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Find number of edges that can be broken in a tree such that Bitwise OR of resulting two trees are equal

  • Difficulty Level : Hard
  • Last Updated : 19 Aug, 2021

Given a tree with n nodes and a number associated with every node. We can break any edge of the tree which will result in the formation of 2 new trees. We have to count the number of edges such that the Bitwise OR of the nodes present in the two trees formed after breaking that edge are equal. The value of every node is ≤ 106.
Examples: 
 

Input: values[]={2, 3, 32, 43, 8}
         1
        / \
       2   3
      /     \
     4       5
Output: 1
If we break the edge between 2 and 4 then the Bitwise OR 
of both the resulting tree will be 43.

Input: values[]={1, 3, 2, 3}
        1
      / | \
     2  3  4
Output: 2
The edge between 1 and 2 can be broken,the Bitwise OR 
of the resulting two trees will be 3.
Similarly, the edge between 1 and 4 can also be broken.

 

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Approach: This problem can be solved using simple DFS. Since the value of the nodes are ≤ 106, it can be represented using 22 binary bits. The Bitwise OR of the value of the nodes can also be represented in 22 binary bits. The approach is to find out the number of times each bit is set in all the values of a sub-tree. For each edge we will check that for each bit from 0 to 21 the numbers with that particular bit as set are either zero in both the resulting trees or greater than zero in both the resulting trees and if the condition is satisfied for all the bits than that edge is counted in the result.
Below is the implementation of the above approach: 
 



C++




// C++ implementation of the approach
#include<bits/stdc++.h>
using namespace std;
 
    int m[1000],x[22];
     
    // Array to store the number of times each bit
    // is set in all the values of a subtree
    int a[1000][22];
     
    vector<vector<int>> g;
    int ans = 0;
 
    // Function to perform simple DFS
    void dfs(int u, int p)
    {
        for (int i=0;i<g[u].size();i++) {
            int v = g[u][i];
            if (v != p) {
                dfs(v, u);
 
                // Finding the number of times each bit is set
                // in all the values of a subtree rooted at v
                for (int i = 0; i < 22; i++)
                    a[u][i] += a[v][i];
            }
        }
 
        // Checking for each bit whether the numbers
        // with that particular bit as set are
        // either zero in both the resulting trees or
        // greater than zero in both the resulting trees
        int pp = 0;
        for (int i = 0; i < 22; i++) {
            if (!((a[u][i] > 0 && x[i] - a[u][i] > 0)
                || (a[u][i] == 0 && x[i] == 0))) {
                pp = 1;
                break;
            }
        }
        if (pp == 0)
            ans++;
    }
 
    // Driver code
    int main()
    {
 
        // Number of nodes
        int n = 4;
 
        // ArrayList to store the tree
        g.resize(n+1);
 
        // Array to store the value of nodes
        m[1] = 1;
        m[2] = 3;
        m[3] = 2;
        m[4] = 3;
 
 
        // Array to store the number of times each bit
        // is set in all the values in complete tree
         
        for (int i = 1; i <= n; i++) {
            int y = m[i];
            int k = 0;
 
            // Finding the set bits in the value of node i
            while (y != 0) {
                int p = y % 2;
                if (p == 1) {
                    x[k]++;
                    a[i][k]++;
                }
                y = y / 2;
                k++;
            }
        }
 
        // push_back edges
        g[1].push_back(2);
        g[2].push_back(1);
        g[1].push_back(3);
        g[3].push_back(1);
        g[1].push_back(4);
        g[4].push_back(1);
        dfs(1, 0);
        cout<<(ans);
}
//contributed by Arnab Kundu

Java




// Java implementation of the approach
import java.util.*;
public class GFG {
    static int m[], a[][], x[];
    static ArrayList<Integer> g[];
    static int ans = 0;
 
    // Function to perform simple DFS
    static void dfs(int u, int p)
    {
        for (int v : g[u]) {
            if (v != p) {
                dfs(v, u);
 
                // Finding the number of times each bit is set
                // in all the values of a subtree rooted at v
                for (int i = 0; i < 22; i++)
                    a[u][i] += a[v][i];
            }
        }
 
        // Checking for each bit whether the numbers
        // with that particular bit as set are
        // either zero in both the resulting trees or
        // greater than zero in both the resulting trees
        int pp = 0;
        for (int i = 0; i < 22; i++) {
            if (!((a[u][i] > 0 && x[i] - a[u][i] > 0)
                  || (a[u][i] == 0 && x[i] == 0))) {
                pp = 1;
                break;
            }
        }
        if (pp == 0)
            ans++;
    }
 
    // Driver code
    public static void main(String args[])
    {
 
        // Number of nodes
        int n = 4;
 
        // ArrayList to store the tree
        g = new ArrayList[n + 1];
 
        // Array to store the value of nodes
        m = new int[n + 1];
        m[1] = 1;
        m[2] = 3;
        m[3] = 2;
        m[4] = 3;
 
        // Array to store the number of times each bit
        // is set in all the values of a subtree
        a = new int[n + 1][22];
 
        // Array to store the number of times each bit
        // is set in all the values in complete tree
        x = new int[22];
        for (int i = 1; i <= n; i++) {
            g[i] = new ArrayList<>();
            int y = m[i];
            int k = 0;
 
            // Finding the set bits in the value of node i
            while (y != 0) {
                int p = y % 2;
                if (p == 1) {
                    x[k]++;
                    a[i][k]++;
                }
                y = y / 2;
                k++;
            }
        }
 
        // Add edges
        g[1].add(2);
        g[2].add(1);
        g[1].add(3);
        g[3].add(1);
        g[1].add(4);
        g[4].add(1);
        dfs(1, 0);
        System.out.println(ans);
    }
}

Python3




# Python3 implementation of the approach
m, x = [0] * 1000, [0] * 22
 
# Array to store the number of times each bit
# is set in all the values of a subtree
a = [[0 for i in range(22)]
        for j in range(1000)]
ans = 0
 
# Function to perform simple DFS
def dfs(u, p):
 
    global ans
    for i in range(0, len(g[u])):
        v = g[u][i]
        if v != p:
            dfs(v, u)
 
            # Finding the number of times
            # each bit is set in all the
            # values of a subtree rooted at v
            for i in range(0, 22):
                a[u][i] += a[v][i]
 
    # Checking for each bit whether the numbers
    # with that particular bit as set are
    # either zero in both the resulting trees or
    # greater than zero in both the resulting trees
    pp = 0
    for i in range(0, 22):
        if (not((a[u][i] > 0 and
                 x[i] - a[u][i] > 0)
            or (a[u][i] == 0 and x[i] == 0))):
            pp = 1
            break
         
    if pp == 0:
        ans += 1
 
# Driver code
if __name__ == "__main__":
 
    # Number of nodes
    n = 4
 
    # ArrayList to store the tree
    g = [[] for i in range(n+1)]
 
    # Array to store the value of nodes
    m[1] = 1
    m[2] = 3
    m[3] = 2
    m[4] = 3
 
    # Array to store the number of times
    # each bit is set in all the values
    # in complete tree
    for i in range(1, n+1):
        y, k = m[i], 0
 
        # Finding the set bits in the
        # value of node i
        while y != 0:
            p = y % 2
            if p == 1:
                x[k] += 1
                a[i][k] += 1
             
            y = y // 2
            k += 1
         
    # append edges
    g[1].append(2)
    g[2].append(1)
    g[1].append(3)
    g[3].append(1)
    g[1].append(4)
    g[4].append(1)
    dfs(1, 0)
    print(ans)
 
# This code is contributed by Rituraj Jain

C#




// C# implementation of the approach
using System;
using System.Collections.Generic;
 
class GFG
{
    static int []m;
    static int [,]a;
    static int []x;
    static List<int> []g;
    static int ans = 0;
 
    // Function to perform simple DFS
    static void dfs(int u, int p)
    {
        foreach (int v in g[u])
        {
            if (v != p)
            {
                dfs(v, u);
 
                // Finding the number of times each bit is set
                // in all the values of a subtree rooted at v
                for (int i = 0; i < 22; i++)
                    a[u,i] += a[v,i];
            }
        }
 
        // Checking for each bit whether the numbers
        // with that particular bit as set are
        // either zero in both the resulting trees or
        // greater than zero in both the resulting trees
        int pp = 0;
        for (int i = 0; i < 22; i++)
        {
            if (!((a[u,i] > 0 && x[i] - a[u,i] > 0)
                || (a[u,i] == 0 && x[i] == 0)))
            {
                pp = 1;
                break;
            }
        }
        if (pp == 0)
            ans++;
    }
 
    // Driver code
    public static void Main(String []args)
    {
 
        // Number of nodes
        int n = 4;
 
        // ArrayList to store the tree
        g = new List<int>[n + 1];
 
        // Array to store the value of nodes
        m = new int[n + 1];
        m[1] = 1;
        m[2] = 3;
        m[3] = 2;
        m[4] = 3;
 
        // Array to store the number of times each bit
        // is set in all the values of a subtree
        a = new int[n + 1,22];
 
        // Array to store the number of times each bit
        // is set in all the values in complete tree
        x = new int[22];
        for (int i = 1; i <= n; i++)
        {
            g[i] = new List<int>();
            int y = m[i];
            int k = 0;
 
            // Finding the set bits in the value of node i
            while (y != 0)
            {
                int p = y % 2;
                if (p == 1)
                {
                    x[k]++;
                    a[i,k]++;
                }
                y = y / 2;
                k++;
            }
        }
 
        // Add edges
        g[1].Add(2);
        g[2].Add(1);
        g[1].Add(3);
        g[3].Add(1);
        g[1].Add(4);
        g[4].Add(1);
        dfs(1, 0);
        Console.WriteLine(ans);
    }
}
 
// This code contributed by Rajput-Ji

Javascript




<script>
 
// JavaScript implementation of the approach
var m = [];
var a = [];
var x = [];
var g = [];
var ans = 0;
 
// Function to perform simple DFS
function dfs(u, p)
{
    for(var v of g[u])
    {
        if (v != p)
        {
            dfs(v, u);
            // Finding the number of times each bit is set
            // in all the values of a subtree rooted at v
            for (var i = 0; i < 22; i++)
                a[u][i] += a[v][i];
        }
    }
    // Checking for each bit whether the numbers
    // with that particular bit as set are
    // either zero in both the resulting trees or
    // greater than zero in both the resulting trees
    var pp = 0;
    for (var i = 0; i < 22; i++)
    {
        if (!((a[u][i] > 0 && x[i] - a[u][i] > 0)
            || (a[u][i] == 0 && x[i] == 0)))
        {
            pp = 1;
            break;
        }
    }
    if (pp == 0)
        ans++;
}
// Driver code
// Number of nodes
var n = 4;
// ArrayList to store the tree
g = Array.from(Array(n+1), ()=>Array().fill(0));
// Array to store the value of nodes
m = Array(n+1);
m[1] = 1;
m[2] = 3;
m[3] = 2;
m[4] = 3;
// Array to store the number of times each bit
// is set in all the values of a subtree
a = Array.from(Array(n+1), ()=>Array(22).fill(0));
// Array to store the number of times each bit
// is set in all the values in complete tree
x = Array(22).fill(0);
for (var i = 1; i <= n; i++)
{
    g[i] = [];
    var y = m[i];
    var k = 0;
    // Finding the set bits in the value of node i
    while (y != 0)
    {
        var p = y % 2;
        if (p == 1)
        {
            x[k]++;
            a[i][k]++;
        }
        y = parseInt(y / 2);
        k++;
    }
}
// push edges
g[1].push(2);
g[2].push(1);
g[1].push(3);
g[3].push(1);
g[1].push(4);
g[4].push(1);
dfs(1, 0);
document.write(ans);
 
</script>
Output: 
2

 

Time Complexity: O(N * log(MAX)), where N is the number of nodes and MAX is the maximum value of the node in the tree.

Auxiliary Space: O(N) 




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