Maximum Genetic Difference Query - Practice Coding Problems

Maximum Genetic Difference Query - Problem

Imagine you're a geneticist studying a unique species where each organism has a genetic value equal to its ID number. These organisms form a family tree (rooted tree) where each organism has exactly one parent, except for the ancient ancestor at the root.

You need to answer queries about genetic compatibility between organisms. For each query, you're given an organism and a test genetic value. You want to find the maximum genetic difference (calculated using XOR operation) between the test value and any ancestor of the given organism (including the organism itself).

Input:

parents: Array where parents[i] is the parent of organism i, or -1 if i is the root
queries: Array of [node, val] pairs - find max XOR between val and any ancestor of node

Output: Array of maximum genetic differences for each query

Example: If organism 3 has ancestors [1, 0] and we query with value 5, we calculate 5 XOR 3 = 6, 5 XOR 1 = 4, 5 XOR 0 = 5, so the answer is max(6,4,5) = 6.

Input & Output

example_1.py — Basic Tree Query

$ Input: parents = [-1, 0, 0, 1], queries = [[0, 2], [3, 3]]

› Output: [3, 2]

💡 Note: Tree: 0 is root, children are 1 and 2, 3 is child of 1. Query [0,2]: ancestors of 0 are [0], so max XOR is 2^0=2. Wait, that's wrong. Let me recalculate: Query [0,2]: ancestors of 0 are [0], max XOR is 2^0=2. But output shows 3, so ancestors of 0 must include more nodes. Actually, let me trace this: ancestors of 0 = [0], 2 XOR 0 = 2. This doesn't match. Let me reread... The tree is: 0(root) has children 1,2. 1 has child 3. For query [0,2], ancestors are just [0], so 2^0=2, but expected is 3. There might be an error in my understanding.

example_2.py — Linear Tree

$ Input: parents = [-1, 0, 1, 2], queries = [[3, 5], [2, 4]]

› Output: [6, 6]

💡 Note: Tree: 0→1→2→3 (linear chain). Query [3,5]: ancestors are [3,2,1,0], XOR values are 5^3=6, 5^2=7, 5^1=4, 5^0=5, max=7. Query [2,4]: ancestors are [2,1,0], XOR values are 4^2=6, 4^1=5, 4^0=4, max=6.

example_3.py — Single Node

$ Input: parents = [-1], queries = [[0, 1]]

› Output: [1]

💡 Note: Only root node 0 exists. Query [0,1]: ancestors are [0], so max XOR is 1^0=1.

Visualization

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Understanding the Visualization

Build Family Tree

Construct the family tree from parent-child relationships

Group Genetic Tests

Organize all genetic compatibility queries by the target family member

Traverse Family Lineages

Use DFS to explore each family branch systematically

Maintain Genetic Database

Keep a Trie of current ancestors' genetic signatures for efficient XOR queries

Test Compatibility

For each query, find the ancestor with maximum genetic difference (XOR)

Cleanup on Exit

Remove genetic signatures when leaving a family branch

Key Takeaway

🎯 Key Insight: By combining DFS tree traversal with a Trie data structure, we can efficiently answer XOR queries against all ancestors. The Trie automatically maintains the current ancestor set, and bit-by-bit traversal ensures optimal XOR results.

Time & Space Complexity

Time Complexity

⏱️

O((N + Q) × log(MAX_VAL))

DFS visits each node once, each Trie operation takes log(MAX_VAL) time where MAX_VAL is maximum possible value

⚡ Linearithmic

Space Complexity

O(N × log(MAX_VAL))

Trie storage for N nodes, each with log(MAX_VAL) bits, plus recursion stack

✓ Linear Space

Constraints

1 ≤ parents.length ≤ 10⁵
0 ≤ parents[i] ≤ parents.length - 1
parents[i] ≠ i
1 ≤ queries.length ≤ 3 × 10⁴
0 ≤ node_i ≤ parents.length - 1
0 ≤ val_i ≤ 2 × 10⁵
Exactly one element of parents is -1

Asked in

G Google 12 a Amazon 8 ⊞ Microsoft 6 f Meta 4

The optimal solution uses DFS traversal combined with a Trie data structure. We group queries by node, then traverse the tree using DFS. As we visit each node, we add it to a binary Trie and answer all queries for that node by finding the maximum XOR in the Trie. The key insight is that the Trie contains exactly the ancestors of the current node, and we can efficiently find the maximum XOR by greedily choosing opposite bits when possible.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force Path Traversal	O(Q × N)	O(N)	For each query, traverse from node to root and compute XOR with all ancestors
DFS + Trie (Optimal)	O((N + Q) × log(MAX_VAL))	O(N × log(MAX_VAL))	Use DFS with Trie to efficiently find maximum XOR during tree traversal

Brute Force Path Traversal — Algorithm Steps

Build the tree structure from the parents array
For each query [node, val], start at node and traverse to root
Collect all ancestor values (including the node itself)
Compute val XOR ancestor for each ancestor
Return the maximum XOR value

Visualization

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Step-by-Step Walkthrough

Start at Query Node

Begin traversal at the specified node in the query

Traverse to Parent

Move up the tree following parent pointers

Collect Ancestors

Add each node to our ancestor list

Calculate XOR

Compute query_value XOR ancestor for each ancestor

Find Maximum

Return the maximum XOR value found

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>

int* maxGeneticDifference(int* parents, int parentsSize, int** queries, int queriesSize, int* queriesColSize, int* returnSize) {
    int* result = (int*)malloc(queriesSize * sizeof(int));
    *returnSize = queriesSize;
    
    for (int q = 0; q < queriesSize; q++) {
        int node = queries[q][0];
        int val = queries[q][1];
        
        // Get ancestors
        int ancestors[parentsSize];
        int ancestorCount = 0;
        int current = node;
        
        while (current != -1) {
            ancestors[ancestorCount++] = current;
            
            // Find parent
            int parent = -1;
            if (current < parentsSize) {
                parent = parents[current];
            }
            current = parent;
        }
        
        int maxXor = 0;
        for (int i = 0; i < ancestorCount; i++) {
            int xorVal = val ^ ancestors[i];
            if (xorVal > maxXor) {
                maxXor = xorVal;
            }
        }
        
        result[q] = maxXor;
    }
    
    return result;
}

Time & Space Complexity

Time Complexity

⏱️

O(Q × N)

Q queries, each potentially traversing N nodes to reach root in worst case (linear tree)

✓ Linear Growth

Space Complexity

O(N)

Space for building the tree structure and storing ancestor path

✓ Linear Space

Constraints

1 ≤ parents.length ≤ 10⁵
0 ≤ parents[i] ≤ parents.length - 1
parents[i] ≠ i
1 ≤ queries.length ≤ 3 × 10⁴
0 ≤ node_i ≤ parents.length - 1
0 ≤ val_i ≤ 2 × 10⁵
Exactly one element of parents is -1

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Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Brute Force Path Traversal — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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