LeetCode Problem Workspace
Minimum Changes to Make K Semi-palindromes
Minimize the number of letter changes to partition a string into k semi-palindromes using dynamic programming and two pointers.
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5
Code langs
3
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Practice Focus
Hard · State transition dynamic programming
Answer-first summary
Minimize the number of letter changes to partition a string into k semi-palindromes using dynamic programming and two pointers.
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To solve the problem, use dynamic programming with a state transition approach. Define dp[i][j] as the minimum number of changes to split the suffix of string s starting from s[i] into j valid parts. Apply two pointers for efficient substring checking and adjust the changes to minimize the number of modifications required to make each substring a semi-palindrome.
Problem Statement
Given a string s and an integer k, partition s into k substrings such that the letter changes needed to make each substring a semi-palindrome are minimized. A semi-palindrome is a string that can be rearranged into palindromes based on a repeating pattern. For each substring, determine the minimum number of changes to transform it into a semi-palindrome.
Return the minimum number of letter changes required to achieve the k-partition. The solution should minimize modifications while ensuring each substring satisfies the semi-palindrome condition.
Examples
Example 1
Input: s = "abcac", k = 2
Output: 1
Divide s into "ab" and "cac" . "cac" is already semi-palindrome. Change "ab" to "aa" , it becomes semi-palindrome with d = 1 .
Example 2
Input: s = "abcdef", k = 2
Output: 2
Divide s into substrings "abc" and "def" . Each needs one change to become semi-palindrome.
Example 3
Input: s = "aabbaa", k = 3
Output: 0
Divide s into substrings "aa" , "bb" and "aa" . All are already semi-palindromes.
Constraints
- 2 <= s.length <= 200
- 1 <= k <= s.length / 2
- s contains only lowercase English letters.
Solution Approach
State Transition Dynamic Programming
Use dynamic programming (DP) to break the problem into smaller subproblems. Define dp[i][j] as the minimum number of changes required to partition the suffix of string s starting from index i into j valid parts. Start by considering the entire string and work backwards to evaluate the minimum changes for each possible partition.
Two Pointers for Efficient Substring Validation
Utilize the two pointers technique to check if a substring can be transformed into a semi-palindrome with minimal changes. This helps reduce the time complexity of validating the substrings compared to brute force checking of each possible combination.
Iterate Over Possible Substring Divisions
Iterate through possible ways to split the string into k parts. For each division, calculate the minimum changes needed to transform the resulting substrings into semi-palindromes, and update the DP table accordingly. Keep track of the best solution by selecting the division with the least number of changes.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | Depends on the final approach |
| Space | Depends on the final approach |
The time complexity depends on the final dynamic programming approach and how the two pointers are used to validate each substring. In general, the approach is expected to be O(n^2 * k), where n is the length of the string and k is the number of partitions. The space complexity is O(n * k) due to the DP table storing results for each substring and partition combination.
What Interviewers Usually Probe
- Look for the candidate's ability to optimize the problem using dynamic programming and two-pointer techniques.
- Check how well the candidate understands the state transition DP approach and its application to partitioning problems.
- Evaluate their understanding of optimizing substring validation to minimize time complexity.
Common Pitfalls or Variants
Common pitfalls
- Misunderstanding the concept of semi-palindromes and overcomplicating substring validation.
- Failing to optimize the validation process, resulting in a brute force solution with high time complexity.
- Incorrectly managing the transitions between substring partitions, leading to higher than necessary letter changes.
Follow-up variants
- Adjusting the value of k to test different partition sizes and their effect on time complexity.
- Varying the types of allowed transformations to include other string operations beyond letter changes.
- Increasing the length of the string to test the scalability of the solution.
FAQ
How do I approach the 'Minimum Changes to Make K Semi-palindromes' problem?
Use dynamic programming with a state transition approach to minimize letter changes. Define dp[i][j] as the minimum changes to split the suffix of s starting from index i into j parts.
What is a semi-palindrome in the context of this problem?
A semi-palindrome is a string that can be rearranged into palindromes based on a repeating pattern. The string needs to meet this condition with minimal letter changes.
What role do two pointers play in this problem?
Two pointers help efficiently check if a substring can be transformed into a semi-palindrome. This reduces the need for brute force checking and speeds up the solution.
What is the time complexity of the optimal solution?
The time complexity is typically O(n^2 * k), where n is the length of the string and k is the number of partitions. The exact time complexity depends on the final DP approach.
How can I optimize my solution for large strings?
Focus on optimizing substring validation using two pointers and manage the DP state transitions carefully to reduce unnecessary computations and improve efficiency.
Solution
Solution 1
#### Python3
class Solution:
def minimumChanges(self, s: str, k: int) -> int:
n = len(s)
g = [[inf] * (n + 1) for _ in range(n + 1)]
for i in range(1, n + 1):
for j in range(i, n + 1):
m = j - i + 1
for d in range(1, m):
if m % d == 0:
cnt = 0
for l in range(m):
r = (m // d - 1 - l // d) * d + l % d
if l >= r:
break
if s[i - 1 + l] != s[i - 1 + r]:
cnt += 1
g[i][j] = min(g[i][j], cnt)
f = [[inf] * (k + 1) for _ in range(n + 1)]
f[0][0] = 0
for i in range(1, n + 1):
for j in range(1, k + 1):
for h in range(i - 1):
f[i][j] = min(f[i][j], f[h][j - 1] + g[h + 1][i])
return f[n][k]Solution 2
#### Java
class Solution:
def minimumChanges(self, s: str, k: int) -> int:
n = len(s)
g = [[inf] * (n + 1) for _ in range(n + 1)]
for i in range(1, n + 1):
for j in range(i, n + 1):
m = j - i + 1
for d in range(1, m):
if m % d == 0:
cnt = 0
for l in range(m):
r = (m // d - 1 - l // d) * d + l % d
if l >= r:
break
if s[i - 1 + l] != s[i - 1 + r]:
cnt += 1
g[i][j] = min(g[i][j], cnt)
f = [[inf] * (k + 1) for _ in range(n + 1)]
f[0][0] = 0
for i in range(1, n + 1):
for j in range(1, k + 1):
for h in range(i - 1):
f[i][j] = min(f[i][j], f[h][j - 1] + g[h + 1][i])
return f[n][k]Continue Topic
two pointers
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State transition dynamic programming
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Hard
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