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 / 9Output: 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 = 0Input:
20 / \ 13 14 / \ \ 10 8 26 / 25Output: 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++
// 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 nodeint sub[100005];// Function to get minimum difference// between the divisors of a numberint 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 sumint 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 codeint 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);} |
Java
// Java program to maximize the sum of// minimum difference of divisors// of nodes in an n-ary treeimport java.util.Vector;class GFG{// Array to store the// result at each nodestatic int []sub = new int[100005];// Function to get minimum difference// between the divisors of a numberstatic 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 sumstatic 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 codepublic 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 |
Python3
# Python3 program to maximize the sum# of minimum difference of divisors# of nodes in an n-ary treeimport math# Array to store the# result at each nodesub = [0 for i in range(100005)] # Function to get minimum difference# between the divisors of a numberdef 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 sumdef 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 codeif __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 |
C#
// C# program to maximize the sum of// minimum difference of divisors// of nodes in an n-ary treeusing System;using System.Collections.Generic;class GFG{// Array to store the// result at each nodestatic int []sub = new int[100005];// Function to get minimum difference// between the divisors of a numberstatic 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 sumstatic 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 codepublic 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 |
Javascript
<script> // Javascript program to maximize the sum of // minimum difference of divisors // Array to store the // result at each node let sub = new Array(100005); sub.fill(0); // Function to get minimum difference // between the divisors of a number function minDivisorDifference(n) { let num1 = 0; let num2 = 0; // Iterate from square // root of N to N for (let i = parseInt(Math.sqrt(n), 10); i <= n; i++) { if (n % i == 0) { num1 = i; num2 = parseInt(n / i, 10); break; } } // return absolute difference return Math.abs(num1 - num2); } // DFS function to calculate // the maximum sum function dfs(g, u, par) { // Store the min difference sub[u] = minDivisorDifference(u); let mx = 0; for (let c = 0; c < g[u].length; c++) { if (g[u] != par) { let ans = dfs(g, g[u], 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]; } let g = new Array(100005); for (let i = 0; i < g.length; i++) g[i] = []; let edges = 6; g[18].push(7); g[7].push(18); g[18].push(15); g[15].push(18); g[15].push(2); g[2].push(15); g[7].push(4); g[4].push(7); g[7].push(12); g[12].push(7); g[12].push(9); g[9].push(12); let root = 18; document.write(dfs(g, root, -1)); // This code is contributed by mukesh07.</script> |
Output:
10
Time Complexity: O(N3/2), where N is the number of nodes.
Auxiliary Space: O(N)
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