LeetCode #2603 — HARD

Collect Coins in a Tree

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

The Problem

Problem Statement

There exists an undirected and unrooted tree with n nodes indexed from 0 to n - 1. You are given an integer n and a 2D integer array edges of length n - 1, where edges[i] = [a_i, b_i] indicates that there is an edge between nodes a_i and b_i in the tree. You are also given an array coins of size n where coins[i] can be either 0 or 1, where 1 indicates the presence of a coin in the vertex i.

Initially, you choose to start at any vertex in the tree. Then, you can perform the following operations any number of times:

Collect all the coins that are at a distance of at most 2 from the current vertex, or
Move to any adjacent vertex in the tree.

Find the minimum number of edges you need to go through to collect all the coins and go back to the initial vertex.

Note that if you pass an edge several times, you need to count it into the answer several times.

Example 1:

Input: coins = [1,0,0,0,0,1], edges = [[0,1],[1,2],[2,3],[3,4],[4,5]]
Output: 2
Explanation: Start at vertex 2, collect the coin at vertex 0, move to vertex 3, collect the coin at vertex 5 then move back to vertex 2.

Example 2:

Input: coins = [0,0,0,1,1,0,0,1], edges = [[0,1],[0,2],[1,3],[1,4],[2,5],[5,6],[5,7]]
Output: 2
Explanation: Start at vertex 0, collect the coins at vertices 4 and 3, move to vertex 2,  collect the coin at vertex 7, then move back to vertex 0.

Constraints:

n == coins.length
1 <= n <= 3 * 10⁴
0 <= coins[i] <= 1
edges.length == n - 1
edges[i].length == 2
0 <= a_i, b_i < n
a_i != b_i
edges represents a valid tree.

Step 01

Brute Force Baseline

Problem summary: There exists an undirected and unrooted tree with n nodes indexed from 0 to n - 1. You are given an integer n and a 2D integer array edges of length n - 1, where edges[i] = [ai, bi] indicates that there is an edge between nodes ai and bi in the tree. You are also given an array coins of size n where coins[i] can be either 0 or 1, where 1 indicates the presence of a coin in the vertex i. Initially, you choose to start at any vertex in the tree. Then, you can perform the following operations any number of times: Collect all the coins that are at a distance of at most 2 from the current vertex, or Move to any adjacent vertex in the tree. Find the minimum number of edges you need to go through to collect all the coins and go back to the initial vertex. Note that if you pass an edge several times, you need to count it into the answer several times.

Baseline thinking

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

Pattern signal: Array · Tree · Topological Sort

Example 1

[1,0,0,0,0,1]
[[0,1],[1,2],[2,3],[3,4],[4,5]]

Example 2

[0,0,0,1,1,0,0,1]
[[0,1],[0,2],[1,3],[1,4],[2,5],[5,6],[5,7]]

Core Insight

What unlocks the optimal approach

All leaves that do not have a coin are redundant and can be deleted from the tree.
Remove the leaves that do not have coins on them, so that the resulting tree will have a coin on every leaf.
In the remaining tree, remove each leaf node and its parent from the tree. The remaining nodes in the tree are the ones that must be visited. Hence, the answer is equal to (# remaining nodes -1) * 2

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
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 #2603: Collect Coins in a Tree
class Solution {
    public int collectTheCoins(int[] coins, int[][] edges) {
        int n = coins.length;
        Set<Integer>[] g = new Set[n];
        Arrays.setAll(g, k -> new HashSet<>());
        for (var e : edges) {
            int a = e[0], b = e[1];
            g[a].add(b);
            g[b].add(a);
        }
        Deque<Integer> q = new ArrayDeque<>();
        for (int i = 0; i < n; ++i) {
            if (coins[i] == 0 && g[i].size() == 1) {
                q.offer(i);
            }
        }
        while (!q.isEmpty()) {
            int i = q.poll();
            for (int j : g[i]) {
                g[j].remove(i);
                if (coins[j] == 0 && g[j].size() == 1) {
                    q.offer(j);
                }
            }
            g[i].clear();
        }
        q.clear();
        for (int k = 0; k < 2; ++k) {
            for (int i = 0; i < n; ++i) {
                if (g[i].size() == 1) {
                    q.offer(i);
                }
            }
            for (int i : q) {
                for (int j : g[i]) {
                    g[j].remove(i);
                }
                g[i].clear();
            }
        }
        int ans = 0;
        for (var e : edges) {
            int a = e[0], b = e[1];
            if (g[a].size() > 0 && g[b].size() > 0) {
                ans += 2;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2603: Collect Coins in a Tree
func collectTheCoins(coins []int, edges [][]int) int {
	n := len(coins)
	g := make([]map[int]bool, n)
	for i := range g {
		g[i] = map[int]bool{}
	}
	for _, e := range edges {
		a, b := e[0], e[1]
		g[a][b] = true
		g[b][a] = true
	}
	q := []int{}
	for i, c := range coins {
		if c == 0 && len(g[i]) == 1 {
			q = append(q, i)
		}
	}
	for len(q) > 0 {
		i := q[0]
		q = q[1:]
		for j := range g[i] {
			delete(g[j], i)
			if coins[j] == 0 && len(g[j]) == 1 {
				q = append(q, j)
			}
		}
		g[i] = map[int]bool{}
	}
	for k := 0; k < 2; k++ {
		q := []int{}
		for i := range coins {
			if len(g[i]) == 1 {
				q = append(q, i)
			}
		}
		for _, i := range q {
			for j := range g[i] {
				delete(g[j], i)
			}
			g[i] = map[int]bool{}
		}
	}
	ans := 0
	for _, e := range edges {
		a, b := e[0], e[1]
		if len(g[a]) > 0 && len(g[b]) > 0 {
			ans += 2
		}
	}
	return ans
}

# Accepted solution for LeetCode #2603: Collect Coins in a Tree
class Solution:
    def collectTheCoins(self, coins: List[int], edges: List[List[int]]) -> int:
        g = defaultdict(set)
        for a, b in edges:
            g[a].add(b)
            g[b].add(a)
        n = len(coins)
        q = deque(i for i in range(n) if len(g[i]) == 1 and coins[i] == 0)
        while q:
            i = q.popleft()
            for j in g[i]:
                g[j].remove(i)
                if coins[j] == 0 and len(g[j]) == 1:
                    q.append(j)
            g[i].clear()
        for k in range(2):
            q = [i for i in range(n) if len(g[i]) == 1]
            for i in q:
                for j in g[i]:
                    g[j].remove(i)
                g[i].clear()
        return sum(len(g[a]) > 0 and len(g[b]) > 0 for a, b in edges) * 2

// Accepted solution for LeetCode #2603: Collect Coins in a Tree
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2603: Collect Coins in a Tree
// class Solution {
//     public int collectTheCoins(int[] coins, int[][] edges) {
//         int n = coins.length;
//         Set<Integer>[] g = new Set[n];
//         Arrays.setAll(g, k -> new HashSet<>());
//         for (var e : edges) {
//             int a = e[0], b = e[1];
//             g[a].add(b);
//             g[b].add(a);
//         }
//         Deque<Integer> q = new ArrayDeque<>();
//         for (int i = 0; i < n; ++i) {
//             if (coins[i] == 0 && g[i].size() == 1) {
//                 q.offer(i);
//             }
//         }
//         while (!q.isEmpty()) {
//             int i = q.poll();
//             for (int j : g[i]) {
//                 g[j].remove(i);
//                 if (coins[j] == 0 && g[j].size() == 1) {
//                     q.offer(j);
//                 }
//             }
//             g[i].clear();
//         }
//         q.clear();
//         for (int k = 0; k < 2; ++k) {
//             for (int i = 0; i < n; ++i) {
//                 if (g[i].size() == 1) {
//                     q.offer(i);
//                 }
//             }
//             for (int i : q) {
//                 for (int j : g[i]) {
//                     g[j].remove(i);
//                 }
//                 g[i].clear();
//             }
//         }
//         int ans = 0;
//         for (var e : edges) {
//             int a = e[0], b = e[1];
//             if (g[a].size() > 0 && g[b].size() > 0) {
//                 ans += 2;
//             }
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #2603: Collect Coins in a Tree
function collectTheCoins(coins: number[], edges: number[][]): number {
    const n = coins.length;
    const g: Set<number>[] = new Array(n).fill(0).map(() => new Set<number>());
    for (const [a, b] of edges) {
        g[a].add(b);
        g[b].add(a);
    }
    let q: number[] = [];
    for (let i = 0; i < n; ++i) {
        if (coins[i] === 0 && g[i].size === 1) {
            q.push(i);
        }
    }
    while (q.length) {
        const i = q.pop()!;
        for (const j of g[i]) {
            g[j].delete(i);
            if (coins[j] === 0 && g[j].size === 1) {
                q.push(j);
            }
        }
        g[i].clear();
    }
    q = [];
    for (let k = 0; k < 2; ++k) {
        for (let i = 0; i < n; ++i) {
            if (g[i].size === 1) {
                q.push(i);
            }
        }
        for (const i of q) {
            for (const j of g[i]) {
                g[j].delete(i);
            }
            g[i].clear();
        }
    }
    let ans = 0;
    for (const [a, b] of edges) {
        if (g[a].size > 0 && g[b].size > 0) {
            ans += 2;
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

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

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(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.

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.