LeetCode #1519 — MEDIUM

Number of Nodes in the Sub-Tree With the Same Label

Move from brute-force thinking to an efficient approach using hash map strategy.

The Problem

Problem Statement

You are given 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. The root of the tree is the node 0, and each node of the tree has a label which is a lower-case character given in the string labels (i.e. The node with the number i has the label labels[i]).

The edges array is given on the form edges[i] = [a_i, b_i], which means there is an edge between nodes a_i and b_i in the tree.

Return an array of size n where ans[i] is the number of nodes in the subtree of the i^th node which have the same label as node i.

A subtree of a tree T is the tree consisting of a node in T and all of its descendant nodes.

Example 1:

Input: n = 7, edges = [[0,1],[0,2],[1,4],[1,5],[2,3],[2,6]], labels = "abaedcd"
Output: [2,1,1,1,1,1,1]
Explanation: Node 0 has label 'a' and its sub-tree has node 2 with label 'a' as well, thus the answer is 2. Notice that any node is part of its sub-tree.
Node 1 has a label 'b'. The sub-tree of node 1 contains nodes 1,4 and 5, as nodes 4 and 5 have different labels than node 1, the answer is just 1 (the node itself).

Example 2:

Input: n = 4, edges = [[0,1],[1,2],[0,3]], labels = "bbbb"
Output: [4,2,1,1]
Explanation: The sub-tree of node 2 contains only node 2, so the answer is 1.
The sub-tree of node 3 contains only node 3, so the answer is 1.
The sub-tree of node 1 contains nodes 1 and 2, both have label 'b', thus the answer is 2.
The sub-tree of node 0 contains nodes 0, 1, 2 and 3, all with label 'b', thus the answer is 4.

Example 3:

Input: n = 5, edges = [[0,1],[0,2],[1,3],[0,4]], labels = "aabab"
Output: [3,2,1,1,1]

Constraints:

1 <= n <= 10⁵
edges.length == n - 1
edges[i].length == 2
0 <= a_i, b_i < n
a_i != b_i
labels.length == n
labels is consisting of only of lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: You are given 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. The root of the tree is the node 0, and each node of the tree has a label which is a lower-case character given in the string labels (i.e. The node with the number i has the label labels[i]). The edges array is given on the form edges[i] = [ai, bi], which means there is an edge between nodes ai and bi in the tree. Return an array of size n where ans[i] is the number of nodes in the subtree of the ith node which have the same label as node i. A subtree of a tree T is the tree consisting of a node in T and all of its descendant nodes.

Baseline thinking

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

Pattern signal: Hash Map · Tree

Example 1

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

Example 2

4
[[0,1],[1,2],[0,3]]
"bbbb"

Example 3

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

Step 02

Core Insight

What unlocks the optimal approach

Start traversing the tree and each node should return a vector to its parent node.
The vector should be of length 26 and have the count of all the labels in the sub-tree of this node.

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

Upper-end input sizes

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 #1519: Number of Nodes in the Sub-Tree With the Same Label
class Solution {
    private List<Integer>[] g;
    private String labels;
    private int[] ans;
    private int[] cnt;

    public int[] countSubTrees(int n, int[][] edges, String labels) {
        g = new List[n];
        Arrays.setAll(g, k -> new ArrayList<>());
        for (int[] e : edges) {
            int a = e[0], b = e[1];
            g[a].add(b);
            g[b].add(a);
        }
        this.labels = labels;
        ans = new int[n];
        cnt = new int[26];
        dfs(0, -1);
        return ans;
    }

    private void dfs(int i, int fa) {
        int k = labels.charAt(i) - 'a';
        ans[i] -= cnt[k];
        cnt[k]++;
        for (int j : g[i]) {
            if (j != fa) {
                dfs(j, i);
            }
        }
        ans[i] += cnt[k];
    }
}

// Accepted solution for LeetCode #1519: Number of Nodes in the Sub-Tree With the Same Label
func countSubTrees(n int, edges [][]int, labels string) []int {
	g := make([][]int, n)
	for _, e := range edges {
		a, b := e[0], e[1]
		g[a] = append(g[a], b)
		g[b] = append(g[b], a)
	}
	ans := make([]int, n)
	cnt := [26]int{}
	var dfs func(int, int)
	dfs = func(i, fa int) {
		k := labels[i] - 'a'
		ans[i] -= cnt[k]
		cnt[k]++
		for _, j := range g[i] {
			if j != fa {
				dfs(j, i)
			}
		}
		ans[i] += cnt[k]
	}
	dfs(0, -1)
	return ans
}

# Accepted solution for LeetCode #1519: Number of Nodes in the Sub-Tree With the Same Label
class Solution:
    def countSubTrees(self, n: int, edges: List[List[int]], labels: str) -> List[int]:
        def dfs(i, fa):
            ans[i] -= cnt[labels[i]]
            cnt[labels[i]] += 1
            for j in g[i]:
                if j != fa:
                    dfs(j, i)
            ans[i] += cnt[labels[i]]

        g = defaultdict(list)
        for a, b in edges:
            g[a].append(b)
            g[b].append(a)
        cnt = Counter()
        ans = [0] * n
        dfs(0, -1)
        return ans

// Accepted solution for LeetCode #1519: Number of Nodes in the Sub-Tree With the Same Label
struct Solution;

impl Solution {
    fn count_sub_trees(n: i32, edges: Vec<Vec<i32>>, labels: String) -> Vec<i32> {
        let n = n as usize;
        let mut adj: Vec<Vec<usize>> = vec![vec![]; n];
        for e in edges {
            let u = e[0] as usize;
            let v = e[1] as usize;
            adj[u].push(v);
            adj[v].push(u);
        }
        let mut visited = vec![false; n];
        let mut counts: Vec<usize> = vec![0; 26];
        let mut res = vec![0; n];
        let labels: Vec<u8> = labels.bytes().collect();
        Self::dfs(0, &mut visited, &mut counts, &mut res, &adj, &labels);
        res
    }

    fn dfs(
        u: usize,
        visited: &mut Vec<bool>,
        counts: &mut Vec<usize>,
        sizes: &mut Vec<i32>,
        adj: &[Vec<usize>],
        labels: &[u8],
    ) {
        visited[u] = true;
        let i = (labels[u] - b'a') as usize;
        let last_count = counts[i];
        counts[i] += 1;
        for &v in adj[u].iter() {
            if !visited[v] {
                Self::dfs(v, visited, counts, sizes, adj, labels);
            }
        }
        sizes[u] = (counts[i] - last_count) as i32;
    }
}

#[test]
fn test() {
    let n = 7;
    let edges = vec_vec_i32![[0, 1], [0, 2], [1, 4], [1, 5], [2, 3], [2, 6]];
    let labels = "abaedcd".to_string();
    let res = vec![2, 1, 1, 1, 1, 1, 1];
    assert_eq!(Solution::count_sub_trees(n, edges, labels), res);
    let n = 4;
    let edges = vec_vec_i32![[0, 1], [1, 2], [0, 3]];
    let labels = "bbbb".to_string();
    let res = vec![4, 2, 1, 1];
    assert_eq!(Solution::count_sub_trees(n, edges, labels), res);
    let n = 5;
    let edges = vec_vec_i32![[0, 1], [0, 2], [1, 3], [0, 4]];
    let labels = "aabab".to_string();
    let res = vec![3, 2, 1, 1, 1];
    assert_eq!(Solution::count_sub_trees(n, edges, labels), res);
    let n = 5;
    let edges = vec_vec_i32![[0, 1], [0, 2], [1, 3], [0, 4]];
    let labels = "aabab".to_string();
    let res = vec![3, 2, 1, 1, 1];
    assert_eq!(Solution::count_sub_trees(n, edges, labels), res);
    let n = 6;
    let edges = vec_vec_i32![[0, 1], [0, 2], [1, 3], [3, 4], [4, 5]];
    let labels = "cbabaa".to_string();
    let res = vec![1, 2, 1, 1, 2, 1];
    assert_eq!(Solution::count_sub_trees(n, edges, labels), res);
    let n = 7;
    let edges = vec_vec_i32![[0, 1], [1, 2], [2, 3], [3, 4], [4, 5], [5, 6]];
    let labels = "aaabaaa".to_string();
    let res = vec![6, 5, 4, 1, 3, 2, 1];
    assert_eq!(Solution::count_sub_trees(n, edges, labels), res);
}

// Accepted solution for LeetCode #1519: Number of Nodes in the Sub-Tree With the Same Label
function countSubTrees(n: number, edges: number[][], labels: string): number[] {
    const dfs = (i: number, fa: number) => {
        const k = labels.charCodeAt(i) - 97;
        ans[i] -= cnt[k];
        cnt[k]++;
        for (const j of g[i]) {
            if (j !== fa) {
                dfs(j, i);
            }
        }
        ans[i] += cnt[k];
    };
    const ans = new Array(n).fill(0),
        cnt = new Array(26).fill(0);
    const g: number[][] = Array.from({ length: n }, () => []);
    for (const [a, b] of edges) {
        g[a].push(b);
        g[b].push(a);
    }
    dfs(0, -1);
    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(h)

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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

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.