The O(n) Club: Palindrome Partitioning II—Fewer Cuts, Less Regret

The O(n) Club: Palindrome Partitioning II—Fewer Cuts, Less Regret

⚡ TL;DR

Chop a string into palindromic slices using as few cuts as possible (think sushi, but for code). Brute force is a crime against runtime; dynamic programming gets you home before dark.

// Brute force? Life’s too short.
String s = "aab";
int cuts = s.length() - 1; // 🙃
// There’s a smarter way. See below.

🧠 How Devs Usually Mess This Up

Most devs wrestle this beast by recalculating every palindrome on the fly, dragging the runtime to O(n³)—great for killing CPUs, bad for passing interviews. Others mix up partitions and cuts, returning the number of palindromic pieces instead of the number of times they actually had to hack the string apart. Bonus points for off-by-one mistakes: returning dp[n] instead of dp[n-1], or forgetting to subtract one from the partition count and confusing both the interviewer and your future self.

🔍 What’s Actually Going On

Pretend you’re a sushi chef with only one trick: you can only slice if the chunk is a palindrome. The goal? Serve the dish in as few breathtaking, mirrored bites as possible. The real play is: precompute every palindromic substring (that’s your menu), then use dynamic programming to find the tiniest set of cuts that gets the job done. No slicing on the fly—it’s all about prepping the kitchen and minimizing your moves. Thank me when you don’t have to burn the midnight oil debugging false negatives.

🛠️ PseudoCode

• Step 1: Precompute palindrome[i][j]
  • palindrome[i][j] is true if s.charAt(i) == s.charAt(j) and the substring between is also a palindrome (or empty).
  • One-letter substrings: always a palindrome. Two same letters: also a palindrome. Sometimes, life is simple.
• Step 2: Fill DP array where dp[i] = min cuts for s[0...i]
  • For every ending index i, check all j ≤ i.
  • If palindrome[j][i]:
    • If j == 0: No cuts needed—it’s a palindrome all the way.
    • Otherwise: dp[i] = Math.min(dp[i], dp[j-1] + 1). Update with minimal regret.
• Step 3: The answer is dp[n-1]—no extra subtraction circus.

💻 The Code

public int minCut(String s) {
    int n = s.length();
    boolean[][] palindrome = new boolean[n][n];
    for (int i = 0; i < n; i++) palindrome[i][i] = true;
    for (int len = 2; len <= n; len++) {
        for (int start = 0; start <= n - len; start++) {
            int end = start + len - 1;
            if (s.charAt(start) == s.charAt(end)) {
                if (len == 2) palindrome[start][end] = true;
                else palindrome[start][end] = palindrome[start + 1][end - 1];
            }
        }
    }
    int[] dp = new int[n];
    for (int i = 0; i < n; i++) {
        if (palindrome[0][i]) {
            dp[i] = 0;
        } else {
            dp[i] = i;
            for (int j = 1; j <= i; j++) {
                if (palindrome[j][i]) {
                    dp[i] = Math.min(dp[i], dp[j - 1] + 1);
                }
            }
        }
    }
    return dp[n - 1];
}

⚠️ Pitfalls, Traps & Runtime Smacks

• Partitions are not cuts. Cuts = pieces – 1. Or just use dp[n-1] and skip the heartbreak.
• Brute force equals timeout. Your interviewer didn’t bring snacks for a reason.
• Edge cases are not optional: empty string returns 0; all same letters, 0 cuts; no palindromes, n-1 cuts (literal one-cut-per-character chaos).
• Space: O(n²) for your palindrome table. Unless you need to run this on a smartwatch, you’re probably fine.
• Off-by-one is not just a meme—it’s real, and it’s coming for you. dp[n-1] please.

🧠 Interviewer Brain-Tattoo

“Counting partitions gives you extra slices nobody ordered—count cuts. Your runtime (and ego) will thank you.”