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Sum of the distances from every node to all other nodes is maximum

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  • Difficulty Level : Hard
  • Last Updated : 14 Jun, 2022

Given a tree with N nodes and N-1 edges with root at 1 and given an array of N-1 integers. The task is to assign weights to the edges in the tree such that the sum of the distances from every node to all other nodes is maximum.

Examples: 

Input: 

Output: 46 
Assign the edge 1-2 with weight 5 
Assign the edge 2-3 with weight 7 
Assign the edge 3-4 with weight 1 
The distance of node 1 from the nodes 2, 3, 4 is {5, 5+7, 5+7+1} 
The distance of node 2 from the nodes 3, 4 is {7, 7+1} 
The distance of node 3 from the node 4 is {1} 

Input: 
 

Output: 94

Approach: The problem can be solved using Combinations, DFS, DP on trees and Greedy logic. Since we need to assign weights to edges in the tree, hence assigning the maximum weight to the edge which occurs the maximum number of times in all the paths will be the way to get the maximum sum. In order to find the number of times an edge occurs in all the paths possible, we need to know the number of nodes on both the side of the edge. Let c1 and c2 be the count of the number of nodes on the left and right side, then the number of times the edge occurs in all paths will be c1 * c2. Sort all the possible values of c1 * c2 in ascending order. Assign the maximum weight to the maximum c1 * c2 value, and to the others in the same way. We can follow the below steps to get the number of nodes on the left side and on the right side of an edge: 

  • Run a dfs starting from the root, and initialize a dp[] array which stores the count of the nodes in the subtree of a given node.
  • Iterate for every possible edge, and find the number of nodes in the both the side of the edges.
  • To find the number of nodes in both the sides, find out the smaller value of dp[node1] or dp[node2], where node1 and node2 are the nodes on the either side of the edge
  • If one side has min(dp[node1], dp[node2]), then the other side will have (N – min(dp[node1], dp[node2])).

Below is the implementation of the above approach: 

C++




// C++ program to implement the
// above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to add an edge to the tree
void addEdge(vector<pair<int, int> >& edges,
             list<int>* tree, int x, int y)
{
    edges.push_back({ x, y });
    tree[x].push_back(y);
    tree[y].push_back(x);
}
 
// Function to run DFS and calculate the
// height of the subtree below it
void dfs(vector<pair<int, int> >& edges, list<int>* tree,
         int node, int parent, int dp[])
{
    // Initially initialize with 1
    dp[node] = 1;
 
    // Traverse for all nodes connected to node
    for (auto it : tree[node]) {
        // If node is not parent
        // then recall dfs
        if (it != parent) {
            dfs(edges, tree, it, node, dp);
 
            // Add the size of the
            // subtree beneath it
            dp[node] += dp[it];
        }
    }
}
 
// Function to assign weights to edges
// to maximize the final sum
int maximizeSum(int a[], vector<pair<int, int> >& edges,
                list<int>* tree, int n)
{
 
    // Initialize it which stores the
    // height of the subtree beneath it
    int dp[n + 1] = { 0 };
 
    // Call the DFS function to
    dfs(edges, tree, 1, 0, dp);
 
    // Sort the given array
    sort(a, a + (n - 1));
 
    // Stores the number of times an
    // edge is part of a path
    vector<int> ans;
 
    // Iterate for all edges and find the
    // number of nodes on the left and on the right
    for (auto it : edges) {
 
        // Node 1
        int x = it.first;
 
        // Node 2
        int y = it.second;
 
        // If the number of nodes below is less
        // then the other will be n - dp[node]
        if (dp[x] < dp[y]) {
            int fi = dp[x];
            int sec = n - dp[x];
            ans.push_back(fi * sec);
        }
 
        // Second condition
        else {
            int fi = dp[y];
            int sec = n - dp[y];
            ans.push_back(fi * sec);
        }
    }
 
    // Sort the number of times
    // an edges occurs in the path
    sort(ans.begin(), ans.end());
    int res = 0;
 
    // Find the summation of all those
    // paths and return
    for (int i = 0; i < n - 1; i++) {
        res += ans[i] * a[i];
    }
 
    return res;
}
 
// Driver code
int main()
{
    int n = 5;
    vector<pair<int, int> > edges;
 
    list<int>* tree = new list<int>[n + 1];
 
    // Add an edge 1-2 in the tree
    addEdge(edges, tree, 1, 2);
 
    // Add an edge 2-3 in the tree
    addEdge(edges, tree, 1, 3);
 
    // Add an edge 3-4 in the tree
    addEdge(edges, tree, 3, 4);
 
    // Add an edge 3-5 in the tree
    addEdge(edges, tree, 3, 5);
 
    // Array which gives the edges weight
    // to be assigned
    int a[] = { 6, 3, 1, 9, 3 };
 
    cout << maximizeSum(a, edges, tree, n);
}

Python3




# Python3 program to implement the
# above approach
 
edges = [[] for i in range(100)]
tree = [[] for i in range(100)]
 
# Function to add an edge to the tree
def addEdge(x, y):
    edges.append([x, y])
    tree[x].append(y)
    tree[y].append(x)
 
# Function to run DFS and calculate the
# height of the subtree below it
def dfs(node, parent, dp):
     
    # Initially initialize with 1
    dp[node] = 1
 
    # Traverse for all nodes connected to node
    for it in tree[node]:
         
        # If node is not parent
        # then recall dfs
        if (it != parent):
            dfs(it, node, dp)
 
            # Add the size of the
            # subtree beneath it
            dp[node] += dp[it]
 
# Function to assign weights to edges
# to maximize the final sum
def maximizeSum(a, n):
 
    # Initialize it which stores the
    # height of the subtree beneath it
    dp = [0 for i in range(n + 1)]
 
    # Call the DFS function to
    dfs(1, 0, dp)
 
    # Sort the given array
    a = sorted(a[:-1])
 
    # Stores the number of times an
    # edge is part of a path
    ans = []
 
    # Iterate for all edges and find the
    # number of nodes on the left and on the right
    for it in edges:
 
        if len(it) > 0:
 
            # Node 1
            x = it[0]
 
            # Node 2
            y = it[1]
 
            # If the number of nodes below is less
            # then the other will be n - dp[node]
            if (dp[x] < dp[y]):
 
                fi = dp[x]
                sec = n - dp[x]
                ans.append(fi * sec)
 
            # Second condition
            else:
                fi = dp[y]
                sec = n - dp[y]
                ans.append(fi * sec)
 
    # Sort the number of times
    # an edges occurs in the path
    ans = sorted(ans)
    res = 0
 
    # Find the summation of all those
    # paths and return
    for i in range(n - 1):
        res += ans[i] * a[i]
 
    return res
 
# Driver code
n = 5
 
# Add an edge 1-2 in the tree
addEdge(1, 2)
 
# Add an edge 2-3 in the tree
addEdge(1, 3)
 
# Add an edge 3-4 in the tree
addEdge(3, 4)
 
# Add an edge 3-5 in the tree
addEdge(3, 5)
 
# Array which gives the edges weight
# to be assigned
a = [6, 3, 1, 9, 3]
print(maximizeSum(a, n))
 
# This code is contributed by Mohit Kumar

Javascript




<script>
 
// JavaScript program to implement the
// above approach
 
let edges = new Array().fill(0).map(()=>new Array())
let tree = new Array(100).fill(0).map(()=>new Array())
 
// Function to add an edge to the tree
function addEdge(x, y){
    edges.push([x, y])
    tree[x].push(y)
    tree[y].push(x)
}
 
// Function to run DFS and calculate the
// height of the subtree below it
function dfs(node, parent, dp){
     
    // Initially initialize with 1
    dp[node] = 1
 
    // Traverse for all nodes connected to node
    for(let it of tree[node]){
         
        // If node is not parent
        // then recall dfs
        if (it != parent){
            dfs(it, node, dp)
 
            // Add the size of the
            // subtree beneath it
            dp[node] += dp[it]
        }
    }
}
 
// Function to assign weights to edges
// to maximize the final sum
function maximizeSum(a, n){
 
    // Initialize it which stores the
    // height of the subtree beneath it
    let dp = new Array(n+1).fill(0)
 
    // Call the DFS function to
    dfs(1, 0, dp)
 
    // Sort the given array
    a = a.slice(0, n-1).sort().concat(a.slice(n-1,));
 
    // Stores the number of times an
    // edge is part of a path
    let ans = []
 
    // Iterate for all edges and find the
    // number of nodes on the left and on the right
    for(let it of edges){
 
        if(it.length > 0){
 
            // Node 1
            let x = it[0]
 
            // Node 2
            let y = it[1]
 
            // If the number of nodes below is less
            // then the other will be n - dp[node]
            if (dp[x] < dp[y]){
 
                let fi = dp[x]
                let sec = n - dp[x]
                ans.push(fi * sec)
            }
 
            // Second condition
            else{
                let fi = dp[y]
                let sec = n - dp[y]
                ans.push(fi * sec)
            }
        }
    }
    // Sort the number of times
    // an edges occurs in the path
    ans.sort()
    let res = 0
 
    // Find the summation of all those
    // paths and return
    for(let i=0;i<n-1;i++)
        res += ans[i] * a[i]
 
    return res
}
 
// Driver code
let n = 5
 
// Add an edge 1-2 in the tree
addEdge(1, 2)
 
// Add an edge 2-3 in the tree
addEdge(1, 3)
 
// Add an edge 3-4 in the tree
addEdge(3, 4)
 
// Add an edge 3-5 in the tree
addEdge(3, 5)
 
// Array which gives the edges weight
// to be assigned
let a = [6, 3, 1, 9, 3]
document.write(maximizeSum(a, n),"</br>")
 
// This code is contributed by shinjanpatra
 
</script>

Output: 

94

 

Time Complexity: O(V+E) + O(n log n) , for doing the dfs and sorting
Auxiliary Space: O(n), as extra spaces are used


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