The O(n) Club: Integer Break: How to Multiply Your Happiness (and Numbers)

The O(n) Club: Integer Break — How to Multiply Your Happiness (and Numbers)

⚡ TL;DR

Split your number n into at least two positive integers. Maximize the product? Think dynamic programming or just spam 3s if you want to show off. Quick Java for the curious:

// Classic DP approach
int integerBreak(int n) {
    int[] dp = new int[n + 1];
    dp[1] = 1;
    for (int i = 2; i <= n; i++)
        for (int j = 1; j < i; j++)
            dp[i] = Math.max(dp[i], Math.max(j * (i - j), j * dp[i - j]));
    return dp[n];
}

🧠 How Devs Usually Mess This Up

We’ve all been that developer: break everything into 1s (hope is free, right?)—then watch your product stubbornly stay at 1. Or you return n itself for n=3, even though the rules say ‘actually break it, please.’ Interviewers sigh, recruiters cry, nobody’s happy. Bonus points for panicking and hardcoding some random cases at midnight.

🔍 What’s Actually Going On

This is the playground of greedy meets DP. Splitting into too many pieces (like all 2s or worse, all 1s) is like trying to cut a pizza into atomic slices—sad and hopeless. The sweet spot? Threes. Chop off as many 3s as you can, unless you’re stuck with leftovers that are, well, socially awkward—like 2 or 4. The magical math behind it? Multiplying 3s gives you quicker exponential growth than lower numbers. It’s not magic, it’s just algebra’s slightly drunk cousin, optimization.

🛠️ PseudoCode

• Initialize: Set dp[1]=1 (it can’t break but we’re polite).
• For each i from 2 to n:
  • For every j in 1 to (i-1) (try every first break):
    • Option one: don’t break further: j * (i-j)
    • Option two: break the rest: j * dp[i-j]
    • Set dp[i] to the maximum you get.
• Return dp[n] and take a victory lap.