Palindrome Partitioning DP Minimum Cuts in DSA Python - Time & Space Complexity
We want to find how fast the palindrome partitioning algorithm runs when finding minimum cuts.
How does the time needed grow as the input string gets longer?
Analyze the time complexity of the following code snippet.
def minCut(s: str) -> int:
n = len(s)
dp = [0] * n
palindrome = [[False] * n for _ in range(n)]
for end in range(n):
min_cuts = end
for start in range(end + 1):
if s[start] == s[end] and (end - start <= 1 or palindrome[start + 1][end - 1]):
palindrome[start][end] = True
min_cuts = 0 if start == 0 else min(min_cuts, dp[start - 1] + 1)
dp[end] = min_cuts
return dp[-1]
This code finds the minimum cuts needed to split a string into palindromes using dynamic programming.
Identify the loops, recursion, array traversals that repeat.
- Primary operation: Nested loops checking all substrings for palindrome property.
- How many times: Outer loop runs n times; inner loop runs up to n times for each outer loop.
As the string length grows, the number of substring checks grows roughly with the square of the length.
| Input Size (n) | Approx. Operations |
|---|---|
| 10 | About 100 checks |
| 100 | About 10,000 checks |
| 1000 | About 1,000,000 checks |
Pattern observation: Operations grow roughly by the square of input size (n²).
Time Complexity: O(n²)
This means the time needed grows roughly with the square of the string length.
[X] Wrong: "The algorithm runs in linear time because it uses dynamic programming."
[OK] Correct: Dynamic programming helps avoid repeated work, but here we still check all substrings, which is about n² operations.
Understanding this time complexity helps you explain how dynamic programming improves efficiency and why some problems naturally require checking many combinations.
"What if we precomputed palindrome substrings using a different method? How would that affect the time complexity?"
Practice
minimum cuts mean in the context of palindrome partitioning of a string?Solution
Step 1: Understand palindrome partitioning
Palindrome partitioning means splitting a string into parts where each part is a palindrome.Step 2: Define minimum cuts
Minimum cuts is the least number of cuts needed to divide the string so every piece is a palindrome.Final Answer:
The smallest number of cuts needed to split the string into palindromic substrings -> Option DQuick Check:
Minimum cuts = smallest cuts for palindrome parts [OK]
- Confusing minimum cuts with longest palindrome length
- Thinking minimum cuts means maximum palindromes
- Counting total substrings instead of cuts
dp of size n x n with False values for palindrome checks?Solution
Step 1: Understand 2D list initialization
To create a 2D list with independent inner lists, use nested list comprehensions.Step 2: Analyze options
dp = [[False for _ in range(n)] for _ in range(n)] uses nested list comprehension correctly; dp = [[False]*n]*n creates shallow copies causing shared references.Final Answer:
dp = [[False for _ in range(n)] for _ in range(n)] -> Option AQuick Check:
Nested list comprehension creates independent rows [OK]
- Using [[False]*n]*n causes shared inner lists
- Using single list for 2D structure
- Confusing list multiplication with nested lists
def minCut(s):
n = len(s)
dp = [0] * n
for i in range(n):
min_cut = i
for j in range(i + 1):
if s[j:i+1] == s[j:i+1][::-1]:
min_cut = 0 if j == 0 else min(min_cut, dp[j-1] + 1)
dp[i] = min_cut
return dp[-1]
print(minCut('aab'))Solution
Step 1: Trace the dp array updates
For 'aab': i=0: 'a' palindrome, dp[0]=0; i=1: 'aa' palindrome, dp[1]=0; i=2: 'aab' not palindrome, check substrings: 'b' palindrome, dp[2]=dp[1]+1=1.Step 2: Final dp array and result
dp = [0, 0, 1], so minimum cuts = 1.Final Answer:
1 -> Option CQuick Check:
Minimum cuts for 'aab' = 1 [OK]
- Not updating dp correctly for substrings
- Assuming entire string is palindrome without check
- Confusing indices in slicing
def minCut(s):
n = len(s)
dp = [0]*n
for i in range(n):
min_cut = i
for j in range(i):
if s[j:i+1] == s[j:i+1][::-1]:
min_cut = 0 if j == 0 else min(min_cut, dp[j-1] + 1)
dp[i] = min_cut
return dp[-1]
print(minCut('aab'))Solution
Step 1: Analyze loop range error
The inner loop runs from 0 to i-1, missing the case j = i, which is needed to check substring s[i:i+1].Step 2: Correct loop range
Changing inner loop to range(i+1) includes the substring ending at i, fixing the bug.Final Answer:
Change inner loop to for j in range(i+1) -> Option AQuick Check:
Include full substring checks with correct loop range [OK]
- Using range(i) misses last substring
- Incorrect dp initialization
- Unnecessary string reversal
s = 'ababbbabbababa', what is the minimum number of cuts needed to partition it into palindromic substrings using dynamic programming?Solution
Step 1: Understand the problem and DP approach
We want to split 'ababbbabbababa' into palindromic parts with minimum cuts. Using DP, we check all substrings and update minimum cuts.Step 2: Known result from palindrome partitioning DP
This classic example has minimum cuts = 3, partitioning as 'a|babbbab|b|ababa' or similar (4 parts, 3 cuts).Final Answer:
3 -> Option BQuick Check:
Minimum cuts for 'ababbbabbababa' = 3 [OK]
- Underestimating cuts by assuming longer palindromes
- Overcounting cuts by splitting too often
- Not using DP to optimize palindrome checks
Ready to prove it?
Pass the test and this pattern moves unknown
