LeetCode #3331 — MEDIUM

Find Subtree Sizes After Changes

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

The Problem

Problem Statement

You are given a tree rooted at node 0 that consists of n nodes numbered from 0 to n - 1. The tree is represented by an array parent of size n, where parent[i] is the parent of node i. Since node 0 is the root, parent[0] == -1.

You are also given a string s of length n, where s[i] is the character assigned to node i.

We make the following changes on the tree one time simultaneously for all nodes x from 1 to n - 1:

Find the closest node y to node x such that y is an ancestor of x, and s[x] == s[y].
If node y does not exist, do nothing.
Otherwise, remove the edge between x and its current parent and make node y the new parent of x by adding an edge between them.

Return an array answer of size n where answer[i] is the size of the subtree rooted at node i in the final tree.

Example 1:

Input: parent = [-1,0,0,1,1,1], s = "abaabc"

Output: [6,3,1,1,1,1]

Explanation:

The parent of node 3 will change from node 1 to node 0.

Example 2:

Input: parent = [-1,0,4,0,1], s = "abbba"

Output: [5,2,1,1,1]

Explanation:

The following changes will happen at the same time:

The parent of node 4 will change from node 1 to node 0.
The parent of node 2 will change from node 4 to node 1.

Constraints:

n == parent.length == s.length
1 <= n <= 10⁵
0 <= parent[i] <= n - 1 for all i >= 1.
parent[0] == -1
parent represents a valid tree.
s consists only of lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: You are given a tree rooted at node 0 that consists of n nodes numbered from 0 to n - 1. The tree is represented by an array parent of size n, where parent[i] is the parent of node i. Since node 0 is the root, parent[0] == -1. You are also given a string s of length n, where s[i] is the character assigned to node i. We make the following changes on the tree one time simultaneously for all nodes x from 1 to n - 1: Find the closest node y to node x such that y is an ancestor of x, and s[x] == s[y]. If node y does not exist, do nothing. Otherwise, remove the edge between x and its current parent and make node y the new parent of x by adding an edge between them. Return an array answer of size n where answer[i] is the size of the subtree rooted at node i in the final tree.

Baseline thinking

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

Pattern signal: Array · Hash Map · Tree

Example 1

[-1,0,0,1,1,1]
"abaabc"

Example 2

[-1,0,4,0,1]
"abbba"

Step 02

Core Insight

What unlocks the optimal approach

Perform a depth-first search on the tree, starting from the root.
During the DFS, keep track of the most recent node where each character from 'a' to 'z' has been seen.

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 #3331: Find Subtree Sizes After Changes
class Solution {
    private List<Integer>[] g;
    private List<Integer>[] d;
    private char[] s;
    private int[] ans;

    public int[] findSubtreeSizes(int[] parent, String s) {
        int n = s.length();
        g = new List[n];
        d = new List[26];
        this.s = s.toCharArray();
        Arrays.setAll(g, k -> new ArrayList<>());
        Arrays.setAll(d, k -> new ArrayList<>());
        for (int i = 1; i < n; ++i) {
            g[parent[i]].add(i);
        }
        ans = new int[n];
        dfs(0, -1);
        return ans;
    }

    private void dfs(int i, int fa) {
        ans[i] = 1;
        int idx = s[i] - 'a';
        d[idx].add(i);
        for (int j : g[i]) {
            dfs(j, i);
        }
        int k = d[idx].size() > 1 ? d[idx].get(d[idx].size() - 2) : fa;
        if (k >= 0) {
            ans[k] += ans[i];
        }
        d[idx].remove(d[idx].size() - 1);
    }
}

// Accepted solution for LeetCode #3331: Find Subtree Sizes After Changes
func findSubtreeSizes(parent []int, s string) []int {
	n := len(s)
	g := make([][]int, n)
	for i := 1; i < n; i++ {
		g[parent[i]] = append(g[parent[i]], i)
	}
	d := [26][]int{}
	ans := make([]int, n)
	var dfs func(int, int)
	dfs = func(i, fa int) {
		ans[i] = 1
		idx := int(s[i] - 'a')
		d[idx] = append(d[idx], i)
		for _, j := range g[i] {
			dfs(j, i)
		}
		k := fa
		if len(d[idx]) > 1 {
			k = d[idx][len(d[idx])-2]
		}
		if k != -1 {
			ans[k] += ans[i]
		}
		d[idx] = d[idx][:len(d[idx])-1]
	}
	dfs(0, -1)
	return ans
}

# Accepted solution for LeetCode #3331: Find Subtree Sizes After Changes
class Solution:
    def findSubtreeSizes(self, parent: List[int], s: str) -> List[int]:
        def dfs(i: int, fa: int):
            ans[i] = 1
            d[s[i]].append(i)
            for j in g[i]:
                dfs(j, i)
            k = fa
            if len(d[s[i]]) > 1:
                k = d[s[i]][-2]
            if k != -1:
                ans[k] += ans[i]
            d[s[i]].pop()

        n = len(s)
        g = [[] for _ in range(n)]
        for i in range(1, n):
            g[parent[i]].append(i)
        d = defaultdict(list)
        ans = [0] * n
        dfs(0, -1)
        return ans

// Accepted solution for LeetCode #3331: Find Subtree Sizes After Changes
// 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 #3331: Find Subtree Sizes After Changes
// class Solution {
//     private List<Integer>[] g;
//     private List<Integer>[] d;
//     private char[] s;
//     private int[] ans;
// 
//     public int[] findSubtreeSizes(int[] parent, String s) {
//         int n = s.length();
//         g = new List[n];
//         d = new List[26];
//         this.s = s.toCharArray();
//         Arrays.setAll(g, k -> new ArrayList<>());
//         Arrays.setAll(d, k -> new ArrayList<>());
//         for (int i = 1; i < n; ++i) {
//             g[parent[i]].add(i);
//         }
//         ans = new int[n];
//         dfs(0, -1);
//         return ans;
//     }
// 
//     private void dfs(int i, int fa) {
//         ans[i] = 1;
//         int idx = s[i] - 'a';
//         d[idx].add(i);
//         for (int j : g[i]) {
//             dfs(j, i);
//         }
//         int k = d[idx].size() > 1 ? d[idx].get(d[idx].size() - 2) : fa;
//         if (k >= 0) {
//             ans[k] += ans[i];
//         }
//         d[idx].remove(d[idx].size() - 1);
//     }
// }

// Accepted solution for LeetCode #3331: Find Subtree Sizes After Changes
function findSubtreeSizes(parent: number[], s: string): number[] {
    const n = parent.length;
    const g: number[][] = Array.from({ length: n }, () => []);
    const d: number[][] = Array.from({ length: 26 }, () => []);
    for (let i = 1; i < n; ++i) {
        g[parent[i]].push(i);
    }
    const ans: number[] = Array(n).fill(1);
    const dfs = (i: number, fa: number): void => {
        const idx = s.charCodeAt(i) - 97;
        d[idx].push(i);
        for (const j of g[i]) {
            dfs(j, i);
        }
        const k = d[idx].length > 1 ? d[idx].at(-2)! : fa;
        if (k >= 0) {
            ans[k] += ans[i];
        }
        d[idx].pop();
    };
    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.

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.

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.