How to Solve: Palindrome Partitioning

How to Solve: Palindrome Partitioning

(Complete Guide for FAANG-Level Interviews)

? Introduction

"Palindrome partitioning appears in 1 out of every 3 onsite interviews—master it, and you’ll prove you can break down complex problems into clean, recursive solutions while optimizing for edge cases."

? WHAT YOU NEED TO KNOW FIRST

Before tackling palindrome partitioning, ensure you understand: 1. Backtracking (DFS): Recursive exploration of all possible partitions. 2. Dynamic Programming (Palindrome Check): Precompute palindrome substrings to avoid redundant checks. 3. String Manipulation: Slicing, indexing, and substring operations.

? KEY TECHNIQUES & PATTERNS

? STEP-BY-STEP FRAMEWORK

(Repeatable mental checklist for every palindrome partitioning problem.)

1. Understand the Problem
2. Input: A string s.
3. Output: All possible partitions where every substring is a palindrome.

4. Example: "aab" → [["a","a","b"], ["aa","b"]].

5. Precompute Palindromes (Optional Optimization)

6. Use DP to store dp[i][j] = True if s[i..j] is a palindrome.

7. Time: O(n²), Space: O(n²).

8. Backtracking (DFS) Setup

9. Define a helper function backtrack(start, path, result).
10. start: Current index in s.
11. path: Current partition being built.

12. result: Final list of valid partitions.

13. Base Case

14. If start == len(s), add path to result.

15. Recursive Case

16. For end from start+1 to len(s):

    • If s[start..end] is a palindrome:
    • Add s[start..end] to path.
    • Recurse with start = end.
    • Backtrack (remove last substring).

17. Optimize (If Needed)

18. Use memoization (DP table) to avoid recomputing palindromes.

19. Edge Cases

20. Empty string: "" → [[]].
21. Single character: "a" → [["a"]].
22. All same characters: "aaa" → [["a","a","a"], ["a","aa"], ["aa","a"], ["aaa"]].

? WORKED EXAMPLES

Example 1 – Basic: "aab"

Problem: Return all palindrome partitions of "aab".

Thought Process: 1. Start at index 0. 2. Check substrings:
- "a" (palindrome) → Recurse on "ab".
- "a" (palindrome) → Recurse on "b".
- "b" (palindrome) → Add ["a","a","b"] to result.
- "ab" (not palindrome) → Skip.
- "aa" (palindrome) → Recurse on "b".
- "b" (palindrome) → Add ["aa","b"] to result.
- "aab" (not palindrome) → Skip.

Code (Python):

def partition(s):

    def is_palindrome(sub):

        return sub == sub[::-1]


    def backtrack(start, path):

        if start == len(s):

            result.append(path.copy())

            return

        for end in range(start + 1, len(s) + 1):

            substring = s[start:end]

            if is_palindrome(substring):

                path.append(substring)

                backtrack(end, path)

                path.pop()


    result = []

    backtrack(0, [])

    return result

Time Complexity: - O(n 2ⁿ) (worst case: all substrings are palindromes, e.g., "aaa"). - Each of the 2ⁿ partitions takes O(n) time to check palindromes.

Space Complexity: - O(n) (recursion stack depth).

Why This Works: - Backtracking explores all possible partitions. - Palindrome checks ensure only valid substrings are included.

Example 2 – Medium: "aabaa" (Optimized with DP)

Problem: Return all palindrome partitions of "aabaa" using DP for optimization.

Thought Process: 1. Precompute dp[i][j] where dp[i][j] = True if s[i..j] is a palindrome. 2. Use backtracking with the DP table to avoid recomputing palindromes.

Code (Python):

def partition(s):

    n = len(s)

    dp = [[False]  n for _ in range(n)]


    # Precompute palindrome substrings

    for i in range(n):

        dp[i][i] = True

    for i in range(n - 1):

        dp[i][i + 1] = (s[i] == s[i + 1])

    for length in range(3, n + 1):

        for i in range(n - length + 1):

            j = i + length - 1

            dp[i][j] = (s[i] == s[j]) and dp[i + 1][j - 1]


    def backtrack(start, path):

        if start == n:

            result.append(path.copy())

            return

        for end in range(start, n):

            if dp[start][end]:

                path.append(s[start:end + 1])

                backtrack(end + 1, path)

                path.pop()


    result = []

    backtrack(0, [])

    return result

Time Complexity: - O(n²) for DP precomputation. - O(n 2ⁿ) for backtracking (same as before, but palindrome checks are O(1)).

Space Complexity: - O(n²) for DP table.

Why This Works: - DP table avoids redundant palindrome checks. - Backtracking remains the same but is now more efficient.

Example 3 – Hard/Follow-up: Minimum Cuts for Palindrome Partitioning

Problem: Given a string s, return the minimum number of cuts needed so that every substring is a palindrome.

Thought Process: 1. Use DP to precompute palindromes (dp[i][j]). 2. Use another DP array cuts[i] to store the minimum cuts for s[0..i]. 3. For each i, if s[0..i] is a palindrome, cuts[i] = 0. 4. Otherwise, for each j from 0 to i-1, if s[j+1..i] is a palindrome, update cuts[i] = min(cuts[i], cuts[j] + 1).

Code (Python):

def minCut(s):

    n = len(s)

    dp = [[False]  n for _ in range(n)]

    cuts = [0]  n


    for i in range(n):

        min_cut = i  # Maximum cuts for s[0..i] is i (single characters)

        for j in range(i + 1):

            if s[i] == s[j] and (i - j <= 2 or dp[j + 1][i - 1]):

                dp[j][i] = True

                min_cut = 0 if j == 0 else min(min_cut, cuts[j - 1] + 1)

        cuts[i] = min_cut


    return cuts[-1]

Time Complexity: - O(n²) (DP precomputation + cuts calculation).

Space Complexity: - O(n²) (DP table).

Why This Works: - cuts[i] is built by checking all possible partitions ending at i. - The DP table ensures we only consider valid palindromes.

❌ Common Mistakes

? INTERVIEW TrapS

? 1-Minute Recap

"Alright, let’s lock this in. Palindrome partitioning is all about backtracking + palindrome checks. Here’s the playbook: 1. Precompute palindromes with DP to avoid redundant checks. 2. Backtrack from the start of the string, adding valid palindromes to your path. 3. Base case: When you reach the end, save the partition. 4. Optimize: Use DP to make palindrome checks O(1). 5. Edge cases: Empty string, single character, all same characters.

For follow-ups like minimum cuts, switch to DP with cuts[i] storing the answer for s[0..i]. Remember: O(n 2ⁿ) for backtracking, O(n²) for DP. Now go crush that interview!

? Final Notes

• Practice: Solve "aab", "aabaa", and "abab" manually first.
• Whiteboard: Draw the recursion tree for "aab".
• Follow-ups: Be ready for "minimum cuts" or "count palindrome partitions."

This framework is directly usable in an interview—no fluff, just actionable steps. Good luck! ?