] = (R > i) ? min(R - i, (*P)[i_mirror]) : 0;
while (T[i + 1 + (*P)[i]] == T[i - 1 - (*P)[i]]) {
++(*P)[i];
}
if (i + (*P)[i] > R) {
C = i;
R = i + (*P)[i];
}
}
}
string preProcess(const string& s) {
if (s.empty()) {
return "^$";
}
string ret = "^";
for (int i = 0; i < s.length(); ++i) {
ret += "#" + s.substr(i, 1);
}
ret += "#$";
return ret;
}
};
// Time: O(n * k^2), n is the number of the words, k is the max length of the words.
// Space: O(n * k)
// Trie solution.
class Solution_MLE {
public:
vector<vector<int>> palindromePairs(vector<string>& words) {
vector<vector<int>> res;
TrieNode trie;
for (int i = 0; i < words.size(); ++i) {
trie.insert(words[i], i);
}
for (int i = 0; i < words.size(); ++i) {
trie.find(words[i], i, &res);
}
return res;
}
private:
struct TrieNode {
int word_idx = -1;
unordered_map<char, TrieNode *> leaves;
void insert(const string& s, int i) {
auto* p = this;
for (const auto& c : s) {
if (p->leaves.find(c) == p->leaves.cend()) {
p->leaves[c] = new TrieNode;
}
p = p->leaves[c];
}
p->word_idx = i;
}
void find(const string& s, int idx, vector<vector<int>> *res) {
auto* p = this;
for (int i = s.length() - 1; i >= 0; --i) { // O(k)
if (p->leaves.find(s[i]) != p->leaves.cend()) {
p = p->leaves[s[i]];
if (p->word_idx != -1 && p->word_idx != idx &&
is_palindrome(s, i - 1)) { // O(k)
res->push_back({p->word_idx, idx});
}
} else {
break;
}
}
}
bool is_palindrome(const string& s, int j) {
int i = 0;
while (i <= j) {
if (s[i++] != s[j--]) {
return false;
}
}
return true;
}
~TrieNode() {
for (auto& kv : leaves) {
if (kv.second) {
delete kv.second;
}
}
}
};
};
Beginner Explanation
What is Palindrome Pairs?
Palindrome Pairs (LeetCode #336) is a Hard problem that primarily trains hash table.
How to think about it
- Restate the goal in your own words before coding.
- Work a tiny example by hand so the invariant becomes obvious.
- Identify the pattern — this problem aligns with hash map.
- Only then translate the idea into code.
Why this problem matters
Hard problems force you to combine patterns and prove complexity carefully — interview gold.
AlgoForge explanations are original teaching notes. Always open the official problem statement on LeetCode for constraints and examples.
Interview Walkthrough
Interview approach for Palindrome Pairs
Opening (30–60 seconds)
- Clarify inputs/outputs and edge cases (empty input, single element, duplicates, overflow).
- State a brute force so the interviewer knows you can solve it naively.
- Propose the optimal direction tied to hash map.
Core solution narrative
- Define the state you track (pointers, DP cell, set membership, stack top, etc.).
- Explain the transition when you process the next element.
- Call out time (O(n * k^2)) and space (O(n * k)) before coding.
- Code cleanly; narrate variable names.
What interviewers listen for
- Correctness on edge cases
- Complexity honesty
- Ability to discuss trade-offs (e.g., hash map space vs. sort + two pointers)
Follow-up questions they may ask
- Can you solve it with less memory?
- What if the input stream is infinite / doesn't fit in RAM?
- How would tests look for adversarial inputs?
Optimized Approach
Optimized solution notes
The reference solutions on AlgoForge target O(n * k^2) time and O(n * k) space.
Pattern focus: hash map
Use the pattern as a checklist:
- hash map — confirm the invariant holds after each step
Multiple methods appear in the source solutions — compare them and explain when each is preferable.
Implementation tips
- Prefer readable names over micro-optimizations in interviews.
- Extract helpers only when they clarify (e.g., expand-around-center, DFS visit).
- After AC-level logic, re-scan for off-by-one and null checks.
Complexity Analysis
Complexity
| Measure | Bound |
|---|---|
| Time | O(n * k^2) |
| Space | O(n * k) |
How to justify this in an interview
- Time: count loops, map/set operations, and recursive branching; state average vs worst case if relevant.
- Space: include hash maps, recursion stack, and output allocation when the problem asks for it.
If your implementation differs from the reference, re-derive big-O from your code — never memorize a complexity you cannot defend.
Common Mistakes
Common mistakes on Palindrome Pairs
- Skipping edge cases — empty collections, single-element inputs, max constraints.
- Wrong invariant for hash map — updating state too early or too late.
- Mutating input unexpectedly when the problem forbids it.
- Off-by-one in windows, ranges, or binary search bounds.
- Ignoring overflow / precision for integer arithmetic problems.
- Overengineering — jumping to an advanced structure when a simpler approach works.
Alternative Approaches
Alternatives
The source file includes more than one method. Compare:
- Primary optimized path — best complexity for typical interviews.
- Secondary approach — often brute force, sorting-based, or space-optimized variant.
Practice articulating when you would pick each (constraints, readability, follow-ups).
Edge Cases
Edge cases checklist
- Minimum input size
- Maximum input size / time limits
- Duplicates and already-sorted input
- Negative numbers / zeros (if applicable)
- Disconnected structures (graphs/trees)
- Single path vs branching recursion depth
Pattern Recognition
Spotting this pattern
Signal phrases that point to hash map:
- Sorted input or ability to sort without changing the answer class
- Need for contiguous subarray / substring → consider sliding window
- Need for O(1) membership → hash set/map
- Optimal substructure + overlapping subproblems → DP
- Connectivity / components → graph DFS/BFS or Union-Find
Primary topics: hash table.
Follow-up Interview Questions
Follow-ups
- How does the solution change if the input is a stream?
- Can you solve it in-place?
- What if duplicates must be handled differently?
- How would you parallelize the approach?
- Design tests that would break a buggy implementation.
Practice Recommendations
What to practice next
- Re-solve Palindrome Pairs in a second language (cpp, python).
- Drill 3–5 more problems tagged hash table.
- Teach the solution out loud in under 5 minutes.
- Add this problem to your revision calendar in 3 days and 14 days.
Visualization
Study checklist
- Read the official problem statement on LeetCode
- Solve on paper / whiteboard first
- Implement the hash map approach
- Verify edge cases from the checklist
- State time and space complexity aloud
- Compare with the AlgoForge reference solution
- Schedule a revision session
Revision notes
Palindrome Pairs (#336) — Hard. Pattern: hash map. Complexity: O(n * k^2) time / O(n * k) space. Re-derive the invariant before coding.
FAQs
What is the time complexity of Palindrome Pairs?+
The reference solutions aim for O(n * k^2) time and O(n * k) space. Always re-derive complexity from the code you write in the interview.
What pattern does Palindrome Pairs use?+
It primarily maps to hash map, within the broader topic of hash table.
Is Palindrome Pairs good for interviews?+
Yes — as a Hard problem it is a solid practice target. Pair it with related problems in the same pattern family for spaced repetition.
Where can I read the official statement?+
Open the official LeetCode page for constraints and examples: https://leetcode.com/problems/palindrome-pairs/