Maximal Bipartite Energy Synthesizer 2 — Problem Statement & Solution Guide

TreesHardTreap Balanced Tree

TimeO(N log N)

SpaceO(N)

Problem Description

Given a high-dimensional input dataset or state graph of length N, calculate the optimal result using the Treap Balanced Tree algorithm. Formally, implement an optimal sub-linear or O(N log N) solution capable of satisfying strict time and space complexity limits under maximum competitive edge cases.

DSA Pattern Breakdown

"Maximal Bipartite Energy Synthesizer 2"

hard

WHY DOES IT MATTER?

Tree DP with naive merging leads to exponential time, which is unacceptable for large N. Using a balanced BST pattern reduces the merge complexity to logarithmic, making the algorithm scalable and predictable.

OPTIMIZATION CHALLENGE

The core insight is to treat each child’s DP value as a key in a Treap and use the heap property to quickly retrieve the best combination, turning an O(2^k) merge into O(k log k).

REAL-WORLD CONNECTION

In distributed systems, merging state from multiple shards often requires efficient aggregation. A Treap‑like structure can be used to maintain the top‑k metrics across shards, similar to how we aggregate DP states across tree children.

Always keep the Treap implementation generic and test its balancing on worst‑case inputs; a single unbalanced Treap can degrade the entire algorithm.

COMPLEXITY AT A GLANCE

⏱ Time:O(N log N)

💾 Space:O(N)

Core Theory — Why This Approach?

The problem reduces to a classic tree dynamic programming (DP) where each node can be in one of two states: selected or not selected. A naive DP that merges child states by iterating over all combinations would lead to an exponential blow‑up, especially when the tree has high branching factor or when we need to maintain additional constraints such as a global energy budget. To achieve sub‑linear or O(N log N) performance, we replace the brute‑force merging with a balanced binary search tree – specifically a Treap – that stores the DP values of processed children. The Treap allows us to insert, delete, and query the maximum prefix sum in logarithmic time, enabling us to merge child sub‑trees efficiently while preserving the optimal choice for each node.

Treaps combine the binary search tree property (ordered by key) with a heap property (priority), ensuring expected O(log N) operations. By treating the DP value of a child as the key and its contribution to the parent’s energy as the priority, we can quickly find the best combination of children that maximizes the bipartite energy. This approach eliminates the need to enumerate all subsets of children, which would otherwise be infeasible for large N. The expected logarithmic time per merge, multiplied by the number of edges, yields an overall O(N log N) algorithm that comfortably meets strict time and space constraints.

Moreover, the Treap’s randomized nature guarantees balanced height on average, avoiding worst‑case scenarios that plague deterministic BSTs. This property is crucial in competitive settings where input can be adversarial. By integrating the Treap into the DP recurrence, we maintain a clean, modular solution that scales to the maximum input sizes specified in the problem statement.

Interview Questions on This Problem

Q1How would you explain the advantage of using a Treap over a standard BST in a tree DP problem?

A Treap maintains expected O(log N) operations by randomizing node priorities, ensuring balanced height on average. This guarantees that insertions, deletions, and queries (like finding the maximum prefix sum) remain efficient even in worst‑case input sequences, which is essential for meeting strict time limits in competitive programming.

Q2What is the key idea behind merging DP states of child subtrees using a balanced BST?

Instead of enumerating all combinations of child states, we store each child’s DP contribution in a balanced BST keyed by its value. We can then query the maximum achievable sum by retrieving the best prefix or suffix in O(log N) time, effectively collapsing the exponential merge into a logarithmic operation.

Q3In a high‑growth fintech startup interview, how would you justify the use of a randomized data structure like a Treap for production code?

I would highlight that Treaps provide deterministic expected performance, are simple to implement, and avoid pathological cases that can occur with deterministic BSTs. For production, we can seed the random generator for reproducibility and add tests to ensure balance, making the structure reliable for real‑time analytics pipelines.

Examples

Example 1

Input

[12, 18, 24, 10]

Output

Explanation: Step-by-step: Given the input array [12, 18, 24, 10], we first calculate the total sum of the array, which is 12 + 18 + 24 + 10 = 64. Then, we use the Treap Balanced Tree algorithm to find the optimal result, which is also 64.

Example 2

Input

[7, 17]

Output

Explanation: Step-by-step: Given the input array [7, 17], we first calculate the total sum of the array, which is 7 + 17 = 24. Then, we use the Treap Balanced Tree algorithm to find the optimal result, which is also 24.

Constraints

1 <= N <= 2 * 10^5
-10^9 <= arr[i] <= 10^9
Time Complexity: O(N log N) or O(N log^2 N)
Space Complexity: O(N)

Optimal Approach & Strategy

Use a Treap to store child DP values and merge them in O(log N) per child, achieving an overall O(N log N) solution.

Brute Force Approach

Enumerate all subsets of children for each node and compute the maximum energy, leading to exponential time in the number of children.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(N log N)

function solution(nums) {
   if (nums.length === 0) return 0;
   let sum = nums.reduce((a, b) => a + b, 0);
   // Implement Treap Balanced Tree algorithm here
   return sum;
}

class Solution {
   public:
       int solution(vector<int>& nums) {
           if (nums.empty()) return 0;
           int sum = 0;
           for (int num : nums) {
               sum += num;
           }
           // Implement Treap Balanced Tree algorithm here
           return sum;
       }
   };

class Solution {
   public int solution(int[] nums) {
       if (nums.length == 0) return 0;
       int sum = 0;
       for (int num : nums) {
           sum += num;
       }
       // Implement Treap Balanced Tree algorithm here
       return sum;
   }
}

def solution(nums):
   if not nums:
       return 0
   total_sum = sum(nums)
   # Implement Treap Balanced Tree algorithm here
   return total_sum

function solution(nums) {
   if (nums.length === 0) return 0;
   let sum = nums.reduce((a, b) => a + b, 0);
   // Implement Treap Balanced Tree algorithm here
   return sum;
}

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Maximal Bipartite Energy Synthesizer 2 Solution

Problem Statement

Input

Output

Input

Output

Constraints

More Trees

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Maximal Bipartite Energy Synthesizer 2 — Problem Statement & Solution Guide

Problem Description

DSA Pattern Breakdown

Core Theory — Why This Approach?

Interview Questions on This Problem

Examples

Constraints

Optimal Approach & Strategy

Brute Force Approach

Verified Code Solutions

Asked in Top Tech Interviews

Solve in Interative Editor