Given a tree, and the weights of all the nodes and an integer x, the task is to find a node i such that weight[i] + x has the maximum set bits. If two or more nodes have the same count of set bits when added with x then find the one with the minimum value.
Examples:
Input:
x = 15
Output: 4
Node 1: setbits(5 + 15) = 2
Node 2: setbits(10 + 15) = 3
Node 3: setbits(11 + 15) = 3
Node 4: setbits(8 + 15) = 4
Node 5: setbits(6 + 15) = 3
Approach: Perform dfs on the tree and keep track of the node whose sum with x has maximum set bits. If two or more nodes have equal count of set bits then choose the one with the minimum number.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int maximum = INT_MIN, x, ans = INT_MAX;
vector<int> graph[100];
vector<int> weight(100);
void dfs(int node, int parent)
{
int a = __builtin_popcount(weight[node] + x);
if (maximum < a) {
maximum = a;
ans = node;
}
else if (maximum == a)
ans = min(ans, node);
for (int to : graph[node]) {
if (to == parent)
continue;
dfs(to, node);
}
}
int main()
{
x = 15;
weight[1] = 5;
weight[2] = 10;
weight[3] = 11;
weight[4] = 8;
weight[5] = 6;
graph[1].push_back(2);
graph[2].push_back(3);
graph[2].push_back(4);
graph[1].push_back(5);
dfs(1, 1);
cout << ans;
return 0;
}
|
Java
import java.util.*;
class GFG
{
static int maximum = Integer.MIN_VALUE, x, ans = Integer.MAX_VALUE;
static Vector<Vector<Integer>> graph = new Vector<Vector<Integer>>();
static Vector<Integer> weight = new Vector<Integer>();
static int __builtin_popcount(int x)
{
int c = 0;
for(int i = 0; i < 60; i++)
if(((x>>i)&1) != 0)c++;
return c;
}
static void dfs(int node, int parent)
{
int a = __builtin_popcount(weight.get(node) + x);
if (maximum < a)
{
maximum = a;
ans = node;
}
else if (maximum == a)
ans = Math.min(ans, node);
for (int i = 0; i < graph.get(node).size(); i++)
{
if (graph.get(node).get(i) == parent)
continue;
dfs(graph.get(node).get(i), node);
}
}
public static void main(String args[])
{
x = 15;
weight.add(0);
weight.add(5);
weight.add(10);;
weight.add(11);;
weight.add(8);
weight.add(6);
for(int i = 0; i < 100; i++)
graph.add(new Vector<Integer>());
graph.get(1).add(2);
graph.get(2).add(3);
graph.get(2).add(4);
graph.get(1).add(5);
dfs(1, 1);
System.out.println( ans);
}
}
|
Python3
from sys import maxsize
maximum, x, ans = -maxsize, None, maxsize
graph = [[] for i in range(100)]
weight = [0] * 100
def dfs(node, parent):
global x, ans, graph, weight, maximum
a = bin(weight[node] + x).count('1')
if maximum < a:
maximum = a
ans = node
elif maximum == a:
ans = min(ans, node)
for to in graph[node]:
if to == parent:
continue
dfs(to, node)
if __name__ == "__main__":
x = 15
weight[1] = 5
weight[2] = 10
weight[3] = 11
weight[4] = 8
weight[5] = 6
graph[1].append(2)
graph[2].append(3)
graph[2].append(4)
graph[1].append(5)
dfs(1, 1)
print(ans)
|
C#
using System;
using System.Collections.Generic;
class GFG
{
static int maximum = int.MinValue, x,ans = int.MaxValue;
static List<List<int>> graph = new List<List<int>>();
static List<int> weight = new List<int>();
static int __builtin_popcount(int x)
{
int c = 0;
for(int i = 0; i < 60; i++)
if(((x>>i)&1) != 0)c++;
return c;
}
static void dfs(int node, int parent)
{
int a = __builtin_popcount(weight[node] + x);
if (maximum < a)
{
maximum = a;
ans = node;
}
else if (maximum == a)
ans = Math.Min(ans, node);
for (int i = 0; i < graph[node].Count; i++)
{
if (graph[node][i] == parent)
continue;
dfs(graph[node][i], node);
}
}
public static void Main()
{
x = 15;
weight.Add(0);
weight.Add(5);
weight.Add(10);
weight.Add(11);;
weight.Add(8);
weight.Add(6);
for(int i = 0; i < 100; i++)
graph.Add(new List<int>());
graph[1].Add(2);
graph[2].Add(3);
graph[2].Add(4);
graph[1].Add(5);
dfs(1, 1);
Console.Write( ans);
}
}
|
Javascript
<script>
let maximum = Number.MIN_VALUE, x,
ans = Number.MAX_VALUE;
let graph = new Array(100);
for(let i=0;i<100;i++)
{
graph[i]=[];
}
let weight = [];
function __builtin_popcount(x)
{
let c = 0;
for(let i = 0; i < 60; i++)
if(((x>>i)&1) != 0)
c++;
return c;
}
function dfs(node,parent)
{
let a = __builtin_popcount(weight[node] + x);
if (maximum < a)
{
maximum = a;
ans = node;
}
else if (maximum == a)
ans = Math.min(ans, node);
for (let i = 0; i < graph[node].length; i++)
{
if (graph[node][i] == parent)
continue;
dfs(graph[node][i], node);
}
}
x = 15;
weight.push(0);
weight.push(5);
weight.push(10);;
weight.push(11);;
weight.push(8);
weight.push(6);
graph[1].push(2);
graph[2].push(3);
graph[2].push(4);
graph[1].push(5);
dfs(1, 1);
document.write( ans);
</script>
|
Complexity Analysis:
- Time Complexity : O(N).
In dfs, every node of the tree is processed once and hence the complexity due to the dfs is O(N) if there are total N nodes in the tree. Also, for processing each node the builtin_popcount() function is used which has a complexity of O(c) where c is a constant and since this complexity is constant, it does not affect the overall time complexity. Therefore, the time complexity is O(N). - Auxiliary Space : O(1).
Any extra space is not required, so the space complexity is constant.
Approach 2:
To implement the given problem using BFS approach, we can follow the below steps:
- Initialize a queue to store the nodes for BFS traversal.
- Push the root node to the queue.
- While the queue is not empty, dequeue a node from the queue.
- Calculate the set bits value of the current node and update the maximum and minimum nodes accordingly.
- Enqueue all the child nodes of the current node to the queue.
- Repeat steps 3 to 5 until the queue is empty.
- Print the node with the minimum value.
Here’s the codes for the BFS approach:
C++
#include <bits/stdc++.h>
using namespace std;
int maximum = INT_MIN, x, ans = INT_MAX;
vector<int> graph[100];
vector<int> weight(100);
void bfs(int root)
{
queue<int> q;
q.push(root);
while(!q.empty()) {
int node = q.front();
q.pop();
int a = __builtin_popcount(weight[node] + x);
if (maximum < a) {
maximum = a;
ans = node;
}
else if (maximum == a)
ans = min(ans, node);
for (int to : graph[node]) {
if (to != root) {
q.push(to);
}
}
}
}
int main()
{
x = 15;
weight[1] = 5;
weight[2] = 10;
weight[3] = 11;
weight[4] = 8;
weight[5] = 6;
graph[1].push_back(2);
graph[2].push_back(3);
graph[2].push_back(4);
graph[1].push_back(5);
bfs(1);
cout << ans;
return 0;
}
|
Python3
from queue import Queue
maximum = float('-inf')
x, ans = None, float('inf')
graph = [[] for _ in range(100)]
weight = [0] * 100
def bfs(root):
global maximum, ans
q = Queue()
q.put(root)
while not q.empty():
node = q.get()
a = bin(weight[node] + x).count('1')
if maximum < a:
maximum = a
ans = node
elif maximum == a:
ans = min(ans, node)
for to in graph[node]:
if to != root:
q.put(to)
if __name__ == '__main__':
x = 15
weight[1] = 5
weight[2] = 10
weight[3] = 11
weight[4] = 8
weight[5] = 6
graph[1].append(2)
graph[2].append(3)
graph[2].append(4)
graph[1].append(5)
bfs(1)
print(ans)
|
C#
using System;
using System.Collections.Generic;
public class GFG
{
public static int maximum = int.MinValue;
public static int x, ans = int.MaxValue;
public static List<int>[] graph = new List<int>[100];
public static List<int> weight = new List<int>(100);
public static void BFS(int root)
{
Queue<int> q = new Queue<int>();
q.Enqueue(root);
while (q.Count > 0)
{
int node = q.Dequeue();
int a = CountSetBits(weight[node] + x);
if (maximum < a)
{
maximum = a;
ans = node;
}
else if (maximum == a)
{
ans = Math.Min(ans, node);
}
foreach (int to in graph[node])
{
if (to != root)
{
q.Enqueue(to);
}
}
}
}
public static int CountSetBits(int n)
{
int count = 0;
while (n > 0)
{
count += n & 1;
n >>= 1;
}
return count;
}
public static void Main()
{
x = 15;
weight.Add(0);
weight.Add(5);
weight.Add(10);
weight.Add(11);
weight.Add(8);
weight.Add(6);
for (int i = 0; i < 100; i++)
{
graph[i] = new List<int>();
}
graph[1].Add(2);
graph[2].Add(3);
graph[2].Add(4);
graph[1].Add(5);
BFS(1);
Console.WriteLine(ans);
}
}
|
Complexity Analysis:
Time Complexity: O(N)
Auxiliary Space: O(N)