Problem Description

You are given a string s containing lowercase letters and an integer k. Your task is to perform two operations:

1. Change characters: First, you can change some characters in s to other lowercase English letters.
2. Partition into palindromes: Then, divide the modified string into exactly k non-empty, non-overlapping substrings where each substring must be a palindrome.

The goal is to find the minimum number of character changes needed to make such a partition possible.

For example, if you have a string that needs to be divided into k palindromic parts, you want to strategically change the fewest characters possible so that when you split the string into k consecutive substrings, each piece forms a palindrome.

A palindrome is a string that reads the same forwards and backwards (like "aba" or "racecar"). The substrings must be disjoint (non-overlapping) and together they must form the entire original string s.

The function should return a single integer representing the minimum number of character changes required to achieve this palindromic partition.

Intuition

The key insight is that this is an optimization problem with two interconnected decisions: where to partition the string and which characters to change. We need to find the best way to split the string into k parts while minimizing character changes.

Let's think about this step by step. If we need to divide a string into k palindromic parts, we're essentially making k-1 cuts in the string. For each resulting substring, we need to figure out how many characters must be changed to make it a palindrome.

For any substring to become a palindrome, we compare characters from both ends moving toward the center. If characters at mirror positions don't match, we need to change one of them. For a substring s[i...j], the minimum changes needed is the number of mismatched pairs when comparing s[i] with s[j], s[i+1] with s[j-1], and so on.

This suggests a dynamic programming approach where we build up solutions for smaller subproblems. We can think of it as: "What's the minimum cost to partition the first i characters into j palindromic parts?"

For each position i and number of parts j, we try all possible positions for the last partition. If we place the (j-1)-th partition at position h, then we need:

• The optimal cost for partitioning the first h characters into j-1 parts
• Plus the cost of making characters from position h to i-1 into a palindrome

By trying all valid positions for h and taking the minimum, we find the optimal solution for each subproblem. The preprocessing step calculates the palindrome conversion cost for all possible substrings, allowing us to quickly look up these values during the main dynamic programming computation.

This bottom-up approach ensures we consider all possible ways to partition the string while reusing previously computed optimal solutions for smaller subproblems.

Learn more about Dynamic Programming patterns.

Solution Approach

The solution uses dynamic programming with preprocessing to efficiently solve this problem.

Step 1: Preprocessing - Calculate palindrome conversion costs

First, we create a 2D array g[i][j] where g[i][j] represents the minimum number of changes needed to convert substring s[i...j] into a palindrome.

g = [[0] * n for _ in range(n)]
for i in range(n - 1, -1, -1):
    for j in range(i + 1, n):
        g[i][j] = int(s[i] != s[j])
        if i + 1 < j:
            g[i][j] += g[i + 1][j - 1]

We iterate from bottom-right to top-left of the matrix:

• For each substring s[i...j], we check if the characters at positions i and j match
• If they don't match, we need 1 change (stored as int(s[i] != s[j]))
• If there are more characters between them (i + 1 < j), we add the cost for the inner substring s[i+1...j-1]

This gives us the palindrome conversion cost for any substring in O(n²) time.

Step 2: Dynamic Programming - Find optimal partitioning

We define f[i][j] as the minimum changes needed to partition the first i characters into j palindromic substrings.

f = [[0] * (k + 1) for _ in range(n + 1)]
for i in range(1, n + 1):
    for j in range(1, min(i, k) + 1):
        if j == 1:
            f[i][j] = g[0][i - 1]
        else:
            f[i][j] = inf
            for h in range(j - 1, i):
                f[i][j] = min(f[i][j], f[h][j - 1] + g[h][i - 1])

The algorithm works as follows:

• For each position i (1 to n) and number of partitions j (1 to min(i, k)):
  • Base case: If j = 1, we need to make the entire substring s[0...i-1] a palindrome, so f[i][j] = g[0][i-1]
  • General case: For j > 1, we try all possible positions h for the (j-1)-th partition:
    • The cost is f[h][j-1] (optimal cost for first h characters with j-1 partitions)
    • Plus g[h][i-1] (cost to make substring from position h to i-1 a palindrome)
    • We take the minimum across all valid h values

The final answer is f[n][k], representing the minimum changes needed to partition all n characters into exactly k palindromic substrings.

Time Complexity: O(n²k) - We have n × k states, and for each state, we try up to n positions for the last partition.

Space Complexity: O(n² + nk) - For the preprocessing array g and the DP array f.

Example Walkthrough

Let's walk through a small example with string s = "abcac" and k = 2 (we need to partition into 2 palindromic substrings).

Step 1: Preprocessing - Calculate palindrome conversion costs

We build the g matrix where g[i][j] = minimum changes to make substring s[i...j] a palindrome.

For s = "abcac":

• g[0][1]: substring "ab" → needs 1 change (a≠b)
• g[0][2]: substring "abc" → check outer chars: a≠c (1 change), no inner substring → total: 1
• g[0][3]: substring "abca" → check outer chars: a=a (0 changes), inner is "bc" which needs g[1][2]=1 → total: 1
• g[0][4]: substring "abcac" → check outer chars: a≠c (1 change), inner is "bca" which needs g[1][3]=1 → total: 2
• g[1][2]: substring "bc" → needs 1 change (b≠c)
• g[1][3]: substring "bca" → check outer chars: b≠a (1 change), no inner substring → total: 1
• g[1][4]: substring "bcac" → check outer chars: b≠c (1 change), inner is "ca" which needs g[2][3]=1 → total: 2
• g[2][3]: substring "ca" → needs 1 change (c≠a)
• g[2][4]: substring "cac" → check outer chars: c=c (0 changes), no inner substring → total: 0
• g[3][4]: substring "ac" → needs 1 change (a≠c)

Complete g matrix:

    0  1  2  3  4
0 [ 0  1  1  1  2 ]
1 [ -  0  1  1  2 ]
2 [ -  -  0  1  0 ]
3 [ -  -  -  0  1 ]
4 [ -  -  -  -  0 ]

Step 2: Dynamic Programming - Find optimal partitioning

We build f[i][j] = minimum changes to partition first i characters into j palindromes.

For i=1, j=1: First 1 character ("a") as 1 palindrome → f[1][1] = g[0][0] = 0

For i=2, j=1: First 2 characters ("ab") as 1 palindrome → f[2][1] = g[0][1] = 1

For i=2, j=2: First 2 characters ("ab") as 2 palindromes

• Try splitting at position 1: "a" | "b"
  • Cost = f[1][1] + g[1][1] = 0 + 0 = 0
• f[2][2] = 0

For i=3, j=1: First 3 characters ("abc") as 1 palindrome → f[3][1] = g[0][2] = 1

For i=3, j=2: First 3 characters ("abc") as 2 palindromes

• Try splitting at position 1: "a" | "bc"
  • Cost = f[1][1] + g[1][2] = 0 + 1 = 1
• Try splitting at position 2: "ab" | "c"
  • Cost = f[2][1] + g[2][2] = 1 + 0 = 1
• f[3][2] = min(1, 1) = 1

For i=4, j=2: First 4 characters ("abca") as 2 palindromes

• Try splitting at position 1: "a" | "bca"
  • Cost = f[1][1] + g[1][3] = 0 + 1 = 1
• Try splitting at position 2: "ab" | "ca"
  • Cost = f[2][1] + g[2][3] = 1 + 1 = 2
• Try splitting at position 3: "abc" | "a"
  • Cost = f[3][1] + g[3][3] = 1 + 0 = 1
• f[4][2] = min(1, 2, 1) = 1

For i=5, j=2: All 5 characters ("abcac") as 2 palindromes

• Try splitting at position 1: "a" | "bcac"
  • Cost = f[1][1] + g[1][4] = 0 + 2 = 2
• Try splitting at position 2: "ab" | "cac"
  • Cost = f[2][1] + g[2][4] = 1 + 0 = 1
• Try splitting at position 3: "abc" | "ac"
  • Cost = f[3][1] + g[3][4] = 1 + 1 = 2
• Try splitting at position 4: "abca" | "c"
  • Cost = f[4][1] + g[4][4] = 1 + 0 = 1
• f[5][2] = min(2, 1, 2, 1) = 1

Final Answer: f[5][2] = 1

The optimal solution requires 1 character change. One optimal partition is "ab" | "cac" where we change "ab" to "aa" (or "bb"), requiring 1 change total.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n^2 × k)

The time complexity is determined by three main parts:

1. Building the cost matrix g: The nested loops iterate from i = n-1 to 0 and j = i+1 to n-1, resulting in O(n^2) operations to calculate the cost of converting each substring s[i:j+1] into a palindrome.

2. Dynamic programming calculation: The main DP section has three nested loops:

   • The outer loop iterates i from 1 to n: O(n)
   • The middle loop iterates j from 1 to min(i, k): O(k) in the worst case
   • The inner loop (when j > 1) iterates h from j-1 to i-1: O(n) in the worst case

   This gives us O(n × k × n) = O(n^2 × k) for the DP computation.

Since the DP computation dominates, the overall time complexity is O(n^2 × k).

Space Complexity: O(n × (n + k))

The space complexity comes from:

1. Matrix g: A 2D array of size n × n, requiring O(n^2) space to store the cost of converting each substring into a palindrome.

2. DP table f: A 2D array of size (n+1) × (k+1), requiring O(n × k) space to store the minimum cost for partitioning the first i characters into j palindromes.

The total space complexity is O(n^2 + n × k) = O(n × (n + k)) since we can factor out n from both terms.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Cost Calculation for Single Character Substrings

A common mistake is not handling single-character substrings correctly in the cost preprocessing table. Single characters are always palindromes and require 0 changes, but the cost calculation logic might incorrectly compute a non-zero value.

Problematic Code:

# This might incorrectly calculate cost for single characters
for start in range(n - 1, -1, -1):
    for end in range(start + 1, n):  # Missing start == end case
        cost[start][end] = int(s[start] != s[end])
        if start + 1 < end:
            cost[start][end] += cost[start + 1][end - 1]

Solution:

# Initialize diagonal (single characters) explicitly
for i in range(n):
    cost[i][i] = 0  # Single character is always a palindrome

for start in range(n - 1, -1, -1):
    for end in range(start + 1, n):
        cost[start][end] = int(s[start] != s[end])
        if start + 1 < end:
            cost[start][end] += cost[start + 1][end - 1]

2. Off-by-One Errors in Index Mapping

The DP table uses 1-based indexing for string length while the cost table uses 0-based indexing. This mismatch often leads to accessing wrong indices.

Problematic Code:

# Incorrect index mapping
dp[length][groups] = cost[0][length]  # Should be length - 1
# Or
dp[split_point][groups - 1] + cost[split_point - 1][length - 1]  # Wrong!

Solution:

# Correct index mapping
dp[length][groups] = cost[0][length - 1]  # length characters means index 0 to length-1
# And
dp[split_point][groups - 1] + cost[split_point][length - 1]  # split_point is already 0-based

3. Insufficient Initialization of DP Table

Not properly initializing the DP table with infinity for impossible states can lead to incorrect minimum calculations.

Problematic Code:

dp = [[0] * (k + 1) for _ in range(n + 1)]  # All initialized to 0
# Later comparison might pick 0 as minimum incorrectly

Solution:

dp = [[float('inf')] * (k + 1) for _ in range(n + 1)]
dp[0][0] = 0  # Base case: 0 characters into 0 groups needs 0 changes

for length in range(1, n + 1):
    for groups in range(1, min(length, k) + 1):
        if groups == 1:
            dp[length][groups] = cost[0][length - 1]
        else:
            for split_point in range(groups - 1, length):
                dp[length][groups] = min(
                    dp[length][groups], 
                    dp[split_point][groups - 1] + cost[split_point][length - 1]
                )

4. Incorrect Loop Bounds for Split Points

The split point iteration must ensure that there are enough characters left for both the previous partitions and the current partition.

Problematic Code:

# Might not leave enough characters for previous groups
for split_point in range(1, length):  # Wrong starting point

Solution:

# Ensure at least (groups - 1) characters for previous (groups - 1) partitions
for split_point in range(groups - 1, length):
    # split_point characters go to first (groups - 1) partitions
    # Remaining (length - split_point) characters form the last partition