How do you solve Minimum Changes to Make K Semi-palindromes on LeetCode?

Minimum Changes to Make K Semi-palindromes (LeetCode #2911) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n^3) time complexity and O(n²) space complexity. Key insight: Understanding how to partition the string is crucial for minimizing changes.

What is the brute force solution for Minimum Changes to Make K Semi-palindromes?

The brute force approach for Minimum Changes to Make K Semi-palindromes is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible way to partition the string into k substrings and calculating the changes needed to make each substring a semi-palindrome. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n^3).

What algorithm or data structure is used to solve Minimum Changes to Make K Semi-palindromes?

Minimum Changes to Make K Semi-palindromes uses the following concepts: Two Pointers, String, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Two Pointers.

What are the interview tips for Minimum Changes to Make K Semi-palindromes?

Explain your thought process clearly while coding. Discuss the trade-offs of different approaches before diving into code. Practice similar problems to become familiar with dynamic programming techniques.

What are common mistakes when solving Minimum Changes to Make K Semi-palindromes?

Not considering all valid divisors when checking for semi-palindromes. Failing to initialize DP arrays properly, leading to incorrect results.

#2911

Minimum Changes to Make K Semi-palindromes

Hard

Two PointersStringDynamic ProgrammingDynamic ProgrammingTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n^3)
Space	O(1)	O(n²)

💡

Intuition

Time O(n^3)Space O(n²)

The optimal approach uses dynamic programming to efficiently calculate the minimum changes needed by storing results of subproblems. This avoids redundant calculations and significantly reduces the time complexity.

⚙️

Algorithm

3 steps

1Step 1: Create a DP table where dp[i][j] represents the minimum changes needed to partition the substring s[i:] into j parts.
2Step 2: Precompute the minimum changes needed to convert any substring s[i..x] into a semi-palindrome for all valid divisors.
3Step 3: Fill the DP table using the relation dp[i][j] = min(dp[x + 1][j - 1] + v[i][x]) for all valid x.

solution.py33 lines

1def minChanges(s, k):
2    n = len(s)
3    dp = [[float('inf')] * (k + 1) for _ in range(n + 1)]
4    v = [[0] * n for _ in range(n)]
5
6    def isSemiPalindrome(sub):
7        changes = 0
8        for d in range(1, len(sub)):
9            groups = [[] for _ in range(d)]
10            for i in range(len(sub)):
11                groups[i % d].append(sub[i])
12            changes_needed = 0
13            for group in groups:
14                count = [0] * 26
15                for char in group:
16                    count[ord(char) - ord('a')] += 1
17                changes_needed += len(group) - max(count)
18            changes = min(changes, changes_needed)
19        return changes
20
21    for i in range(n):
22        for j in range(i, n):
23            v[i][j] = isSemiPalindrome(s[i:j + 1])
24
25    for i in range(n):
26        dp[i][1] = v[i][n - 1]
27
28    for j in range(2, k + 1):
29        for i in range(n):
30            for x in range(i, n):
31                dp[i][j] = min(dp[i][j], dp[x + 1][j - 1] + v[i][x])
32
33    return dp[0][k]

ℹ

Complexity note: The time complexity is O(n^3) due to the nested loops for filling the DP table and calculating semi-palindrome changes, while space complexity is O(n²) for storing the DP and precomputed values.

1Understanding how to partition the string is crucial for minimizing changes.
2Precomputing semi-palindrome changes for substrings can save time in dynamic programming.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.