Find the root of the sub-tree whose weighted sum is minimum

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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 itself
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
void 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 code
int 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 approach
ans = 0
mini = 2**32
 
graph = [[] 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 itself
def 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 sum
def 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 code
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].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 itself
function 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 sum
function 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 code
let 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(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:

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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