MANG Problem #45: Maximize Score After N Operations (Hard)

1. Problem Statement

Mental Model

Breaking down a complex problem into its most efficient algorithmic primitive.

You are given nums, an array of positive integers of size 2 * n. You must perform n operations.

In the i-th operation (1-indexed), you will:

Choose two elements, x and y.
Receive a score of i * gcd(x, y).
Remove x and y from nums.

Return the maximum score you can receive after performing n operations.

2. Approach: DP + Bitmasking

We have an array of up to 14 elements (7 operations). A brute-force exploration of all pairs would be $O((2N)!)$, which is too slow.

Because $N$ is very small (max 14), this is a classic Bitmask DP problem. We use an integer's bits to represent the state of the array (e.g., 1001 means the first and fourth elements are available).

State: dp(mask) returns the max score for the remaining operations, given the current available elements (mask).
Operation Count: The operation number i can be derived from the number of set bits in the mask. i = (TotalElements - SetBits) / 2 + 1.
Transitions: For the current mask, try every possible pair of available elements. Find their GCD, multiply by i, and add the result of dp(new_mask). Return the maximum.

3. Java Implementation

public int maxScore(int[] nums) {
    int n = nums.length;
    // Memoization table for all 2^n possible bitmasks
    int[] dp = new int[1 << n];
    return solve(nums, (1 << n) - 1, dp);
}

private int solve(int[] nums, int mask, int[] dp) {
    if (mask == 0) return 0; // All elements used
    if (dp[mask] != 0) return dp[mask]; // Return cached result

    int n = nums.length;
    // Calculate which operation number we are on
    int opNum = (n - Integer.bitCount(mask)) / 2 + 1;
    int maxScore = 0;

    // Try all pairs of available elements
    for (int i = 0; i < n; i++) {
        if ((mask & (1 << i)) != 0) { // If element i is available
            for (int j = i + 1; j < n; j++) {
                if ((mask & (1 << j)) != 0) { // If element j is available
                    // Calculate score for this pair
                    int score = opNum * gcd(nums[i], nums[j]);
                    // Create new mask with elements i and j removed
                    int newMask = mask ^ (1 << i) ^ (1 << j);
                    
                    // Recursive call
                    maxScore = Math.max(maxScore, score + solve(nums, newMask, dp));
                }
            }
        }
    }
    dp[mask] = maxScore;
    return maxScore;
}

private int gcd(int a, int b) {
    return b == 0 ? a : gcd(b, a % b);
}

4. 5-Minute "Video-Style" Walkthrough

The "Aha!" Moment: Any array length $\le 20$ combined with a combinatorial optimization is almost always Bitmask DP. A 14-bit integer represents every combination of available elements effortlessly.
The State Space: dp[mask] caches the best score for a specific set of remaining numbers. It doesn't matter how we reached this mask (which pairs were picked previously), the optimal future score only depends on what's left.
Bit Operations:
- (mask & (1 << i)) != 0 checks if the $i$-th bit is 1 (element is available).
- mask ^ (1 << i) ^ (1 << j) toggles the $i$-th and $j$-th bits to 0 (removes the elements).

5. Interview Discussion

Interviewer: "What is the time complexity?"
You: "O(2^N * N^2) where N is the total number of elements. There are 2^N states, and for each state, we use two loops to try all O(N^2) pairs."
Interviewer: "Could we pre-compute the GCDs?"
You: "Yes, we could compute the GCD of all pairs once and store it in an $N \times N$ matrix to avoid redundant calculations inside the loops, which speeds up the constant factor significantly."

5. Verbal Interview Script (Staff Tier)

Interviewer: "Walk me through your optimization strategy for this problem."

You: "When approaching this type of challenge, my primary objective is to identify the underlying Monotonicity or Optimal Substructure that allow us to bypass a naive brute-force search. In my implementation of 'MANG Problem #45: Maximize Score After N Operations (Hard)', I focused on reducing the time complexity by leveraging a Dynamic Programming state transition. This allows us to handle input sizes that would typically cause a standard O(N^2) approach to fail. Furthermore, I prioritized memory efficiency by optimizing the DP state to use only a 1D array. This ensures that the application remains performant even under heavy garbage collection pressure in a high-concurrency Java environment."

6. Staff-Level Interview Follow-Ups

Once you provide the optimized solution, a senior interviewer at Google or Meta will likely push you further. Here is how to handle the most common follow-ups:

Follow-up 1: "How does this scale to a Distributed System?"

If the input data is too large to fit on a single machine (e.g., billions of records), we would move from a single-node algorithm to a MapReduce or Spark-based approach. We would shard the data based on a consistent hash of the keys and perform local aggregations before a global shuffle and merge phase, similar to the logic used in External Merge Sort.

Follow-up 2: "What are the Concurrency implications?"

In a multi-threaded Java environment, we must ensure that our state (e.g., the DP table or the frequency map) is thread-safe. While we could use synchronized blocks, a higher-performance approach would be to use AtomicVariables or ConcurrentHashMap. For problems involving shared arrays, I would consider a Work-Stealing pattern where each thread processes an independent segment of the data to minimize lock contention.

7. Performance Nuances (The Java Perspective)

Autoboxing Overhead: When using HashMap<Integer, Integer>, Java performs autoboxing which creates thousands of Integer objects on the heap. In a performance-critical system, I would use a primitive-specialized library like fastutil or Trove to use Int2IntMap, significantly reducing GC pauses.
Recursion Depth: As discussed in the code, recursive solutions are elegant but risky for deep inputs. I always ensure the recursion depth is bounded, or I rewrite the logic to be Iterative using an explicit stack on the heap to avoid StackOverflowError.

Key Takeaways

Choose two elements, x and y.
Receive a score of i * gcd(x, y).
Remove x and y from nums.