The O(n) Club: Partition Array for Maximum Sum – Your Brain vs. Brute Force

The O(n) Club: Partition Array for Maximum Sum – Your Brain vs. Brute Force

⚡ TL;DR

Split your array into chunks of up to k—not exactly k, not “something that sort of feels like k”—and in each chunk, every value is replaced by the max of that chunk (because fairness is overrated). Your job: find the partitioning that brings the biggest possible sum. Brute force? Sure, if you want your machine to sound like a 90s dial-up modem.

// Brute force – looks fun until your laptop combusts
int maxSum = 0;
for (every legal chunking up to k) {
    int localSum = 0;
    for (each group) {
        int maxVal = group max;
        localSum += maxVal * group size;
    }
    maxSum = Math.max(maxSum, localSum);
}
// Spoiler: don't brute force.

🧠 How Devs Usually Mess This Up

A classic: assume “at most k” means “exactly k,” then stuff numbers into whatever chunk crosses your mind. Or, go fully greedy—always picking the local max—then act shocked when your answer is off by a mile. Most tragic: forgetting to replace every number in a chunk with the chunk’s max, as if your array will fix itself. (It will not.)

🔍 What’s Actually Going On

Imagine giving out bonuses at a company, but your policy is: group anyone you want (up to k per group), and every worker in that group gets whatever the highest-paid person gets. Grouping well means more happy employees and, in this case, a max sum. The trick? For every ending position in the array, try all group lengths up to k. For each, take the chunk, cram everyone up to the local max, and stack up your running best using—what else—dynamic programming. No misfit left behind. Only max value, every time.

🛠️ PseudoCode

• Set up dp[i] as: max sum you can achieve using first i elements.
• For each index i from 1 to n:
• Walk backwards up to k steps (call it group size j).
• Don’t step past the start. (Arrays don’t like negative indices.)
• Track current group max while expanding the window.
• Set dp[i] = max(dp[i], dp[i-j] + max_in_group * j).
• After all positions, dp[n] holds the optimal sum.

💻 The Code

public int maxSumAfterPartitioning(int[] arr, int k) {
    int n = arr.length;
    int[] dp = new int[n + 1];
    for (int i = 1; i <= n; i++) {
        int max = 0;
        for (int j = 1; j <= k && i - j >= 0; j++) {
            max = Math.max(max, arr[i - j]);
            dp[i] = Math.max(dp[i], dp[i - j] + max * j);
        }
    }
    return dp[n];
}

⚠️ Pitfalls, Traps & Runtime Smacks

• “At most k” means 1 up to k—don’t force chunks to all be size k like a robot with a for-loop for a brain.
• Transform each chunk: sum the chunk’s max times its length, not the original values (unless you like bad results).
• Think about weird arrays: all numbers equal? Tiny arrays? k bigger than the array? Yes, you still have to handle those.
• Time: O(nk). Space: O(n). Unless n is your AWS bill, relax.

🧠 Interviewer Brain-Tattoo

Greedy gets you grumpy; DP gets you glory. Partition like you mean it—or explain to your interviewer how “at most k” somehow meant “at most my best guess.”