LeetCode #1766 — HARD

Tree of Coprimes

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode
The Problem

Problem Statement

There is a tree (i.e., a connected, undirected graph that has no cycles) consisting of n nodes numbered from 0 to n - 1 and exactly n - 1 edges. Each node has a value associated with it, and the root of the tree is node 0.

To represent this tree, you are given an integer array nums and a 2D array edges. Each nums[i] represents the ith node's value, and each edges[j] = [uj, vj] represents an edge between nodes uj and vj in the tree.

Two values x and y are coprime if gcd(x, y) == 1 where gcd(x, y) is the greatest common divisor of x and y.

An ancestor of a node i is any other node on the shortest path from node i to the root. A node is not considered an ancestor of itself.

Return an array ans of size n, where ans[i] is the closest ancestor to node i such that nums[i] and nums[ans[i]] are coprime, or -1 if there is no such ancestor.

Example 1:

Input: nums = [2,3,3,2], edges = [[0,1],[1,2],[1,3]]
Output: [-1,0,0,1]
Explanation: In the above figure, each node's value is in parentheses.
- Node 0 has no coprime ancestors.
- Node 1 has only one ancestor, node 0. Their values are coprime (gcd(2,3) == 1).
- Node 2 has two ancestors, nodes 1 and 0. Node 1's value is not coprime (gcd(3,3) == 3), but node 0's
  value is (gcd(2,3) == 1), so node 0 is the closest valid ancestor.
- Node 3 has two ancestors, nodes 1 and 0. It is coprime with node 1 (gcd(3,2) == 1), so node 1 is its
  closest valid ancestor.

Example 2:

Input: nums = [5,6,10,2,3,6,15], edges = [[0,1],[0,2],[1,3],[1,4],[2,5],[2,6]]
Output: [-1,0,-1,0,0,0,-1]

Constraints:

  • nums.length == n
  • 1 <= nums[i] <= 50
  • 1 <= n <= 105
  • edges.length == n - 1
  • edges[j].length == 2
  • 0 <= uj, vj < n
  • uj != vj
Patterns Used
🌳Tree Traversal→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: There is a tree (i.e., a connected, undirected graph that has no cycles) consisting of n nodes numbered from 0 to n - 1 and exactly n - 1 edges. Each node has a value associated with it, and the root of the tree is node 0. To represent this tree, you are given an integer array nums and a 2D array edges. Each nums[i] represents the ith node's value, and each edges[j] = [uj, vj] represents an edge between nodes uj and vj in the tree. Two values x and y are coprime if gcd(x, y) == 1 where gcd(x, y) is the greatest common divisor of x and y. An ancestor of a node i is any other node on the shortest path from node i to the root. A node is not considered an ancestor of itself. Return an array ans of size n, where ans[i] is the closest ancestor to node i such that nums[i] and nums[ans[i]] are coprime, or -1 if there is no such ancestor.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Math · Tree

Example 1

[2,3,3,2]
[[0,1],[1,2],[1,3]]

Example 2

[5,6,10,2,3,6,15]
[[0,1],[0,2],[1,3],[1,4],[2,5],[2,6]]
Step 02

Core Insight

What unlocks the optimal approach

  • Note that for a node, it's not optimal to consider two nodes with the same value.
  • Note that the values are small enough for you to iterate over them instead of iterating over the parent nodes.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Largest constraint values
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1766: Tree of Coprimes
class Solution {
    private List<Integer>[] g;
    private List<Integer>[] f;
    private Deque<int[]>[] stks;
    private int[] nums;
    private int[] ans;

    public int[] getCoprimes(int[] nums, int[][] edges) {
        int n = nums.length;
        g = new List[n];
        Arrays.setAll(g, k -> new ArrayList<>());
        for (var e : edges) {
            int u = e[0], v = e[1];
            g[u].add(v);
            g[v].add(u);
        }
        f = new List[51];
        stks = new Deque[51];
        Arrays.setAll(f, k -> new ArrayList<>());
        Arrays.setAll(stks, k -> new ArrayDeque<>());
        for (int i = 1; i < 51; ++i) {
            for (int j = 1; j < 51; ++j) {
                if (gcd(i, j) == 1) {
                    f[i].add(j);
                }
            }
        }
        this.nums = nums;
        ans = new int[n];
        dfs(0, -1, 0);
        return ans;
    }

    private void dfs(int i, int fa, int depth) {
        int t = -1, k = -1;
        for (int v : f[nums[i]]) {
            var stk = stks[v];
            if (!stk.isEmpty() && stk.peek()[1] > k) {
                t = stk.peek()[0];
                k = stk.peek()[1];
            }
        }
        ans[i] = t;
        for (int j : g[i]) {
            if (j != fa) {
                stks[nums[i]].push(new int[] {i, depth});
                dfs(j, i, depth + 1);
                stks[nums[i]].pop();
            }
        }
    }

    private int gcd(int a, int b) {
        return b == 0 ? a : gcd(b, a % b);
    }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n × M)
Space
O(M^2 + n)

Approach Breakdown

LEVEL ORDER
O(n) time
O(n) space

BFS with a queue visits every node exactly once — O(n) time. The queue may hold an entire level of the tree, which for a complete binary tree is up to n/2 nodes = O(n) space. This is optimal in time but costly in space for wide trees.

DFS TRAVERSAL
O(n) time
O(h) space

Every node is visited exactly once, giving O(n) time. Space depends on tree shape: O(h) for recursive DFS (stack depth = height h), or O(w) for BFS (queue width = widest level). For balanced trees h = log n; for skewed trees h = n.

Shortcut: Visit every node once → O(n) time. Recursion depth = tree height → O(h) space.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

Forgetting null/base-case handling

Wrong move: Recursive traversal assumes children always exist.

Usually fails on: Leaf nodes throw errors or create wrong depth/path values.

Fix: Handle null/base cases before recursive transitions.

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