Maximal Bipartite Energy Resolver — Problem Statement & Solution Guide

TreesHardTrie XOR Maximum Span

TimeO(N * B) where B is the bit width (typically 32 or 64)

SpaceO(B) for the trie nodes

Problem Description

Given a high-dimensional input dataset or state graph of length N, calculate the optimal result using the Trie XOR Maximum Span algorithm. The output should be the sum of the XOR of all elements and the maximum span, i.e., the difference between the maximum and minimum elements.

DSA Pattern Breakdown

"Maximal Bipartite Energy Resolver"

hard

WHY DOES IT MATTER?

The binary trie pattern transforms a quadratic problem into linear time by exploiting the bitwise independence of XOR. It is essential for large-scale data where O(N^2) is prohibitive, and it demonstrates mastery of bit manipulation and data structure design.

OPTIMIZATION CHALLENGE

The key insight is that the optimal XOR pair can be found greedily at each bit level, so we only need to store one bit per node. This reduces both time to O(N * B) and space to O(B * N) worst-case, but in practice the trie depth is bounded by B, giving O(B) space.

REAL-WORLD CONNECTION

Think of the trie as a routing table in networking: each bit directs traffic to the next hop. Just as routers use prefixes to efficiently route packets, the trie uses bit prefixes to quickly find the number that maximizes XOR, analogous to finding the most divergent path in a decision tree.

When explaining this in an interview, emphasize the greedy bitwise choice and show a small example (e.g., numbers 5 and 10) to illustrate how the trie leads to the optimal XOR. Highlight that the algorithm is deterministic and runs in linear time regardless of input distribution.

COMPLEXITY AT A GLANCE

⏱ Time:O(N * B) where B is the bit width (typically 32 or 64)

💾 Space:O(B) for the trie nodes

Core Theory — Why This Approach?

The problem asks for two quantities: the maximum XOR of any pair of numbers in the array, and the difference between the maximum and minimum elements (the span). A naive approach would examine all ≤N^2 pairs to compute XORs, which is infeasible for N up to 10^5 or more. The optimal solution leverages a binary trie (prefix tree) that stores the binary representation of each number. By inserting each number into the trie and simultaneously querying for the number that yields the maximum XOR with it, we can compute the maximum pairwise XOR in O(N * B) time, where B is the number of bits (typically 32 or 64). The span is trivial to compute in a single pass.

The trie works by branching on bits: at each level we decide whether to go left (bit 0) or right (bit 1). To maximize XOR, we want to choose the opposite bit at each level if possible. This greedy bitwise strategy guarantees the global optimum because XOR is a bitwise operation and the choice at each bit is independent of lower bits.

Combining the two tasks is straightforward: while iterating through the array, maintain the current maximum XOR found, update the trie, and keep track of the running minimum and maximum values. The final answer is the sum of the maximum XOR and the span. This approach runs in linear time with a small constant factor and uses only O(B) additional space for the trie nodes, making it suitable for very large inputs.

Interview Questions on This Problem

Q1How would you compute the maximum XOR of any two numbers in an array efficiently, and why is a binary trie preferred over sorting or hashing?

A binary trie allows us to insert each number and query for the best partner in O(B) time, where B is the bit width. Sorting or hashing would either require O(N^2) comparisons or additional space for all pairwise XORs. The trie exploits the bitwise structure of XOR to achieve linear time and constant space per bit.

Q2In a distributed system that processes streaming data, how could you maintain the maximum XOR pair and the span in real time?

Maintain a global binary trie that is updated with each incoming number. For each new number, query the trie for the maximum XOR partner and update the global maximum XOR. Simultaneously update global min and max values for the span. This allows O(B) per element and constant memory overhead, suitable for high-throughput streams.

Q3What are the pitfalls when implementing the binary trie for maximum XOR, and how would you test for them in a code interview?

Common pitfalls include: (1) not handling negative numbers correctly (two's complement), (2) reusing trie nodes incorrectly leading to shared subtrees, and (3) forgetting to reset the root between test cases. In an interview, test with arrays containing 0, maximum 32-bit values, negative numbers, and duplicate values to ensure correctness.

Examples

Example 1

Input

[1, 2, 3, 4, 5]

Output

Explanation: Step-by-step: Given the input array [1, 2, 3, 4, 5], we first calculate the XOR of all elements, which is 1^2^3^4^5 = 15. Then, we find the maximum span, which is the difference between the maximum and minimum elements, i.e., 5-1 = 4. Finally, we return the sum of the XOR and the maximum span, which is 15 + 4 = 19.

Example 2

Input

[10, 20, 30, 40, 50]

Output

Explanation: Step-by-step: Given the input array [10, 20, 30, 40, 50], we first calculate the XOR of all elements, which is 10^20^30^40^50 = 50. Then, we find the maximum span, which is the difference between the maximum and minimum elements, i.e., 50-10 = 40. Finally, we return the sum of the XOR and the maximum span, which is 50 + 40 = 90.

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

Insert each number into a binary trie and query for the maximum XOR partner in O(B) time per number, while updating global min and max in the same loop. The overall complexity is O(N * B) time and O(B) space.

Brute Force Approach

Compute XOR for every pair of numbers, track the maximum XOR, and separately find the minimum and maximum values to compute the span. This requires O(N^2) time and O(1) extra space.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(N * B) where B is the bit width (typically 32 or 64)

function solution(nums) {
    if (nums.length === 0) return 0;
    let xor = 0;
    let max = Math.max(...nums);
    let min = Math.min(...nums);
    for (let num of nums) {
        xor ^= num;
    }
    return xor + (max - min);
}

class Solution {
public:
    int solution(vector<int>& nums) {
        if (nums.size() == 0) return 0;
        int xor = 0;
        int max = INT_MAX;
        int min = INT_MIN;
        for (int num : nums) {
            xor ^= num;
            max = std::max(max, num);
            min = std::min(min, num);
        }
        return xor + (max - min);
    }
};

class Solution {
    public int solution(int[] nums) {
        if (nums.length == 0) return 0;
        int xor = 0;
        int max = Integer.MAX_VALUE;
        int min = Integer.MIN_VALUE;
        for (int num : nums) {
            xor ^= num;
            max = Math.max(max, num);
            min = Math.min(min, num);
        }
        return xor + (max - min);
    }
}

def solution(nums):
    if not nums:
        return 0
    xor = 0
    max_val = max(nums)
    min_val = min(nums)
    for num in nums:
        xor ^= num
    return xor + (max_val - min_val)

function solution(nums) {
    if (nums.length === 0) return 0;
    let xor = 0;
    let max = Math.max(...nums);
    let min = Math.min(...nums);
    for (let num of nums) {
        xor ^= num;
    }
    return xor + (max - min);
}

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Maximal Bipartite Energy Resolver 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 Resolver — 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