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1519. Number of Nodes in the Sub-Tree With the Same Label

SourceWeekly Contest 198 Q2DifficultyMediumRating1808

Description

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:

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 <= ai, bi < n
  • ai != bi
  • labels.length == n
  • labels is consisting of only of lowercase English letters.

Solutions

Solution 1

Thinking

For every node we need how many nodes in its subtree share its label. \(n\le 10^5\), so a fresh walk from each node is quadratic. There are only \(26\) labels, which a single DFS can count globally.

Before entering \(i\), remember the current frequency of \(\textit{labels}[i]\); after leaving, read it again. The difference is the count inside that subtree. Storing the entry snapshot in \(\textit{ans}[i]\) (subtract first, add later) avoids an extra array; extra memory is just the alphabet.

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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

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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];
    }
}

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class Solution {
public:
    vector<int> countSubTrees(int n, vector<vector<int>>& edges, string labels) {
        vector<vector<int>> g(n);
        for (auto& e : edges) {
            int a = e[0], b = e[1];
            g[a].push_back(b);
            g[b].push_back(a);
        }
        vector<int> ans(n);
        int cnt[26]{};
        function<void(int, int)> 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;
    }
};

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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
}

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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;
}

Solution 2: Explicit Stack

Thinking

For every node we need how many nodes in its subtree share its label. \(n\le 10^5\), so a fresh walk from each node is quadratic. On a chain the first recursive call always follows the only child, so the depth is \(n\) and exceeds the default recursion limit.

There are only \(26\) labels. The increase in a label's count between entering a node and leaving it is exactly the count inside that subtree.

An explicit stack walks the tree in postorder. On entry, subtract the current count from \(\textit{ans}[i]\), increment the count, then push the exit marker and the children. On exit, add the new count back to \(\textit{ans}[i]\). The difference is the number of matching labels in the subtree.

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class Solution:
    def countSubTrees(self, n: int, edges: list[list[int]], labels: str) -> list[int]:
        g = [[] for _ in range(n)]
        for a, b in edges:
            g[a].append(b)
            g[b].append(a)
        ans = [0] * n
        cnt = Counter()
        stk = [(0, -1, 0)]
        while stk:
            i, fa, state = stk.pop()
            if state == 0:
                ans[i] -= cnt[labels[i]]
                cnt[labels[i]] += 1
                stk.append((i, fa, 1))
                for j in g[i]:
                    if j != fa:
                        stk.append((j, i, 0))
            else:
                ans[i] += cnt[labels[i]]
        return ans

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class Solution {
    public int[] countSubTrees(int n, int[][] edges, String labels) {
        List<Integer>[] 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);
        }
        int[] ans = new int[n];
        int[] cnt = new int[26];
        Deque<int[]> stk = new ArrayDeque<>();
        stk.push(new int[] {0, -1, 0});
        while (!stk.isEmpty()) {
            int[] cur = stk.pop();
            int i = cur[0], fa = cur[1], state = cur[2];
            int k = labels.charAt(i) - 'a';
            if (state == 0) {
                ans[i] -= cnt[k];
                cnt[k]++;
                stk.push(new int[] {i, fa, 1});
                for (int j : g[i]) {
                    if (j != fa) {
                        stk.push(new int[] {j, i, 0});
                    }
                }
            } else {
                ans[i] += cnt[k];
            }
        }
        return ans;
    }
}

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class Solution {
public:
    vector<int> countSubTrees(int n, vector<vector<int>>& edges, string labels) {
        vector<vector<int>> g(n);
        for (auto& e : edges) {
            int a = e[0], b = e[1];
            g[a].push_back(b);
            g[b].push_back(a);
        }
        vector<int> ans(n);
        int cnt[26]{};
        vector<array<int, 3>> stk{{0, -1, 0}};
        while (!stk.empty()) {
            auto cur = stk.back();
            stk.pop_back();
            int i = cur[0], fa = cur[1], state = cur[2];
            int k = labels[i] - 'a';
            if (state == 0) {
                ans[i] -= cnt[k];
                cnt[k]++;
                stk.push_back({i, fa, 1});
                for (int j : g[i]) {
                    if (j != fa) {
                        stk.push_back({j, i, 0});
                    }
                }
            } else {
                ans[i] += cnt[k];
            }
        }
        return ans;
    }
};

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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{}
    stk := [][3]int{{0, -1, 0}}
    for len(stk) > 0 {
        cur := stk[len(stk)-1]
        stk = stk[:len(stk)-1]
        i, fa, state := cur[0], cur[1], cur[2]
        k := labels[i] - 'a'
        if state == 0 {
            ans[i] -= cnt[k]
            cnt[k]++
            stk = append(stk, [3]int{i, fa, 1})
            for _, j := range g[i] {
                if j != fa {
                    stk = append(stk, [3]int{j, i, 0})
                }
            }
        } else {
            ans[i] += cnt[k]
        }
    }
    return ans
}

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function countSubTrees(n: number, edges: number[][], labels: string): number[] {
    const g: number[][] = Array.from({ length: n }, () => []);
    for (const [a, b] of edges) {
        g[a].push(b);
        g[b].push(a);
    }
    const ans: number[] = Array(n).fill(0);
    const cnt: number[] = Array(26).fill(0);
    const stk: number[][] = [[0, -1, 0]];
    while (stk.length) {
        const [i, fa, state] = stk.pop()!;
        const k = labels.charCodeAt(i) - 97;
        if (state === 0) {
            ans[i] -= cnt[k];
            cnt[k]++;
            stk.push([i, fa, 1]);
            for (const j of g[i]) {
                if (j !== fa) {
                    stk.push([j, i, 0]);
                }
            }
        } else {
            ans[i] += cnt[k];
        }
    }
    return ans;
}

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