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Maximize sum of minimum difference of divisors of nodes in N-ary tree
  • Last Updated : 24 Nov, 2020

Given a n-ary tree having nodes with a particular weight, our task is to find out the maximum sum of the minimum difference of divisors of each node from root to leaf.

Examples: 

Input: 

              18
             /  \
            7    15
          /  \     \
         4   12     2
             /
            9

Output: 10 
Explanation: 
The maximum sum is along the path 18 -> 7 -> 12 -> 9 
Minimum difference between divisors of 18 = 6 – 3 = 3 
Minimum difference between divisors of 7 = 7 – 1 = 6 
Minimum difference between divisors of 12 = 4 – 3 = 1 
Minimum difference between divisors of 9 = 3 – 3 = 0

Input: 
 



           20
          /  \
         13   14
        /  \    \
       10   8   26
      /
     25 

Output: 17 
Explanation: 
The maximum sum is along the path 20 -> 14 -> 26 

Approach: 
To solve the problem mentioned above we can store the minimum difference between the divisors at each node in an array using the depth first traversal. Now, the task is to find out the minimum difference between the divisors.  

  • We can observe that for any number N, the smallest divisor x in the range [√N, N] will have the least difference between x and N/x.
  • At each node, calculate the minimum difference between divisors, and finally start filling the array using the results of the children and calculate the maximum sum possible.

Below is the implementation of the above approach: 

C++

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// C++ program to maximize the sum of
// minimum difference of divisors
// of nodes in an n-ary tree
 
#include <bits/stdc++.h>
using namespace std;
 
// Array to store the
// result at each node
int sub[100005];
 
// Function to get minimum difference
// between the divisors of a number
int minDivisorDifference(int n)
{
    int num1;
    int num2;
 
    // Iterate from square
    // root of N to N
    for (int i = sqrt(n); i <= n; i++) {
        if (n % i == 0) {
            num1 = i;
            num2 = n / i;
            break;
        }
    }
    // return absolute difference
    return abs(num1 - num2);
}
 
// DFS function to calculate the maximum sum
int dfs(vector<int> g[], int u, int par)
{
 
    // Store the min difference
    sub[u] = minDivisorDifference(u);
 
    int mx = 0;
    for (auto c : g[u]) {
        if (c != par) {
            int ans = dfs(g, c, u);
            mx = max(mx, ans);
        }
    }
    // Add the maximum of
    // all children to sub[u]
    sub[u] += mx;
 
    // Return maximum sum of
    // node 'u' to its parent
    return sub[u];
}
 
// Driver code
int main()
{
 
    vector<int> g[100005];
 
    int edges = 6;
 
    g[18].push_back(7);
    g[7].push_back(18);
 
    g[18].push_back(15);
    g[15].push_back(18);
 
    g[15].push_back(2);
    g[2].push_back(15);
 
    g[7].push_back(4);
    g[4].push_back(7);
 
    g[7].push_back(12);
    g[12].push_back(7);
 
    g[12].push_back(9);
    g[9].push_back(12);
 
    int root = 18;
 
    cout << dfs(g, root, -1);
}

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Java

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// Java program to maximize the sum of
// minimum difference of divisors
// of nodes in an n-ary tree
import java.util.Vector;
class GFG{
 
// Array to store the
// result at each node
static int []sub = new int[100005];
 
// Function to get minimum difference
// between the divisors of a number
static int minDivisorDifference(int n)
{
  int num1 = 0;
  int num2 = 0;
 
  // Iterate from square
  // root of N to N
  for (int i = (int) Math.sqrt(n);
           i <= n; i++)
  {
    if (n % i == 0)
    {
      num1 = i;
      num2 = n / i;
      break;
    }
  }
  // return absolute difference
  return Math.abs(num1 - num2);
}
 
// DFS function to calculate
// the maximum sum
static int dfs(Vector<Integer> g[],
               int u, int par)
{
  // Store the min difference
  sub[u] = minDivisorDifference(u);
 
  int mx = 0;
  for (int c : g[u])
  {
    if (c != par)
    {
      int ans = dfs(g, c, u);
      mx = Math.max(mx, ans);
    }
  }
   
  // Add the maximum of
  // all children to sub[u]
  sub[u] += mx;
 
  // Return maximum sum of
  // node 'u' to its parent
  return sub[u];
}
 
// Driver code
public static void main(String[] args)
{
  @SuppressWarnings("unchecked")
  Vector<Integer> []g =
         new Vector[100005];
   
  for (int i = 0; i < g.length; i++)
    g[i] = new Vector<Integer>();
   
  int edges = 6;
   
  g[18].add(7);
  g[7].add(18);
   
  g[18].add(15);
  g[15].add(18);
   
  g[15].add(2);
  g[2].add(15);
   
  g[7].add(4);
  g[4].add(7);
 
  g[7].add(12);
  g[12].add(7);
 
  g[12].add(9);
  g[9].add(12);
 
  int root = 18;
  System.out.print(dfs(g, root, -1));
}
}
 
// This code is contributed by Princi Singh

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Python3

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# Python3 program to maximize the sum
# of minimum difference of divisors
# of nodes in an n-ary tree
import math
 
# Array to store the
# result at each node
sub = [0 for i in range(100005)]
  
# Function to get minimum difference
# between the divisors of a number
def minDivisorDifference(n):
 
    num1 = 0
    num2 = 0
  
    # Iterate from square
    # root of N to N
    for i in range(int(math.sqrt(n)), n + 1):
        if (n % i == 0):
            num1 = i
            num2 = n // i
            break
         
    # Return absolute difference
    return abs(num1 - num2)
 
# DFS function to calculate the maximum sum
def dfs(g, u, par):
     
    # Store the min difference
    sub[u] = minDivisorDifference(u)
  
    mx = 0
     
    for c in g[u]:
        if (c != par):
            ans = dfs(g, c, u)
            mx = max(mx, ans)
         
    # Add the maximum of
    # all children to sub[u]
    sub[u] += mx
  
    # Return maximum sum of
    # node 'u' to its parent
    return sub[u]
  
# Driver code
if __name__=='__main__':
  
    g = [[] for i in range(100005)]
  
    edges = 6
  
    g[18].append(7)
    g[7].append(18)
  
    g[18].append(15)
    g[15].append(18)
  
    g[15].append(2)
    g[2].append(15)
  
    g[7].append(4)
    g[4].append(7)
  
    g[7].append(12)
    g[12].append(7)
  
    g[12].append(9)
    g[9].append(12)
  
    root = 18
  
    print(dfs(g, root, -1))
 
# This code is contributed by rutvik_56

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C#

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// C# program to maximize the sum of
// minimum difference of divisors
// of nodes in an n-ary tree
using System;
using System.Collections.Generic;
class GFG{
 
// Array to store the
// result at each node
static int []sub = new int[100005];
 
// Function to get minimum difference
// between the divisors of a number
static int minDivisorDifference(int n)
{
  int num1 = 0;
  int num2 = 0;
 
  // Iterate from square
  // root of N to N
  for (int i = (int) Math.Sqrt(n);
           i <= n; i++)
  {
    if (n % i == 0)
    {
      num1 = i;
      num2 = n / i;
      break;
    }
  }
   
  // return absolute difference
  return Math.Abs(num1 - num2);
}
 
// DFS function to calculate
// the maximum sum
static int dfs(List<int> []g,
               int u, int par)
{
  // Store the min difference
  sub[u] = minDivisorDifference(u);
 
  int mx = 0;
  foreach (int c in g[u])
  {
    if (c != par)
    {
      int ans = dfs(g, c, u);
      mx = Math.Max(mx, ans);
    }
  }
 
  // Add the maximum of
  // all children to sub[u]
  sub[u] += mx;
 
  // Return maximum sum of
  // node 'u' to its parent
  return sub[u];
}
 
// Driver code
public static void Main(String[] args)
{
  List<int> []g =
       new List<int>[100005];
 
  for (int i = 0; i < g.Length; i++)
    g[i] = new List<int>();
 
  int edges = 6;
 
  g[18].Add(7);
  g[7].Add(18);
 
  g[18].Add(15);
  g[15].Add(18);
 
  g[15].Add(2);
  g[2].Add(15);
 
  g[7].Add(4);
  g[4].Add(7);
 
  g[7].Add(12);
  g[12].Add(7);
 
  g[12].Add(9);
  g[9].Add(12);
 
  int root = 18;
  Console.Write(dfs(g, root, -1));
}
}
 
// This code is contributed by Rajput-Ji

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Output: 

10









 

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