How Do You Solve the Palindrome Partitioning Problem?
Learn the two-layer DP solution for minimum-cut palindrome partitioning in O(n squared) time.
Expected Interview Answer
Palindrome partitioning — splitting a string into the minimum number of cuts so every resulting piece is a palindrome — is solved with dynamic programming in O(n²) time by first precomputing which substrings are palindromes in an O(n²) table, then using a second O(n²) pass where minCuts[i] tracks the fewest cuts needed for the prefix ending at i, checking every possible last palindrome piece.
The first pass builds isPalindrome[i][j] using the classic recurrence that s[i..j] is a palindrome if s[i] equals s[j] and the inner substring s[i+1..j-1] is also a palindrome (or the inner span is empty), filling the table by increasing substring length. The second pass computes minCuts[i], the minimum cuts needed for s[0..i], as 0 if s[0..i] is itself a palindrome, otherwise as the minimum over every j less than i where s[j+1..i] is a palindrome of minCuts[j] + 1. This is a classic DP-on-DP pattern: one table answers a yes/no subproblem (is this a palindrome) and a second table uses those answers to solve an optimization subproblem (fewest cuts). Enumerating all partitions instead of all palindromic substrings (as in the "return every valid partition" variant) uses backtracking pruned by the same palindrome table, since checking palindrome-ness during backtracking without the table degrades to exponential time with redundant checks.
- O(n²) time by precomputing the palindrome table once
- Reuses the palindrome table across the cuts computation
- Generalizes to the "return all partitions" backtracking variant
- Classic two-layer DP pattern applicable to many string problems
AI Mentor Explanation
Palindrome partitioning is like slicing a long highlight reel of shots into the fewest possible clips such that each clip, played backward, looks identical to played forward — a mirrored shot sequence. First you build a table marking every possible clip start-and-end pair as "mirrored" or not, using the fact that a clip is mirrored if its first and last shots match and everything between them is also mirrored. Then, for each point in the reel, you find the fewest total clips needed by checking every mirrored clip that could end there and adding one to whatever the best cut count was just before it started. This two-stage approach — mark all mirrored clips first, then optimize cuts using that map — avoids re-checking mirror symmetry over and over.
Step-by-Step Explanation
Step 1
Precompute the palindrome table
Fill isPalindrome[i][j] bottom-up by increasing length using s[i]==s[j] and isPalindrome[i+1][j-1].
Step 2
Initialize the cuts array
minCuts[i] defaults to i (worst case: cut before every character).
Step 3
Fill minCuts using the palindrome table
For each i, check every j <= i where s[j..i] is a palindrome; take minCuts[j-1] + 1 (or 0 if j is 0).
Step 4
Return minCuts at the last index
minCuts[n-1] is the minimum number of cuts for the whole string.
What Interviewer Expects
- Identify this as a two-layer DP problem: palindrome table, then cuts table
- State O(n²) time and O(n²) space for the palindrome table, O(n) extra for cuts
- Explain why precomputing palindromes avoids redundant O(n) checks during the cuts pass
- Connect to the related "return all partitions" backtracking variant
Common Mistakes
- Recomputing palindrome checks from scratch during the cuts pass instead of reusing the table
- Off-by-one errors in the minCuts recurrence indices
- Confusing "minimum cuts" with "minimum pieces" (cuts = pieces - 1)
- Not handling the case where the whole string is already a palindrome (0 cuts)
Best Answer (HR Friendly)
“I would first mark every possible chunk of the string as a palindrome or not, using a table, since a chunk is a palindrome if its ends match and the middle is also a palindrome. Then I would sweep through the string once more, using that table to figure out, for each position, the fewest cuts needed to get there — reusing the palindrome answers instead of rechecking symmetry every time.”
Code Example
def min_cut(s):
n = len(s)
is_pal = [[False] * n for _ in range(n)]
for i in range(n):
is_pal[i][i] = True
for length in range(2, n + 1):
for i in range(n - length + 1):
j = i + length - 1
if s[i] == s[j] and (length == 2 or is_pal[i + 1][j - 1]):
is_pal[i][j] = True
min_cuts = [0] * n
for i in range(n):
if is_pal[0][i]:
min_cuts[i] = 0
continue
min_cuts[i] = i # worst case
for j in range(1, i + 1):
if is_pal[j][i]:
min_cuts[i] = min(min_cuts[i], min_cuts[j - 1] + 1)
return min_cuts[n - 1]Follow-up Questions
- How would you return all valid partitions instead of just the minimum cut count?
- How would you reduce the space complexity of the palindrome table?
- How does this problem relate to the longest palindromic substring problem?
- How would you handle very long strings where O(n²) space becomes a concern?
MCQ Practice
1. What is the time complexity of the full palindrome-partitioning minimum-cuts solution?
Both the palindrome table and the cuts table are filled in O(n²), giving O(n²) overall.
2. Why is the palindrome table precomputed before the cuts table?
Precomputing lets each palindrome check during the cuts pass be an O(1) table lookup instead of an O(n) re-scan.
3. If minCuts[n-1] equals 0, what does that mean?
Zero cuts means the whole string can stand as a single palindromic piece.
Flash Cards
What two tables does the standard palindrome partitioning solution use? — A palindrome-existence table (isPalindrome[i][j]) and a minimum-cuts table (minCuts[i]).
What is the overall time complexity? — O(n²) — O(n²) for the palindrome table plus O(n²) for the cuts table.
How is "minimum cuts" related to "minimum pieces"? — Minimum pieces = minimum cuts + 1.
What backtracking optimization reuses the palindrome table? — The "return all partitions" variant prunes recursive branches using the precomputed isPalindrome table instead of rechecking symmetry.