Understanding the bottom-up solution to the palindrome partitioning problem

To solve the palindrome partitioning problem using a bottom-up dynamic programming approach, we fill a table iteratively beginning from the smallest subproblems and building towards the final answer. As with any bottom-up solution, we process subproblems in an order that guarantees every result we depend on has already been computed and stored before we need it.

The bottom-up solution

We solve this problem in two stages. First, we precompute a 2D table that records, for every pair of indices (i, j), whether the substring s[i...j] is a palindrome. Then, we use this table to build a 1D array that records the minimum number of cuts needed for each substring of the form s[i...n-1], that is, each substring that starts at some index i and extends all the way to the last index of the string.

We first precompute all palindromic substrings and then use it to find the minimum cuts needed to partition the string.

To determine whether each substring of s is a palindrome, we create a 2D boolean array isPalindrome of dimensions n × n, where isPalindrome[i][j] stores whether the substring from index i to j in the string s is a palindrome.

The isPalindrome array has a size n x n and is initialized with false.

We initialize isPalindrome[i][i] to true for all 0 <= i < n since every single character is a palindrome, and this serves as the base case for the problem.

Why is only the diagonal initialized to true, and everything else to false?

The entries isPalindrome[i][i] correspond to substrings of length 1, which are always palindromes, so we set them to true upfront. All other entries correspond to substrings of length 2 or more whose palindrome status depends on the characters and possibly on inner substrings. These are computed during the main loops, and initializing them to false gives a safe default so that if an entry is never explicitly updated, it correctly reflects that the corresponding substring is not a palindrome.