Given a tree, and the weights of all the nodes, the task is to find the root of the sub-tree whose weighted sum is minimum.
Examples:
Input:
Output: 5
Weight of sub-tree for parent 1 = ((-1) + (5) + (-2) + (-1) + (3)) = 4
Weight of sub-tree for parent 2 = ((5) + (-1) + (3)) = 7
Weight of sub-tree for parent 3 = -1
Weight of sub-tree for parent 4 = 3
Weight of sub-tree for parent 5 = -2
Node 5 gives the minimum sub-tree weighted sum.
Approach: Perform dfs on the tree, and for every node calculate the sub-tree weighted sum rooted at the current node then find the minimum sum value for a node.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;int ans = 0, mini = INT_MAX;vector<int> graph[100];vector<int> weight(100);// Function to perform dfs and update the tree// such that every node's weight is the sum of// the weights of all the nodes in the sub-tree// of the current node including itselfvoid dfs(int node, int parent){ for (int to : graph[node]) { if (to == parent) continue; dfs(to, node); // Calculating the weighted // sum of the subtree weight[node] += weight[to]; }}// Function to find the node// having minimum sub-tree sumvoid findMin(int n){ // For every node for (int i = 1; i <= n; i++) { // If current node's weight // is minimum so far if (mini > weight[i]) { mini = weight[i]; ans = i; } }}// Driver codeint main(){ int n = 5; // Weights of the node weight[1] = -1; weight[2] = 5; weight[3] = -1; weight[4] = 3; weight[5] = -2; // 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); findMin(n); cout << ans; return 0;} |
Java
// Java implementation of the approach import java.util.*; class GFG { static int ans = 0, mini = Integer.MAX_VALUE; @SuppressWarnings("unchecked") static Vector<Integer>[] graph = new Vector[100]; static Integer[] weight = new Integer[100]; // Function to perform dfs and update the tree // such that every node's weight is the sum of // the weights of all the nodes in the sub-tree // of the current node including itself static void dfs(int node, int parent) { for (int to : graph[node]) { if (to == parent) continue; dfs(to, node); // Calculating the weighted // sum of the subtree weight[node] += weight[to]; } } // Function to find the node // having minimum sub-tree sum x static void findMin(int n) { // For every node for (int i = 1; i <= n; i++) { // If current node's weight x // is minimum so far if (mini > weight[i]) { mini = weight[i]; ans = i; } } } // Driver code public static void main(String[] args) { int n = 5; for (int i = 0; i < 100; i++) graph[i] = new Vector<Integer>(); // Weights of the node weight[1] = -1; weight[2] = 5; weight[3] = -1; weight[4] = 3; weight[5] = -2; // Edges of the tree graph[1].add(2); graph[2].add(3); graph[2].add(4); graph[1].add(5); dfs(1, 1); findMin(n); System.out.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, mini = int.MaxValue; static List<int>[] graph = new List<int>[100]; static int[] weight = new int[100]; // Function to perform dfs and update the tree // such that every node's weight is the sum of // the weights of all the nodes in the sub-tree // of the current node including itself static void dfs(int node, int parent) { foreach (int to in graph[node]) { if (to == parent) continue; dfs(to, node); // Calculating the weighted // sum of the subtree weight[node] += weight[to]; } } // Function to find the node // having minimum sub-tree sum x static void findMin(int n) { // For every node for (int i = 1; i <= n; i++) { // If current node's weight x // is minimum so far if (mini > weight[i]) { mini = weight[i]; ans = i; } } } // Driver code public static void Main(String[] args) { int n = 5; for (int i = 0; i < 100; i++) graph[i] = new List<int>(); // Weights of the node weight[1] = -1; weight[2] = 5; weight[3] = -1; weight[4] = 3; weight[5] = -2; // Edges of the tree graph[1].Add(2); graph[2].Add(3); graph[2].Add(4); graph[1].Add(5); dfs(1, 1); findMin(n); Console.Write(ans); } } // This code is contributed by Rajput-Ji |
Python3
# Python3 implementation of the approachans = 0mini = 2**32graph = [[] for i in range(100)] weight = [0]*100# Function to perform dfs and update the tree# such that every node's weight is the sum of# the weights of all the nodes in the sub-tree# of the current node including itselfdef dfs(node, parent): global mini, graph, weight, ans for to in graph[node]: if (to == parent): continue dfs(to, node) # Calculating the weighted # sum of the subtree weight[node] += weight[to] # Function to find the node# having minimum sub-tree sumdef findMin(n): global mini, graph, weight, ans # For every node for i in range(1, n + 1): # If current node's weight # is minimum so far if (mini > weight[i]): mini = weight[i] ans = i# Driver coden = 5# Weights of the nodeweight[1] = -1weight[2] = 5weight[3] = -1weight[4] = 3weight[5] = -2# Edges of the treegraph[1].append(2)graph[2].append(3)graph[2].append(4)graph[1].append(5)dfs(1, 1)findMin(n)print(ans)# This code is contributed by SHUBHAMSINGH10 |
Javascript
<script> // Javascript implementation of the approach let ans = 0;let mini = Number.MAX_VALUE;let graph = new Array(100);let weight = new Array(100);for(let i = 0; i < 100; i++){ graph[i] = []; weight[i] = 0;}// Function to perform dfs and update the tree// such that every node's weight is the sum of// the weights of all the nodes in the sub-tree// of the current node including itselffunction dfs(node, parent){ for(let to = 0; to < graph[node].length; to++) { if (graph[node][to] == parent) continue dfs(graph[node][to], node); // Calculating the weighted // sum of the subtree weight[node] += weight[graph[node][to]]; }}// Function to find the node// having minimum sub-tree sumfunction findMin(n){ // For every node for(let i = 1; i <= n; i++) { // If current node's weight // is minimum so far if (mini > weight[i]) { mini = weight[i]; ans = i; } }}// Driver codelet n = 5;// Weights of the nodeweight[1] = -1;weight[2] = 5;weight[3] = -1;weight[4] = 3;weight[5] = -2;// Edges of the treegraph[1].push(2);graph[2].push(3);graph[2].push(4);graph[1].push(5);dfs(1, 1);findMin(n);document.write(ans);// This code is contributed by Dharanendra L V. </script> |
Output:
5
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. Therefore, the time complexity is O(N). - Auxiliary Space : O(n).
Recursion stack.
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