# Maximize sum of minimum difference of divisors of nodes in N-ary tree

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++

 // C++ program to maximize the sum of // minimum difference of divisors // of nodes in an n-ary tree   #include 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 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 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

Output:

10

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