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How to Fix Palindrome Partitioning Errors

DodaTech Updated 2026-06-26 1 min read

In this tutorial, you'll learn about How to Fix Palindrome Partitioning Errors. We cover key concepts, practical examples, and best practices.

Fix palindrome partitioning errors when minimum cuts DP incorrect or palindrome check repeats O(n^3).

Quick Fix

Wrong

def min_cut(s):
    n=len(s); dp=list(range(n))
    for i in range(1,n):
        for j in range(i+1):
            if s[j:i+1]==s[j:i+1][::-1]:
                dp[i]=min(dp[i], (dp[j-1] if j>0 else 0)+1)
    return dp[-1]

Repeated palindrome checks O(n^3). Long substring checks are expensive.

def min_cut(s):
    n=len(s); dp=list(range(n))
    pal=[[False]*n for _ in range(n)]
    for i in range(n): 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 pal[i+1][j-1]): pal[i][j]=True
    for i in range(1,n):
        if pal[0][i]: dp[i]=0
        else:
            for j in range(i):
                if pal[j+1][i]: dp[i]=min(dp[i],dp[j]+1)
    return dp[-1]
s='aab' -> 1 ('aa','b'). O(n^2).

Prevention

Precompute palindrome table O(n^2). Then compute min cuts using palindrome lookups.

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FAQ

What is palindrome partition?

Minimum cuts to partition string so every substring is palindrome.

Why precompute?

Avoids O(n) palindrome check per DP transition. palindrome[i][j] precomputed O(1) lookup.

DP?

dp[i] = min cuts for s[0:i+1]. If pal[0][i]: 0 cuts. Else: min(dp[j]+1) for pal[j+1][i].

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