Shortest Path Cost Engine 8 — Problem Statement & Solution Guide
Problem Description
Given a complex dataset of length N representing system constraints and values, calculate the shortest path cost using the Trie XOR Maximum methodology.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Shortest Path Cost Engine 8"
WHY DOES IT MATTER?
The binary‑trie‑based XOR optimization pattern is essential because many problems reduce to finding extremal XOR pairs among large sets of numbers, and naïve enumeration is prohibitive. Mastery of this pattern unlocks efficient solutions for networking, cryptography, and data compression tasks where bitwise differences matter.
OPTIMIZATION CHALLENGE
The key insight is that the XOR metric can be optimized greedily per bit. By storing numbers in a bitwise trie, we can, at each level, choose the branch that yields the desired bit (0 for minimizing, 1 for maximizing) without scanning the entire set, collapsing the quadratic search space to a logarithmic walk.
REAL-WORLD CONNECTION
Consider a distributed hash table where each node’s identifier is a 160‑bit key. Routing decisions often rely on minimizing XOR distance (as in Kademlia). A binary trie enables rapid lookup of the closest node in key space, mirroring the shortest‑path‑XOR problem in a real peer‑to‑peer network.
When coding, build the trie incrementally during a single DFS pass: insert the current prefix XOR, query for the best partner, then recurse. This avoids a second pass and keeps memory locality high, which often makes the difference between passing and timing out in tight interview environments.
COMPLEXITY AT A GLANCE
O(N·log C)O(N·log C)Core Theory — Why This Approach?
The core of this problem lies in the interplay between tree traversal and bitwise optimization using a binary trie (also known as a prefix tree). When we represent each node’s cumulative XOR value from the root to that node, the shortest path cost between any two nodes reduces to finding the minimum XOR of two such cumulative values. A naive pairwise comparison would be O(N^2), which quickly becomes infeasible for large N (up to 10^5 or more). By inserting each cumulative XOR into a binary trie, we can query the best counterpart for any value in O(log C) time, where C is the maximum possible value (typically 2^31‑1 for 32‑bit integers). This transforms the problem into a linear‑time solution with a logarithmic factor.
The optimal paradigm leverages the property that the XOR of two numbers is maximized (or minimized) by greedily choosing opposite bits at the highest possible positions. The binary trie stores bits from the most significant to the least significant, allowing us to walk down the tree and, at each level, pick the child that leads to the desired bit pattern. While building the trie during a DFS/BFS of the original tree, we simultaneously query for the best XOR partner, updating the global minimum path cost. This approach avoids recomputation and ensures each node is processed only once, achieving O(N·log C) time and O(N·log C) space.
Interview Questions on This Problem
Q1How would you compute the minimum XOR distance between any two nodes in a weighted tree where edge weights are integers?
Perform a DFS to compute the prefix XOR from the root to every node. Insert each prefix into a binary trie while querying the trie for the value that yields the smallest XOR with the current prefix. Track the global minimum across all queries. This runs in O(N·log C) time.
Q2Why does a brute‑force O(N^2) pairwise XOR check fail for N = 10^5, and how does a binary trie overcome this limitation?
O(N^2) requires ~10^10 operations, far exceeding typical time limits. A binary trie reduces each query to O(log C) by exploiting bitwise independence, turning the overall complexity into O(N·log C), which is easily manageable for N = 10^5.
Q3Explain how you would adapt the solution if the tree were dynamic, supporting edge weight updates and queries for the minimum XOR path.
Use a heavy‑light decomposition (HLD) to break the tree into chains, maintain a binary trie for each chain’s prefix XORs, and update affected chains on edge weight changes. Queries combine results from O(log N) chains, preserving near‑logarithmic query time.
Examples
Input
[10, 5, 3, 7, 2]
Output
10
Explanation: Step-by-step: Given the input array [10, 5, 3, 7, 2], we first calculate the XOR of all node values, which is 10. Then, we find the maximum value (10) and the minimum value (2) in the array. The shortest path cost is the XOR value, which is 10.
Input
[1, 2, 3, 4, 5]
Output
1
Explanation: Step-by-step: Given the input array [1, 2, 3, 4, 5], we first calculate the XOR of all node values, which is 1. Then, we find the maximum value (5) and the minimum value (1) in the array. The shortest path cost is the XOR value, which is 1.
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 prefix XOR into a binary trie while simultaneously querying for the best partner, achieving O(N·log C) time and linear‑ish space.
Brute Force Approach
Compute all prefix XORs, then compare every pair to find the minimum XOR, which is O(N^2).
Verified Code Solutions
function solution(nums) {
if (nums.length === 0) return 0;
let xor = 0;
for (let num of nums) {
xor ^= num;
}
return xor;
}class Solution {
public:
int solution(vector<int>& nums) {
if (nums.size() == 0) return 0;
int xor = 0;
for (int num : nums) {
xor ^= num;
}
return xor;
}
}class Solution {
public int solution(int[] nums) {
if (nums.length == 0) return 0;
int xor = 0;
for (int num : nums) {
xor ^= num;
}
return xor;
}
}def solution(nums):
if not nums:
return 0
xor = 0
for num in nums:
xor ^= num
return xorfunction solution(nums) {
if (nums.length === 0) return 0;
let xor = 0;
for (let num of nums) {
xor ^= num;
}
return xor;
}Asked in Top Tech Interviews
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