DrawCode Algorithm Visualizer • Dynamic Programming • Hard

Palindrome Partitioning

Tags: DP, 2D, Strings, Palindrome

1. One-Liner

Palindrome partitioning minimizes cuts so every substring is a palindrome—dp[i] stores the best for prefix ending at i, using palindrome checks on intervals.


2. The Problem It Solves

Given s, split it into palindromic chunks with fewest separators (or enumerate all valid partitions in another variant). Used in text analytics and advanced string DP.


3. The Core Idea

Precompute is_pal[i][j] or expand while DP-ing; then dp[i]=min over j<i+1 pal[j..i] of dp[j-1]+1—try every last palindrome chunk ending at i.


4. How It Works (Step-by-Step)

StepWhat Happens
1Palindrome table for all substrings O()
2dp[0]=0, dp[i] min cuts for s[0..i)
3Transition over palindromic suffixes
4Return dp[n]-1 cuts or dp[n] per indexing

5. Dry Run Example

"aab" → partition "aa"|"b"1 cut (0-based definition dependent).


6. Key Properties

PropertyValue
TimeO() or O() naive
RelatedPalindrome centers

7. Where It Is Used

DomainUse
NLPTokenization heuristics
InterviewsHard string DP

8. Interview Tips

Also know generate all partitions via backtracking + palindrome memo.


9. Comparison with Other Algorithms

ProblemGoal
Min cutsOptimization
List allBacktracking

10. Complexity

MetricValue
TimeO() with palindrome preprocessing
SpaceO() for table

Implementation Example (PYTHON)

def min_palindrome_cuts(s: str) -> int:
    n = len(s)
    pal = [[False] * n for _ in range(n)]
    for i in range(n - 1, -1, -1):
        for j in range(i, n):
            pal[i][j] = s[i] == s[j] and (j - i < 2 or pal[i + 1][j - 1])
    dp = [0] * (n + 1)
    for i in range(1, n + 1):
        dp[i] = i
        for j in range(i):
            if pal[j][i - 1]:
                dp[i] = min(dp[i], dp[j] + 1)
    return dp[n] - 1

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