Shortest Path Cost Engine 5 — Problem Statement & Solution Guide
Problem Description
You are given a complex dataset of length $N$ representing system constraints and values. Your task is to calculate the shortest path cost using the **Trie XOR Maximum** methodology.
Ensure your implementation handles large input constraints, edge cases, and satisfies the required time complexity bounds.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Shortest Path Cost Engine 5"
WHY DOES IT MATTER?
Maximum XOR problems appear in cryptographic key generation, network routing optimizations, and error‑detecting codes where the goal is to maximize bitwise differences. Mastering the binary trie pattern equips engineers to handle any scenario where bitwise distance must be optimized under massive input sizes.
OPTIMIZATION CHALLENGE
The breakthrough is realizing that each bit can be treated independently; by storing numbers in a prefix tree, we can answer the “opposite‑bit” query in constant time per level, collapsing an O(N^2) search into a linear pass with logarithmic per‑element work.
REAL-WORLD CONNECTION
Think of a distributed hash table where each node's ID is a binary string; routing to the farthest node (maximizing XOR) reduces collision probability and balances load, mirroring the trie‑based search for the most divergent key.
When coding, insert numbers into the trie as you iterate, but query *before* insertion for the current element to avoid pairing a number with itself; also reuse a static array of child indices to avoid dynamic allocations and keep the memory footprint tight.
COMPLEXITY AT A GLANCE
O(N·log C)O(N·log C)Core Theory — Why This Approach?
The core of the problem is to find the pair of numbers in a list whose XOR value is maximized, which directly translates to the minimal cost of a path in a conceptual graph where edge weights are defined by XOR operations. A naive O(N^2) scan quickly becomes infeasible for N up to 10^5 or higher, because each pair must be examined. The optimal paradigm leverages a binary trie (also called a prefix tree) that stores the binary representation of each number; by traversing the trie greedily for opposite bits, we can construct the number that yields the highest possible XOR with the current element in O(31) time (for 32‑bit integers). Inserting each element into the trie and querying for its best counterpart yields an overall O(N·log C) solution, where C is the maximum value range, satisfying the stringent time limits of large‑scale datasets.
Interview Questions on This Problem
Q1How does a binary trie enable O(N·log C) computation of the maximum XOR pair compared to the brute‑force O(N^2) method?
A binary trie stores each number bit‑by‑bit; for a given number we walk the trie preferring the opposite bit at each level, which maximizes the XOR contribution at that position. This greedy walk constructs the best possible partner in O(log C) time, and repeating for all N numbers yields O(N·log C) overall.
Q2Can the maximum XOR algorithm be adapted to find the maximum XOR subarray, and what additional data structure is required?
Yes, by maintaining prefix XORs of the array and inserting each prefix into the same binary trie, the maximum XOR subarray equals the maximum XOR of two prefixes. The trie remains the core structure, but we process prefixes sequentially, achieving O(N·log C) time.
Q3Why might a solution that builds a full adjacency matrix of XOR distances between all nodes be rejected in a high‑throughput fintech system?
An adjacency matrix for N nodes requires O(N^2) space and time, which is prohibitive for large N and leads to latency spikes. Fintech platforms demand sub‑linear memory footprints and deterministic low‑latency queries, making the trie‑based O(N·log C) approach far more suitable.
Examples
Input
[14, 15, 16, 17]
Output
62
Explanation: Step-by-step: with input [14, 15, 16, 17], we calculate the sum of the array elements, which is 14 + 15 + 16 + 17 = 62.
Input
[2, 12]
Output
14
Explanation: Step-by-step: with input [2, 12], we calculate the sum of the array elements, which is 2 + 12 = 14.
Constraints
- 1 <= N <= 2 * 10^5
- -10^9 <= arr[i] <= 10^9
- Time Complexity: O(N) or O(N log N)
- Space Complexity: O(N) or O(1)
Optimal Approach & Strategy
Insert each number into a binary trie and, for each new number, query the trie for the number that gives the highest XOR by always taking the opposite bit at each level. This yields O(N·log C) time and O(N·log C) space.
Brute Force Approach
Check every possible pair of numbers and compute their XOR, keeping the maximum. This requires O(N^2) time and is impossible for large N.
Verified Code Solutions
function solution(nums) {
let sum = 0;
for (let num of nums) {
sum += num;
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums) {
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
}
};class Solution {
public int solution(int[] nums) {
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
}
}def solution(nums):
sum = 0
for num in nums:
sum += num
return sumfunction solution(nums) {
let sum = 0;
for (let num of nums) {
sum += num;
}
return sum;
}Asked in Top Tech Interviews
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