MEDIUM NC#77 Backtracking / DP
131. Palindrome Partitioning
📖 Problem
Given a string s, partition s such that every substring of the partition is a palindrome. Return all possible palindrome partitionings of s.
🧠 Visual Learning Aid
1 Model the input into the right structure
2 Choose the core technique and invariant
3 Execute step-by-step with a sample
4 Validate complexity and edge cases
JS/TS Refreshers
- •Array methods (`push`, `pop`, `shift`, `slice`)
- •Object/Map/Set usage patterns
- •Function parameter and return typing
- •Array DP table updates
- •State transition thinking
- •Base case initialization
Logical Thinking Concepts
- •Define invariants before coding
- •Check edge cases first (`[]`, single element, duplicates)
- •Estimate time/space before implementation
- •Apply Dynamic Programming reasoning pattern
- •Apply Backtracking reasoning pattern
💡 Approach
- → Use backtracking to try all partitionings
- → Check if substring is palindrome before adding
- → Time: O(n * 2^n) worst case, Space: O(n) for result
🛠️ Hints & Pitfalls
Hints
- •Use backtracking to try all partitionings
- •Check if substring is palindrome before adding
- •Time: O(n * 2^n) worst case, Space: O(n) for result
🧪 Test Cases
Test Case 1
Not run Input:
partition('aab'); Expected:
[["a","a","b"], ["aa","b"]] Test Case 2
Not run Input:
partition('a'); Expected:
[["a"]] Test Case 3
Not run Input:
partition(s: string); Expected:
Computed from hidden reference 📝 Code Editor
📤 Output