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Find MEX of every subtree in given Tree

  • Difficulty Level : Hard
  • Last Updated : 08 Sep, 2021
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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 <bits/stdc++.h>
using namespace std;
 
// Stores the edges of the tree
vector<vector<int> > edges;
 
// Function to add edges
void add_edge(int x, int y)
{
    edges.push_back({ x, y });
}
 
// Function to merge two sorted vectors
vector<int> merge(vector<int>& a,
                  vector<int>& b)
{
    // To store the result
    vector<int> 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 manner
vector<int> help(vector<int> tree[], int x,
                 int p, vector<int>& c,
                 vector<int>& sol)
{
    vector<int> 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<int> 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] != 0)
        ans = 0;
 
    // Update the MEX for the current
    // tree node
    sol[x] = ans;
 
    return res;
}
 
// Function to find MEX of each
// subtree of tree
void solve(int A, vector<int> C)
{
    int n = A;
    vector<int> tree[n + 1];
    for (auto i : edges) {
        tree[i[0]].push_back(i[1]);
        tree[i[1]].push_back(i[0]);
    }
    vector<int> sol(n, 0);
 
    // Function Call
    help(tree, 0, -1, C, sol);
 
    // Printe the ans for each nodes
    for (auto i : sol)
        cout << i << " ";
}
 
// Driver Code
int 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<int> val = { 4, 3, 5, 1, 0, 2 };
    solve(N, val);
 
    return 0;
}

Javascript




<script>
// Javascript program for the above approach
 
// Stores the edges of the tree
let edges = [];
 
// Function to add edges
function add_edge(x, y) {
  edges.push([x, y]);
}
 
// Function to merge two sorted vectors
function merge(a, b) {
  // To store the result
  let res = [];
 
  let i = 0,
    j = 0;
  let n = a.length,
    m = b.length;
 
  // Iterating both vectors
  while (i < n && j < m) {
    if (a[i] < b[j]) res.push(a[i++]);
    else if (b[j] < a[i]) res.push(b[j++]);
  }
 
  // Pushing remaining elements of
  // vector a
  while (i < n) res.push(a[i++]);
 
  // Pushing remaining elements of
  // vector b
  while (j < m) res.push(b[j++]);
 
  return res;
}
 
// Function to perform the DFS Traversal
// that returns the subtree of node
// in sorted manner
function help(tree, x, p, c, sol) {
  let res = [];
  res.push(c[x]);
 
  // Iterate the childrens
  for (let i of tree[x]) {
    // All values of subtree
    // i in sorted manner
    if (i != p) {
      let tmp = help(tree, i, x, c, sol);
      res = merge(res, tmp);
    }
  }
 
  let l = 0,
    r = res.length - 1;
  let ans = res.length;
 
  // Binary search to find MEX
  while (l <= r) {
    // Find the mid
    let mid = Math.floor((l + r) / 2);
 
    // Update the ranges
    if (res[mid] > mid) r = mid - 1;
    else {
      ans = mid + 1;
      l = mid + 1;
    }
  }
  if (res[0] != 0) ans = 0;
 
  // Update the MEX for the current
  // tree node
  sol[x] = ans;
 
  return res;
}
 
// Function to find MEX of each
// subtree of tree
function solve(A, C) {
  let n = A;
  let tree = new Array(n + 1).fill(0).map(() => []);
  for (let i of edges) {
    tree[i[0]].push(i[1]);
    tree[i[1]].push(i[0]);
  }
  let sol = new Array(n).fill(0);
 
  // Function Call
  help(tree, 0, -1, C, sol);
 
  // Printe the ans for each nodes
  for (let i of sol) document.write(i + " ");
}
 
// Driver Code
 
let N = 6;
add_edge(0, 1);
add_edge(1, 2);
add_edge(0, 3);
add_edge(3, 4);
add_edge(3, 5);
 
let val = [4, 3, 5, 1, 0, 2];
solve(N, val);
 
// This code is contributed by _saurabh_jaiswal.
</script>
Output: 
6 0 0 3 1 0

 

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

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