# Hackerrank - Maximum Disjoint Subtree Product

Problem statement

We define an unrooted tree, T, with the following properties:

T is a connected graph with nodes connected by n-1 edges. Each node has a distinct ID number from 1 to n, and each node ID i has a value, wi.

We define a subtree of T to be a connected part of T. Two subtrees, Ta and Tb, are disjoint if they don't contain any common nodes. For example:

Sum of a subtree is the sum of the wi values for each node i belonging to the subtree.

Given the configuration of tree T, find and print the maximum possible product of the sums of two disjoint subtrees in T (i.e., sum(Ta) · sum(Tb).

My introduction of the algorithm

The algorithm is the hard level algorithm in hackerrank world codesprint 10 in April 2017. I did write a recursive depth first search tree algorithm in the contest, passed the sample test cases but failed all other test cases with wrong answer errors. So I spent hours to study one of code submissions and put together a C# solution after the contest.

The algorithm turned out to me a simple depth first search(DFS) after hours study, debugging and walked through the sample test case. My understanding of DFS solution here is that the base case in the sample test case shown in the graph is the node with one connected edge, for example, starting from left to right, node 4 with weight 9 and node 3 with weight -1. For any edge in the graph, for example, edge 1:6 1, node 6 starts a DFS search until it reaches node 4 whereas node 3 starts a DFS search ended at itself.

The dynamic programming part is not easy to come out and it takes some time to build the recurrence formula. Base case is easy to figure out, node with one connected edges. For any edge to serve each of two nodes, it has to calculate the maximum/ minimum value include/ exclude itself within all connected edges.

The C# code passes all test cases. Depth first search is my favorite algorithm, sometimes I forgot that recursive function is the economical choice for DFS compared to iterative one using stack.

using System;
using System.Collections.Generic;
using System.IO;
class Solution
{
/// <summary>
/// https://www.hackerrank.com/contests/world-codesprint-10/challenges/maximum-disjoint-subtree-product
/// </summary>
public class Graph
{
private Dictionary<long, Node> nodes { set; get; }
private Dictionary<long, Edge> edges { set; get; }

//one edge with one edge id, but use negative edge id
//as well to accommodate two nodes in the edge.
private Dictionary<long, long> edgesWithMaxWeight { set; get; }
private Dictionary<long, long> edgesWithMinimumWeight { set; get; }
private Dictionary<long, long> edgesWithMaxWeightInclusive { set; get; }
private Dictionary<long, long> edgesWithMinWeightInclusive { set; get; }

public Graph()
{
nodes = new Dictionary<long, Node>();
edges = new Dictionary<long, Edge>();

edgesWithMaxWeight = new Dictionary<long, long>();
edgesWithMinimumWeight = new Dictionary<long, long>();
edgesWithMaxWeightInclusive = new Dictionary<long, long>();
edgesWithMinWeightInclusive = new Dictionary<long, long>();
}

/// <summary>
/// Node class - ID, ConnectedEdges, weight,
/// using LinkedList for ConnectedEdges
/// kind of smart using LinkedList for ConnectedEdges
/// How to define connected edges for each node?
/// Go over the sample test case to figure out,
/// node 1 - weight -9, connected edges
/// is a linked list with 2 nodes, where first node is edge 1
/// which are connected by Id 1 and Id 6,
/// and the second node is edge 5 which are connected by
/// Id 1 and Id 2.
/// </summary>
internal class Node
{
public long Id { get; private set; }
public long Weight { get; private set; }

public Node(long id, long weight)
{
Id = id;
Weight = weight;
ConnectedEdges = new LinkedList<Edge>();
}
}

internal class Edge
{
public long ID { get; private set; }
public long Node1 { get; private set; }
public long Node2 { get; private set; }

public Edge(long id, long node1, long node2)
{
ID = id;
Node1 = node1;
Node2 = node2;
}

/// <summary>
/// one edge has one edge id and two nodes with two node's
/// Ids, then the node with large node id
/// takes the positive edge id whereas the small one
/// uses negative edge id.
/// </summary>
/// <param name="signedEdgeId"></param>
/// <returns></returns>
public static long GetEdgeId(long signedEdgeId)
{
return Math.Abs(signedEdgeId);
}

/// <summary>
/// one edge has two node ids, and then one edge id and
/// its negative value both
/// are used for edge id.
/// </summary>
/// <param name="signedEdgeId"></param>
/// <returns></returns>
public long GetNodeId(long signedEdgeId)
{
return signedEdgeId > 0 ? GetMaxOne() : GetMinimumOne();
}

public long GetMaxOne()
{
return Math.Max(Node1, Node2);
}

public long GetMinimumOne()
{
return Math.Min(Node1, Node2);
}
}

/// <summary>
///
/// </summary>
/// <param name="edgeID"></param>
/// <param name="id1"></param>
/// <param name="id2"></param>
public void AddEdge(long edgeID, long id1, long id2)
{
var edge = new Edge(edgeID, id1, id2);

var node1 = nodes[id1];
var node2 = nodes[id2];

// each node has connected edges stored as a linked list
}

public void AddNode(long id, long weight)
{
nodes.Add(id, new Node(id, weight));
}

/// <summary>
/// minimum maximum value tree -
/// </summary>
private void fillMinMaxTrees()
{
foreach (var edge in edges.Values)
{
var id = edge.ID;

if (!edgesWithMaxWeight.ContainsKey(id))
{
fillMinMaxTreeForEdge_DFS(id);
}

if (!edgesWithMaxWeight.ContainsKey(-id))
{
fillMinMaxTreeForEdge_DFS(-id);
}
}
}

/// <summary>
/// how to understand the max tree for the edge?
/// Take the sample test case, and understand it.
/// It is a DFS search, for each edge, try to find max/ min value.
/// First edge with edgeId = 1, node Id 6 connects to node Id 1,
/// so node id 6 only has one extra connected edge to node 3,
///  which only has one connected edge.
/// Node Id = 1, which has connected edge 5(1-2), then DFS
/// search to edge 4(5-2), then DFS
/// search edge 2 (4-5), node 4 has only one connected edge
/// which is the base case.
/// </summary>
/// <param name="signedEdgeId"></param>
private void fillMinMaxTreeForEdge_DFS(long signedEdgeId)
{
long edgeId = Edge.GetEdgeId(signedEdgeId);
var edge = edges[edgeId];
long nodeId = edge.GetNodeId(signedEdgeId);

var node = nodes[nodeId];
var connectedEdges = node.ConnectedEdges;

// base case - node with one connected edge.
if (connectedEdges.Count == 1)
{
var weight = node.Weight;

return;
}

long max = long.MinValue;
long min = long.MaxValue;

long maxInclusive = node.Weight;
long minInclusive = node.Weight;

foreach (var item in connectedEdges)
{
var current = item.ID;
var node1 = item.Node1;
var node2 = item.Node2;

if (current == edgeId)
{
continue;
}

long id = node1 == nodeId ? node2 : node1;
long signedId = id > nodeId ? current : -current;

if (!edgesWithMaxWeight.ContainsKey(signedId))
{
fillMinMaxTreeForEdge_DFS(signedId);
}

max = Math.Max(edgesWithMaxWeight[signedId], max);
min = Math.Min(edgesWithMinimumWeight[signedId], min);

maxInclusive = Math.Max(edgesWithMaxWeightInclusive[signedId] + maxInclusive, maxInclusive);
minInclusive = Math.Min(edgesWithMinWeightInclusive[signedId] + minInclusive, minInclusive);
}

max = Math.Max(max, maxInclusive);
min = Math.Min(min, minInclusive);

}

/// <summary>
/// Need to find two disjoint set with maximum value of
/// product of sum. Each disjoint set has
/// its weight to sum all the nodes's weight.
/// </summary>
/// <returns></returns>
public long FindMaxProductTwoDisjointSets()
{
fillMinMaxTrees();

long maxProduct = long.MinValue;
foreach (var edge in edges.Values)
{
var id = edge.ID;
var set1Value = edgesWithMaxWeight[id];
var set2Value = edgesWithMaxWeight[-id];
maxProduct = Math.Max(set1Value * set2Value, maxProduct);

set1Value = edgesWithMinimumWeight[id];
set2Value = edgesWithMinimumWeight[-id];
maxProduct = Math.Max(set1Value * set2Value, maxProduct);
}

return maxProduct;
}
}

static void Main(String[] args)
{
ProcessInput();
//RunTestcase();
}

public static void RunTestcase()
{
int n = 6;

int[] weights = new int[] { -9, -6, -1, 9, -2, 0 };

int[][] edgeInfo = new int[5][];

edgeInfo[0] = new int[] { 6, 1 };
edgeInfo[1] = new int[] { 4, 5 };
edgeInfo[2] = new int[] { 6, 3 };
edgeInfo[3] = new int[] { 5, 2 };
edgeInfo[4] = new int[] { 1, 2 };

var graph = new Graph();
for (long i = 0; i < n; i++)
{
graph.AddNode(i + 1, weights[i]);
}

for (int i = 0; i < n - 1; i++)
{
graph.AddEdge(i + 1, edgeInfo[i][0], edgeInfo[i][1]);
}

Console.WriteLine(graph.FindMaxProductTwoDisjointSets());
}

public static void ProcessInput()
{
long n = Convert.ToInt32(Console.ReadLine());

// The respective weights of each node:
string[] w_temp = Console.ReadLine().Split(' ');
int[] w = Array.ConvertAll(w_temp, Int32.Parse);

var graph = new Graph();
for (long i = 0; i < n; i++)
{
graph.AddNode(i + 1, w[i]);
}

for (long i = 0; i < n - 1; i++)
{
// Node IDs 'u' and 'v' are connected by an edge:
string[] tokens_n = Console.ReadLine().Split(' ');
long u = Convert.ToInt32(tokens_n[0]);
long v = Convert.ToInt32(tokens_n[1]);

graph.AddEdge(i + 1, u, v);
}

Console.WriteLine(graph.FindMaxProductTwoDisjointSets());
}
}

• Share some statistics to encourage more comments here. The contest bronze medal minimum score is 36 points in this contest. And this algorithm counts 60 points. Top 25% players will be bronze medals, silver medal or gold medal. In other words, this algorithm is supposed to be hard, and we like to make it easy by any means. – Jianmin Chen May 6 '17 at 2:34
• I spent a few hours in May 31, 2018 to study the algorithm 12 months after I posted the algorithm. Actually it is a good drill to solve the algorithm like "Find maximum disjoint two subarray product" first, using dynamic programming with time complexity O(N) where N is the size of the array. – Jianmin Chen May 31 '18 at 21:53
• I wrote the partial solution using the dynamic programming solution, the code is gist.github.com/jianminchen/c7d9a7157365fb2b3df8106ff2322d62 – Jianmin Chen May 31 '18 at 21:54