Find MEX of every subtree in given Tree

• Difficulty Level : Hard
• Last Updated : 21 Jan, 2022

Given a Generic Tree consisting of N nodes numbered from 0 to N – 1 which is rooted at node 0 and an array val[] such that the value at each node is represented by val[i],  the task for each node is to find the value of MEX of its subtree.

The MEX value of node V is defined as the smallest missing positive number in a tree rooted at node V.

Examples:

Input: N = 6, edges = {{0, 1}, {1, 2}, {0, 3}, {3, 4}, {3, 5}}, val[] = {4, 3, 5, 1, 0, 2}
Output: [6, 0, 0, 3, 1, 0]
Explanation:
0(4)
/    \
1(3)    3(1)
/         /    \
2(5)     4(0)   5(2)

In the subtrees of:
Node 0: All the values in range [0, 5] are present, hence the smallest non-negative value not present is 6.
Node 1: The smallest non-negative value not present in subtree of node 1 is 0.
Node 2: The smallest non-negative value not present in subtree of node 2 absent is 0.
Node 3: All the values in range [0, 2] are present, hence the smallest non-negative value not present in subtree of node 3 is 3.
Node 4: The smallest non-negative value not present in subtree of node 4 is 1.
Node 5: The smallest non-negative value not present in subtree of node 5 is 0.

Approach: The given problem can be solved using DFS Traversal on the given Tree and performing the Binary Search to find the missing minimum positive integers in each node subtree. Follow the steps below to solve the problem:

Below is the implementation of the above approach:

C++14

 // C++ program for the above approach #include using namespace std; // Stores the edges of the treevector > edges; // Function to add edgesvoid add_edge(int x, int y){    edges.push_back({ x, y });} // Function to merge two sorted vectorsvector merge(vector& a,                  vector& b){    // To store the result    vector res;     int i = 0, j = 0;    int n = a.size(), m = b.size();     // Iterating both vectors    while (i < n && j < m) {        if (a[i] < b[j])            res.push_back(a[i++]);        else if (b[j] < a[i])            res.push_back(b[j++]);    }     // Pushing remaining elements of    // vector a    while (i < n)        res.push_back(a[i++]);     // Pushing remaining elements of    // vector b    while (j < m)        res.push_back(b[j++]);     return res;} // Function to perform the DFS Traversal// that returns the subtree of node// in sorted mannervector help(vector tree[], int x,                 int p, vector& c,                 vector& sol){    vector res;    res.push_back(c[x]);     // Iterate the childrens    for (auto i : tree[x]) {         // All values of subtree        // i in sorted manner        if (i != p) {            vector tmp                = help(tree, i, x, c, sol);            res = merge(res, tmp);        }    }     int l = 0, r = res.size() - 1;    int ans = res.size();     // Binary search to find MEX    while (l <= r) {        // Find the mid        int mid = (l + r) / 2;         // Update the ranges        if (res[mid] > mid)            r = mid - 1;        else {            ans = mid + 1;            l = mid + 1;        }    }    if (res != 0)        ans = 0;     // Update the MEX for the current    // tree node    sol[x] = ans;     return res;} // Function to find MEX of each// subtree of treevoid solve(int A, vector C){    int n = A;    vector tree[n + 1];    for (auto i : edges) {        tree[i].push_back(i);        tree[i].push_back(i);    }    vector sol(n, 0);     // Function Call    help(tree, 0, -1, C, sol);     // Print the ans for each nodes    for (auto i : sol)        cout << i << " ";} // Driver Codeint main(){    int N = 6;    add_edge(0, 1);    add_edge(1, 2);    add_edge(0, 3);    add_edge(3, 4);    add_edge(3, 5);     vector val = { 4, 3, 5, 1, 0, 2 };    solve(N, val);     return 0;}

Javascript


Output:
6 0 0 3 1 0

Time Complexity: O(N*(N + log N))
Auxiliary Space: O(N)

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