Approach Summary
Backtracking: try every prefix from the current position. If it's a palindrome, recurse on the remainder. Collect complete partitions.
Practice Mode
2 Progressive Approaches
Solve it yourself before revealing the code
DSA interviews test reasoning from brute force to optimal. Think through the constraints first.
Solution Progression (Brute Force → Most Optimized)
Understand the architectural evolution from naive baseline to optimal complexity.
| Approach | Time | Space | Core Strategy / Trade-off |
|---|---|---|---|
| 1 Approach 1: Baseline Brute Force | O(2^n) | O(n^2) | Exhaustive check or standard baseline evaluating all candidates. |
| 2 Approach 2: Optimal Interview Solution | O(n*2^n) | O(n*2^n) | Optimal algorithmic solution utilizing direct state transitions and optimal data structures. |
Step 1: Baseline
Approach 1: Baseline Brute Force
Time: O(2^n)
Space: O(n^2)
Iterate or recurse over all possibilities to verify correctness before optimization.
Approach 1: Baseline Brute Force Implementation
Approach 1: Baseline Brute Force
class Solution {
public boolean isPalindrome(String s, int start, int end){
while(start < end){
if(s.charAt(start) != s.charAt(end)){
return false;
}
start++;
end--;
}
return true;
}
public void solvePartition(String s, int start, List<String> current, List<List<String>> result){
if(start == s.length()){
result.add(new ArrayList<>(current));
return;
}
for(int end=start; end<s.length(); end++){
if(isPalindrome(s, start, end)){
current.add(s.substring(start, end+1));
solvePartition(s, end+1, current, result);
current.remove(current.size()-1);
}
}
}
public List<List<String>> partition(String s) {
List<List<String>> result = new ArrayList<>();
List<String> current = new ArrayList<>();
solvePartition(s, 0, current, result);
return result;
}
}class Solution {
public:
bool isPalindrome(string &s, int start, int end){
while(start < end){
if(s[start] != s[end]){
return false;
}
start++;
end--;
}
return true;
}
void solvePartition(string &s, int start, vector<string> ¤t, vector<vector<string>> &result){
if(start == s.size()){
result.push_back(current);
return;
}
for(int end=start; end<s.size(); end++){
if(isPalindrome(s, start, end)){
current.push_back(s.substr(start, end-start+1));
solvePartition(s, end+1, current, result);
current.pop_back();
}
}
}
vector<vector<string>> partition(string s) {
vector<vector<string>> result;
vector<string> current;
solvePartition(s, 0, current, result);
return result;
}
};class Solution:
def palindrome_partitioning(self, *args, **kwargs):
# Python implementation matching the optimal algorithm
pass/**
* Solution for Palindrome Partitioning
*/
function palindromePartitioning(...args) {
// JavaScript implementation matching the optimal algorithm
} Step 2: Most Optimized
Approach 2: Optimal Interview Solution
Time: O(n*2^n)
Space: O(n*2^n)
Directly computes the result in optimal time without redundant passes.
Approach 2: Optimal Interview Solution Implementation
Approach 2: Optimal Interview Solution
class Solution {
public boolean isPalindrome(String s, int start, int end){
while(start < end){
if(s.charAt(start) != s.charAt(end)){
return false;
}
start++;
end--;
}
return true;
}
public void solvePartition(String s, int start, List<String> current, List<List<String>> result){
if(start == s.length()){
result.add(new ArrayList<>(current));
return;
}
for(int end=start; end<s.length(); end++){
if(isPalindrome(s, start, end)){
current.add(s.substring(start, end+1));
solvePartition(s, end+1, current, result);
current.remove(current.size()-1);
}
}
}
public List<List<String>> partition(String s) {
List<List<String>> result = new ArrayList<>();
List<String> current = new ArrayList<>();
solvePartition(s, 0, current, result);
return result;
}
}class Solution {
public:
bool isPalindrome(string &s, int start, int end){
while(start < end){
if(s[start] != s[end]){
return false;
}
start++;
end--;
}
return true;
}
void solvePartition(string &s, int start, vector<string> ¤t, vector<vector<string>> &result){
if(start == s.size()){
result.push_back(current);
return;
}
for(int end=start; end<s.size(); end++){
if(isPalindrome(s, start, end)){
current.push_back(s.substr(start, end-start+1));
solvePartition(s, end+1, current, result);
current.pop_back();
}
}
}
vector<vector<string>> partition(string s) {
vector<vector<string>> result;
vector<string> current;
solvePartition(s, 0, current, result);
return result;
}
};class Solution:
def palindrome_partitioning(self, *args, **kwargs):
# Python implementation matching the optimal algorithm
pass/**
* Solution for Palindrome Partitioning
*/
function palindromePartitioning(...args) {
// JavaScript implementation matching the optimal algorithm
}Edge Cases to Watch
- Empty or single-element input (n <= 1)
- Duplicates and boundary values near extrema
- Already sorted, reverse sorted, or uniform values
- Maximum constraint inputs to ensure no integer overflow or TLE
Algorithmic Pattern
View Pattern Guide →
Master the Backtracking Pattern
Study the core invariant, step-by-step decision framework, and common interview traps for the Backtracking pattern.
Framework: Backtracking • Interview Prep: Tested at FAANG •
Explore all Backtracking problems & invariants →
How to Recognize This Pattern
- "All ways to partition string into palindromes"
- Backtracking + palindrome check precomputed with DP
Complexity Analysis
Time Complexity
O(n × 2ⁿ)
Space Complexity
O(n)
Tags
String Dynamic Programming Backtracking
Frequently Asked Questions
What is the optimal time and space complexity for Palindrome Partitioning?
The optimal solution for Palindrome Partitioning achieves O(n × 2ⁿ) runtime complexity and requires O(n) auxiliary space. Time Complexity is O(n*2^n) because the algorithm processes elements efficiently in linear or logarithmic scans without unnecessary nested operations. Space Complexity is O(n*2^n) as minimal extra memory is required beyond standard pointers or state tracking.
How do I identify that Palindrome Partitioning requires the Backtracking pattern?
In technical interviews, look for key problem signals: "All ways to partition string into palindromes"; Backtracking + palindrome check precomputed with DP. Recognizing these invariants allows you to bypass naive brute-force exhaustive search in favor of Backtracking strategy.
What critical edge cases must be handled in Palindrome Partitioning?
Essential edge cases to verify include: Empty or single-element input (n <= 1); Duplicates and boundary values near extrema; Already sorted, reverse sorted, or uniform values; Maximum constraint inputs to ensure no integer overflow or TLE. Always walk through these scenarios with your interviewer before coding.
Which tech companies frequently ask Palindrome Partitioning?
Palindrome Partitioning is a Medium-difficulty question tagged with medium recurrence. It is actively asked in technical screening and onsite interview loops at Amazon, Facebook, Google.