KPALINDROME Partitioning
December 21, 2023
Problem #
Given a string s, partition the string into k nonempty substrings such that every substring is a palindrome. Return the minimum number of cuts needed to make this possible.
Solution #
Here’s the Python code for the problem statement:
def is_palindrome(s):
"""Check if the given string is a palindrome."""
return s == s[::1]
def min_cuts_for_palindromes(s, k):
"""
Function to find the minimum number of cuts needed to partition the string into k palindromic substrings.
"""
n = len(s)
# Edge cases
if n == 0 or k == 0 or k > n:
return 1 # Impossible to partition
if k == n:
return 0 # Each character is a palindrome, so no cuts needed
if k == 1:
return 0 if is_palindrome(s) else 1 # One cut needed if not already a palindrome
# Initialize a 2D array for dynamic programming
dp = [[float('inf')] * n for _ in range(k)]
# Fill the first row (k=1) of the DP table
for j in range(n):
dp[0][j] = 0 if is_palindrome(s[:j+1]) else 1
# Fill the DP table for k > 1
for i in range(1, k):
for j in range(n):
if is_palindrome(s[:j+1]):
dp[i][j] = 0
else:
for l in range(j):
if is_palindrome(s[l+1:j+1]):
dp[i][j] = min(dp[i][j], dp[i1][l] + 1)
return dp[1][1] if dp[1][1] != float('inf') else 1
# Example usage
s = "ababbbabbababa"
k = 3
result = min_cuts_for_palindromes(s, k)
print(f"Minimum number of cuts: {result}")
This code defines two functions:
is_palindrome
to check if a given string is a palindrome.min_cuts_for_palindromes
to find the minimum number of cuts needed to partition a string intok
palindromic substrings using dynamic programming.
You can use the min_cuts_for_palindromes
function by passing the string and the desired number of palindromic substrings (k
) as arguments. The function will return the minimum number of cuts required. The example usage at the end demonstrates how to use these functions with the string "ababbbabbababa"
and k = 3
.
Solution Analysis: Step by Step #
The code you’re looking at is designed to solve a specific problem: Given a string s
, it partitions this string into k
nonempty substrings in such a way that each substring is a palindrome. The goal is to do this partitioning with the minimum number of cuts. The solution is implemented using dynamic programming, a method that solves complex problems by breaking them down into simpler subproblems.
Here’s a stepbystep explanation of the code:
Function: is_palindrome(s)
#
 Purpose: This helper function checks if a given string
s
is a palindrome (reads the same forward and backward).  Implementation: It simply compares the string with its reverse (
s[::1]
). If they are the same, it returnsTrue
, indicating the string is a palindrome.
Function: min_cuts_for_palindromes(s, k)
#
 Purpose: This is the main function that calculates the minimum number of cuts required to partition the string
s
intok
palindromic substrings.  Parameters: It takes two parameters:
s
(the input string) andk
(the number of palindromic substrings desired).
Inside the Function: #

Initial Checks and Edge Cases:
 Check the length of the string (
n = len(s)
), and handle edge cases like ifs
is empty, ifk
is 0 or greater thann
, etc.
 Check the length of the string (

Dynamic Programming Table Initialization:
 Create a 2D list
dp
withk
rows andn
columns, filled initially withfloat('inf')
. This table is used for dynamic programming, wheredp[i][j]
represents the minimum cuts needed to partition the substrings[:j+1]
intoi+1
palindromic substrings.
 Create a 2D list

Filling the First Row of DP Table (
k=1
): For each position
j
in the string, check if the substrings[:j+1]
is a palindrome. If it is, no cut is needed (dp[0][j] = 0
); otherwise, one cut is needed (dp[0][j] = 1
).
 For each position

Filling the Rest of the DP Table (
k > 1
): For each
i
(from 1 tok1
) and for eachj
(from 0 ton1
), the algorithm checks ifs[:j+1]
is a palindrome. If it is, no further cuts are needed (dp[i][j] = 0
).  Otherwise, it iterates through all possible earlier cuts (up to position
j
) and checks if the resulting substring is a palindrome. It then selects the minimum cuts needed among these options.
 For each

Returning the Result:
 The value at
dp[1][1]
will be the minimum number of cuts needed for the entire string. If this value isfloat('inf')
, it means it’s not possible to partition the string as desired, so1
is returned.
 The value at
Example Usage: #
 The example provided at the end (
s = "ababbbabbababa", k = 3
) demonstrates how to use the function. It calculates and prints the minimum number of cuts required for the given string andk
value.
This code effectively uses dynamic programming to optimize the process of finding the minimum cuts, as it reuses the results of smaller subproblems to solve larger ones.