# Number of Nodes in the Sub-Tree With the Same Label **Difficulty:** MEDIUM [External](https://leetcode.com/problems/number-of-nodes-in-the-sub-tree-with-the-same-label) Canonical: https://scaleengineer.com/dsa/problems/number-of-nodes-in-the-sub-tree-with-the-same-label **Patterns:** [Counting](https://scaleengineer.com/dsa/patterns/counting) **Algorithms:** [Depth-First Search](https://scaleengineer.com/algorithms/depth-first-search), [Breadth-First Search](https://scaleengineer.com/algorithms/breadth-first-search) **Data structures:** Hash Table, Tree **Companies:** [Samsung](https://scaleengineer.com/companies/samsung) --- ## Problem 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. **Example 1:** ![](https://assets.glich.co/dsa/number-of-nodes-in-the-sub-tree-with-the-same-label/image0.jpg) **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:** ![](https://assets.glich.co/dsa/number-of-nodes-in-the-sub-tree-with-the-same-label/image1.jpg) **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:** ![](https://assets.glich.co/dsa/number-of-nodes-in-the-sub-tree-with-the-same-label/image2.jpg) **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 <= ai, bi < n` * `ai != bi` * `labels.length == n` * `labels` is consisting of only of lowercase English letters. # Approaches ## Brute Force: Subtree Traversal for Each Node This approach iterates through every node in the tree. For each node, it performs a separate traversal (like DFS or BFS) to visit all nodes in its subtree. During this traversal, it counts the nodes that have the same label as the starting node of the subtree. **Time:** O(N^2). Building the directed tree takes O(N). The main loop runs N times. Inside the loop, we traverse the subtree of node `i`. The size of the subtree can be up to O(N). In a skewed tree (like a path), the sum of subtree sizes is `N + (N-1) + ... + 1 = O(N^2)`. Therefore, the total time complexity is dominated by these traversals, leading to O(N^2). · **Space:** O(N). We use O(N) space for the adjacency lists. The queue used for BFS in both the tree-building step and the per-node traversal can hold up to O(N) nodes in the worst case (e.g., a star graph). **Pros:** Simple to understand and implement.; It correctly solves the problem by directly simulating the definition of a subtree count. **Cons:** Highly inefficient due to redundant computations. The traversal for a parent node re-traverses the subtrees of all its children.; Will result in a "Time Limit Exceeded" (TLE) error for larger inputs as specified in the constraints (n up to 10^5). ### Explanation First, we need to represent the tree in a way that's easy to traverse, like an adjacency list. Since the input `edges` are undirected, we build an undirected graph representation. The problem defines a rooted tree at node 0. To correctly identify subtrees, we need parent-child relationships. We can establish these with a preliminary traversal (e.g., BFS) starting from the root (node 0). This allows us to build a directed version of the tree where edges go from parent to child. The main part of the algorithm is a loop that iterates from `i = 0` to `n-1`. Inside the loop, for each node `i`, we start a new traversal (e.g., DFS) from `i`. This traversal will only move from a parent to its children, effectively exploring the subtree of `i`. We maintain a counter for this traversal. Whenever we visit a node `j` in the subtree of `i`, we check if `labels[j]` is the same as `labels[i]`. If they match, we increment the counter. After the traversal for node `i` is complete, the value of the counter is the answer for `i`, so we store it in `ans[i]`. This process is repeated for all `n` nodes. ```java import java.util.*; class Solution { public int[] countSubTrees(int n, int[][] edges, String labels) { // Build adjacency list for the directed tree List> adj = new ArrayList<>(n); for (int i = 0; i < n; i++) { adj.add(new ArrayList<>()); } // We need to know parent-child relationships. // Let's build an undirected graph first, then do a BFS to establish parent-child. List> undirectedAdj = new ArrayList<>(n); for (int i = 0; i < n; i++) { undirectedAdj.add(new ArrayList<>()); } for (int[] edge : edges) { undirectedAdj.get(edge[0]).add(edge[1]); undirectedAdj.get(edge[1]).add(edge[0]); } Queue q = new LinkedList<>(); q.offer(0); boolean[] visited = new boolean[n]; visited[0] = true; while (!q.isEmpty()) { int u = q.poll(); for (int v : undirectedAdj.get(u)) { if (!visited[v]) { visited[v] = true; adj.get(u).add(v); // Add directed edge from parent u to child v q.offer(v); } } } int[] ans = new int[n]; for (int i = 0; i < n; i++) { char targetLabel = labels.charAt(i); int count = 0; // Traverse subtree of i Queue subtreeQueue = new LinkedList<>(); subtreeQueue.offer(i); while (!subtreeQueue.isEmpty()) { int curr = subtreeQueue.poll(); if (labels.charAt(curr) == targetLabel) { count++; } for (int child : adj.get(curr)) { subtreeQueue.offer(child); } } ans[i] = count; } return ans; } } ``` ### Algorithm - Build an undirected adjacency list from the `edges` array. - Perform a Breadth-First Search (BFS) starting from the root (node 0) to build a directed adjacency list representing parent-to-child relationships. This defines the subtrees. - Initialize an answer array `ans` of size `n`. - Iterate through each node `i` from `0` to `n-1`. - For each `i`, perform a traversal (like BFS or DFS) starting from `i` and visiting only its descendants. - During this traversal, count how many nodes have the same label as node `i`. - Store this count in `ans[i]`. - Return the `ans` array. ## Optimal: Single DFS with Label Count Aggregation This efficient approach avoids re-computation by using a single Depth-First Search (DFS) traversal from the root of the tree. It works in a post-order fashion: for any node, it first recursively processes all its children and then computes its own result. Each recursive call returns the frequency of all labels ('a' through 'z') in the subtree it just processed. The parent node then aggregates these frequencies from all its children to determine the frequencies for its own subtree. **Time:** O(N). Each node is visited once. At each node, we perform a constant number of operations plus merging count arrays from its children. The merging operation takes O(26) time. Since each child's count array is merged exactly once (into its parent's), the total time for all merge operations across the entire tree is O(N * 26). Thus, the total time complexity is O(N). · **Space:** O(N). O(N) for the adjacency list. The recursion stack can go up to O(H) deep, where H is the height of the tree. In the worst case (a skewed tree), H=N, so the stack depth is O(N). At each level of recursion, we create a new count array of size 26. So, the space used by the recursion stack and the count arrays is O(H * 26), which is O(N) in the worst case. The `ans` array also takes O(N) space. Total space is O(N). **Pros:** Very efficient, solving the problem in a single pass over the tree.; Optimal time complexity for this problem. **Cons:** The logic is slightly more complex than the brute-force approach, involving recursion and passing data up the call stack.; The space complexity for the recursion stack can be O(N) in the worst case for a skewed tree. ### Explanation First, we build an adjacency list for the undirected graph from the `edges` array. We don't need to build a directed tree explicitly; we can handle parent-child relationships during the DFS traversal by passing the parent node as an argument. We define a recursive DFS function, say `dfs(node, parent, adj, labels, ans)`. This function will traverse the subtree rooted at `node` and return an array of size 26, representing the counts of each character label in that subtree. The `ans` array, which will store the final result, is passed by reference to the DFS function. Inside the `dfs(node, parent, ...)` function: 1. Create a new integer array `nodeCounts` of size 26 to store label frequencies for the subtree rooted at `node`. 2. Increment the count for the label of the current `node`: `nodeCounts[labels.charAt(node) - 'a'] = 1`. 3. Iterate through all neighbors of `node` in the adjacency list. 4. If a neighbor is the `parent` of the current `node`, skip it to avoid going up the tree. 5. For each child neighbor, make a recursive call: `childCounts = dfs(neighbor, node, ...)`. 6. The returned `childCounts` array contains label frequencies for the child's subtree. Merge these counts into `nodeCounts` by adding them element-wise: `for (int i = 0; i < 26; i++) { nodeCounts[i] += childCounts[i]; }`. 7. After iterating through all children, `nodeCounts` now contains the complete label frequencies for the subtree rooted at `node`. 8. The answer for the current `node` is the count of its own label in its subtree, which is `nodeCounts[labels.charAt(node) - 'a']`. Store this in `ans[node]`. 9. Finally, return the `nodeCounts` array to the caller (the parent of `node`). The process starts by calling `dfs(0, -1, adj, labels, ans)`. The `-1` indicates that the root (node 0) has no parent. ```java import java.util.*; class Solution { public int[] countSubTrees(int n, int[][] edges, String labels) { List> adj = new ArrayList<>(n); for (int i = 0; i < n; i++) { adj.add(new ArrayList<>()); } for (int[] edge : edges) { adj.get(edge[0]).add(edge[1]); adj.get(edge[1]).add(edge[0]); } int[] ans = new int[n]; dfs(0, -1, adj, labels, ans); return ans; } private int[] dfs(int node, int parent, List> adj, String labels, int[] ans) { // Store counts of all characters in the subtree of the current node. int[] nodeCounts = new int[26]; char label = labels.charAt(node); nodeCounts[label - 'a'] = 1; for (int neighbor : adj.get(node)) { if (neighbor == parent) { continue; } // Get counts from the child's subtree. int[] childCounts = dfs(neighbor, node, adj, labels, ans); // Add child's counts to the current node's counts. for (int i = 0; i < 26; i++) { nodeCounts[i] += childCounts[i]; } } // The answer for the current node is the count of its own label. ans[node] = nodeCounts[label - 'a']; return nodeCounts; } } ``` ### Algorithm - Build an undirected adjacency list representation of the tree from the `edges`. - Initialize an answer array `ans` of size `n`. - Define a recursive DFS function that takes the current node, its parent, the adjacency list, labels, and the `ans` array as arguments. This function will return an array of size 26 with label counts for the current node's subtree. - Start the traversal from the root: `dfs(0, -1, ...)`. - In the DFS function for a `node`: - Create a frequency array `counts` of size 26 for the node's subtree. Initialize the count for the node's own label to 1. - For each `neighbor` of the `node` (if it's not the parent): - Recursively call DFS on the `neighbor`. - Add the returned frequency array from the child's subtree to the current `node`'s `counts` array. - After visiting all children, the `counts` array holds the total frequencies for the `node`'s subtree. - Set `ans[node]` to `counts[labels.charAt(node) - 'a']`. - Return the `counts` array. - After the initial DFS call completes, return the `ans` array. # Solutions ### Java ```java class Solution { private List[] 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]; } } ``` ### CPP ```cpp class Solution { public: vector countSubTrees(int n, vector> &edges, string labels) { vector> g(n); for (auto &e : edges) { int a = e[0], b = e[1]; g[a].push_back(b); g[b].push_back(a); } vector ans(n); int cnt[26]{}; function dfs = [&](int i, int fa) { int k = labels[i] - 'a'; ans[i] -= cnt[k]; cnt[k]++; for (int &j : g[i]) { if (j != fa) { dfs(j, i); } } ans[i] += cnt[k]; }; dfs(0, -1); return ans; } }; ``` ### Python ```python 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 ```