Dado un árbol y los pesos de todos los Nodes, la tarea es contar el número de Nodes cuya suma de dígitos de pesos es impar.
Ejemplos:
Aporte:
Salida: 3
Node 1: digitSum(144) = 1 + 4 + 4 = 9
Node 2: digitSum(1234) = 1 + 2 + 3 + 4 = 10
Node 3: digitSum(21) = 2 + 1 = 3
Node 4 : digitSum(5) = 5
Node 5: digitSum(77) = 7 + 7 = 14
Solo la suma de dígitos de los pesos de los Nodes 1, 3 y 4 son impares.
Enfoque: Realice dfs en el árbol y para cada Node, verifique si la suma de los dígitos de su peso es impar. Si es así, entonces incremente el conteo.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
int ans = 0;
vector<int> graph[100];
vector<int> weight(100);
// Function to return the
// sum of the digits of n
int digitSum(int n)
{
int sum = 0;
while (n) {
sum += n % 10;
n = n / 10;
}
return sum;
}
// Function to perform dfs
void dfs(int node, int parent)
{
// If sum of the digits of current node's
// weight is odd then increment ans
int sum = digitSum(weight[node]);
if (sum % 2 == 1)
ans += 1;
for (int to : graph[node]) {
if (to == parent)
continue;
dfs(to, node);
}
}
// Driver code
int main()
{
// Weights of the node
weight[1] = 144;
weight[2] = 1234;
weight[3] = 21;
weight[4] = 5;
weight[5] = 77;
// 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 ans = 0;
static Vector<Integer>[] graph = new Vector[100];
static Integer[] weight = new Integer[100];
// Function to return the
// sum of the digits of n
static int digitSum(int n)
{
int sum = 0;
while (n > 0)
{
sum += n % 10;
n = n / 10;
}
return sum;
}
// Function to perform dfs
static void dfs(int node, int parent)
{
// If sum of the digits of current node's
// weight is odd then increment ans
int sum = digitSum(weight[node]);
if (sum % 2 == 1)
ans += 1;
for (int to : graph[node])
{
if (to == parent)
continue;
dfs(to, node);
}
}
// Driver code
public static void main(String[] args)
{
for (int i = 0; i < 100; i++)
graph[i] = new Vector<Integer>();
// Weights of the node
weight[1] = 144;
weight[2] = 1234;
weight[3] = 21;
weight[4] = 5;
weight[5] = 77;
// Edges of the tree
graph[1].add(2);
graph[2].add(3);
graph[2].add(4);
graph[1].add(5);
dfs(1, 1);
System.out.print(ans);
}
}
// This code is contributed by Rajput-Ji
Python3
# Python3 implementation of the approach ans = 0 graph = [[] for i in range(100)] weight = [0] * 100 # Function to return the # sum of the digits of n def digitSum(n): sum = 0 while (n): sum += n % 10 n = n // 10 return sum # Function to perform dfs def dfs(node, parent): global ans # If sum of the digits of current node's # weight is odd then increment ans sum = digitSum(weight[node]) if (sum % 2 == 1): ans += 1 for to in graph[node]: if (to == parent): continue dfs(to, node) # Driver code # Weights of the node weight[1] = 144 weight[2] = 1234 weight[3] = 21 weight[4] = 5 weight[5] = 77 # 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 SHUBHAMSINGH10
C#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
static int ans = 0;
static List<int>[] graph = new List<int>[100];
static int[] weight = new int[100];
// Function to return the
// sum of the digits of n
static int digitSum(int n)
{
int sum = 0;
while (n > 0)
{
sum += n % 10;
n = n / 10;
}
return sum;
}
// Function to perform dfs
static void dfs(int node, int parent)
{
// If sum of the digits of current node's
// weight is odd then increment ans
int sum = digitSum(weight[node]);
if (sum % 2 == 1)
ans += 1;
foreach (int to in graph[node])
{
if (to == parent)
continue;
dfs(to, node);
}
}
// Driver code
public static void Main(String[] args)
{
for (int i = 0; i < 100; i++)
graph[i] = new List<int>();
// Weights of the node
weight[1] = 144;
weight[2] = 1234;
weight[3] = 21;
weight[4] = 5;
weight[5] = 77;
// 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 PrinciRaj1992
Javascript
<script>
// JavaScript implementation of the approach
var ans = 0;
var graph = Array.from(Array(100), ()=>Array());
var weight = Array(100);
// Function to return the
// sum of the digits of n
function digitSum(n)
{
var sum = 0;
while (n > 0)
{
sum += n % 10;
n = parseInt( n / 10);
}
return sum;
}
// Function to perform dfs
function dfs(node, parent)
{
// If sum of the digits of current node's
// weight is odd then increment ans
var sum = digitSum(weight[node]);
if (sum % 2 == 1)
ans += 1;
for(var to of graph[node])
{
if (to == parent)
continue;
dfs(to, node);
}
}
// Driver code
for (var i = 0; i < 100; i++)
graph[i] = [];
// Weights of the node
weight[1] = 144;
weight[2] = 1234;
weight[3] = 21;
weight[4] = 5;
weight[5] = 77;
// 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);
</script>
Producción:
3
Análisis de Complejidad:
- Complejidad temporal: O(N).
En DFS, cada Node del árbol se procesa una vez y, por lo tanto, la complejidad debida al dfs es O(N) para N Nodes en el árbol. 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