Palindrome Partition Cost Minimiser

Given a string `text`, partition it into one or more substrings such that every substring is a palindrome. The cost of a partition is the number of cuts made. Return the **minimum number of cuts** needed to partition `text` into palindromic substrings.

A single character is always a palindrome. If the whole string is already a palindrome, no cuts are needed (cost = 0).

**Example 1:**
```
Input: text = "acdkdca"
Output: 1
Explanation: Cut once → ["acd", "kdca"]? Not palindromes.
Actually: ["a", "cdkdc", "a"] costs 2 cuts.
Or: ["acdkdca"] — is it a palindrome? a==a, c==c, d==d, k middle. Yes! 0 cuts.
Output: 0
```

Hmm, let me use a clearer example:

**Example 1:**
```
Input: text = "abcba"
Output: 0
Explanation: The entire string is a palindrome, so 0 cuts needed.
```

**Example 2:**
```
Input: text = "abfcba"
Output: 2
Explanation: No single cut yields two palindromes.
One optimal: ["a", "bfcb", "a"] — "bfcb" is not a palindrome.
Another: ["abf", "cba"] — neither is a palindrome.
Best: ["a", "b", "fcba"]? "fcba" not palindrome.
Best: ["a", "bfcb", "a"] — not valid.
Try: ["ab", "fcb", "a"] — none palindromes.
["a","b","f","c","b","a"] = 5 cuts.
["a","b","fcbf"... no.
Actually minimum: ["a","bfcb","a"] — "bfcb" reversed is "bcfb" ≠ "bfcb". Not palindrome.
["abfcba"] reversed = "abcfba" ≠ "abfcba".
Minimum is ["a","b","f","c","b","a"] = 5 or ["a","bfcb","a"]=2 cuts but invalid.
Let me use text = "abacaba" → whole string is palindrome → 0 cuts.
```

**Example 1:**
```
Input: text = "xyzyx"
Output: 0
Explanation: "xyzyx" is a palindrome → 0 cuts.
```

**Example 2:**
```
Input: text = "abcde"
Output: 4
Explanation: No substring of length > 1 is a palindrome except single chars,
so minimum cuts = 4: ["a","b","c","d","e"].
```