Find the node whose sum with X has maximum set bits
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++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;int maximum = INT_MIN, x, ans = INT_MAX;vector<int> graph[100];vector<int> weight(100);// Function to perform dfs to find// the maximum set bits valuevoid dfs(int node, int parent){ // If current set bits value is greater than // the current maximum int a = __builtin_popcount(weight[node] + x); if (maximum < a) { maximum = a; ans = node; } // If count is equal to the maximum // then choose the node with minimum value else if (maximum == a) ans = min(ans, node); for (int to : graph[node]) { if (to == parent) continue; dfs(to, node); }}// Driver codeint main(){ x = 15; // Weights of the node weight[1] = 5; weight[2] = 10; weight[3] = 11; weight[4] = 8; weight[5] = 6; // Edges of the tree 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
// Java implementation of the approachimport 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>();//number of set bitsstatic int __builtin_popcount(int x){ int c = 0; for(int i = 0; i < 60; i++) if(((x>>i)&1) != 0)c++; return c;}// Function to perform dfs to find// the maximum valuestatic void dfs(int node, int parent){ // If current set bits value is greater than // the current maximum int a = __builtin_popcount(weight.get(node) + x); if (maximum < a) { maximum = a; ans = node; } // If count is equal to the maximum // then choose the node with minimum value 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); }}// Driver codepublic static void main(String args[]){ x = 15; // Weights of the node 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>()); // Edges of the tree 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);}}// This code is contributed by Arnab Kundu |
Python3
# Python implementation of the approachfrom sys import maxsizemaximum, x, ans = -maxsize, None, maxsizegraph = [[] for i in range(100)]weight = [0] * 100# Function to perform dfs to find# the maximum set bits valuedef dfs(node, parent): global x, ans, graph, weight, maximum # If current set bits value is greater than # the current maximum a = bin(weight[node] + x).count('1') if maximum < a: maximum = a ans = node # If count is equal to the maximum # then choose the node with minimum value elif maximum == a: ans = min(ans, node) for to in graph[node]: if to == parent: continue dfs(to, node)# Driver Codeif __name__ == "__main__": x = 15 # Weights of the node weight[1] = 5 weight[2] = 10 weight[3] = 11 weight[4] = 8 weight[5] = 6 # Edges of the tree graph[1].append(2) graph[2].append(3) graph[2].append(4) graph[1].append(5) dfs(1, 1) print(ans)# This code is contributed by# sanjeev2552 |
C#
// C# implementation of the approachusing 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>();// number of set bitsstatic int __builtin_popcount(int x){ int c = 0; for(int i = 0; i < 60; i++) if(((x>>i)&1) != 0)c++; return c;}// Function to perform dfs to find// the maximum valuestatic void dfs(int node, int parent){ // If current set bits value is greater than // the current maximum int a = __builtin_popcount(weight[node] + x); if (maximum < a) { maximum = a; ans = node; } // If count is equal to the maximum // then choose the node with minimum value 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); }}// Driver codepublic static void Main(){ x = 15; // Weights of the node 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>()); // Edges of the tree graph[1].Add(2); graph[2].Add(3); graph[2].Add(4); graph[1].Add(5); dfs(1, 1); Console.Write( ans);}}// This code is contributed by mits |
Javascript
<script>// Javascript implementation of the approach 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 = []; //number of set bits function __builtin_popcount(x) { let c = 0; for(let i = 0; i < 60; i++) if(((x>>i)&1) != 0) c++; return c; } // Function to perform dfs to find // the maximum value function dfs(node,parent) { // If current set bits value is greater than // the current maximum let a = __builtin_popcount(weight[node] + x); if (maximum < a) { maximum = a; ans = node; } // If count is equal to the maximum // then choose the node with minimum value 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); } } // Driver code x = 15; // Weights of the node weight.push(0); weight.push(5); weight.push(10);; weight.push(11);; weight.push(8); weight.push(6); // Edges of the tree graph[1].push(2); graph[2].push(3); graph[2].push(4); graph[1].push(5); dfs(1, 1); document.write( ans); // This code is contributed by unknown2108 </script> |
Output:
4
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.
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