Dado un árbol, y los pesos de todos los Nodes y un número entero x , la tarea es encontrar un Node i tal que peso[i] + x tenga el conjunto máximo de bits. Si dos o más Nodes tienen la misma cantidad de bits establecidos cuando se agregan con x , encuentre el que tiene el valor mínimo.
Ejemplos:
Aporte:
x = 15
Salida: 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: conjunto de bits (6 + 15) = 3
Enfoque: Realice dfs en el árbol y realice un seguimiento del Node cuya suma con x tiene el máximo de bits establecidos. Si dos o más Nodes tienen el mismo número de bits establecidos, elija el que tenga el número mínimo.
A continuación se muestra la implementación del enfoque anterior:
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 value 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 = min(ans, node); for (int to : graph[node]) { if (to == parent) continue; dfs(to, node); } } // Driver code int 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 approach 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>(); //number of set bits static 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 value static 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 code public 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 approach from sys import maxsize maximum, x, ans = -maxsize, None, maxsize graph = [[] for i in range(100)] weight = [0] * 100 # Function to perform dfs to find # the maximum set bits value def 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 Code if __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 approach 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>(); // number of set bits static 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 value static 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 code public 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>
4
Análisis de Complejidad:
- Complejidad temporal: O(N).
En dfs, cada Node del árbol se procesa una vez y, por lo tanto, la complejidad debida a dfs es O(N) si hay un total de N Nodes en el árbol. Además, para procesar cada Node se usa la función builtin_popcount() que tiene una complejidad de O(c) donde c es una constante y dado que esta complejidad es constante, no afecta la complejidad temporal general. Por lo tanto, la complejidad del tiempo es O(N). - Espacio Auxiliar : O(1).
No se requiere ningún espacio adicional, por lo que la complejidad del espacio es constante.
Publicación traducida automáticamente
Artículo escrito por mohit kumar 29 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA