Maximum product of two non-intersecting paths in a tree
Given an undirected connected tree with N nodes (and N-1 edges), we need to find two paths in this tree such that they are non-intersecting and the product of their length is maximum.
Examples:
In first tree two paths which are non-intersecting and have highest product are, 1-2 and 3-4, so answer is 1*1 = 1 In second tree two paths which are non-intersecting and has highest product are, 1-3-5 and 7-8-6-2 (or 4-8-6-2), so answer is 3*2 = 6
We can solve this problem by depth first search of tree by proceeding as follows, Since tree is connected and paths are non-intersecting, If we take any pair of such paths there must be a third path, connecting these two and If we remove an edge from the third path then tree will be divided into two components — one containing the first path, and the other containing the second path. This observation suggests us the algorithm: iterate over the edges; for each edge remove it, find the length of the path in both connected components and multiply the lengths of these paths. The length of the path in a tree can be found by modified depth first search where we will call for maximum path at each neighbor and we will add two maximum lengths returned, which will be the maximum path length at subtree rooted at current node.
Implementation Details:
Input is a tree, but there is no specified root in it as we have only collection of edges. The tree is represented as undirected graph. We traverse adjacency list. For every edge, we find maximum length paths on both sides of it (after removing the edge). We keep track of maximum product caused by an edge removal.
C++
// C++ program to find maximum product of two // non-intersecting paths #include <bits/stdc++.h> using namespace std; /* Returns maximum length path in subtree rooted at u after removing edge connecting u and v */ int dfs(vector< int > g[], int & curMax, int u, int v) { // To find lengths of first and second maximum // in subtrees. currMax is to store overall // maximum. int max1 = 0, max2 = 0, total = 0; // loop through all neighbors of u for ( int i = 0; i < g[u].size(); i++) { // if neighbor is v, then skip it if (g[u][i] == v) continue ; // call recursively with current neighbor as root total = max(total, dfs(g, curMax, g[u][i], u)); // get max from one side and update if (curMax > max1) { max2 = max1; max1 = curMax; } else max2 = max(max2, curMax); } // store total length by adding max // and second max total = max(total, max1 + max2); // update current max by adding 1, i.e. // current node is included curMax = max1 + 1; return total; } // method returns maximum product of length of // two non-intersecting paths int maxProductOfTwoPaths(vector< int > g[], int N) { int res = INT_MIN; int path1, path2; // one by one removing all edges and calling // dfs on both subtrees for ( int i = 1; i < N+2; i++) { for ( int j = 0; j < g[i].size(); j++) { // calling dfs on subtree rooted at // g[i][j], excluding edge from g[i][j] // to i. int curMax = 0; path1 = dfs(g, curMax, g[i][j], i); // calling dfs on subtree rooted at // i, edge from i to g[i][j] curMax = 0; path2 = dfs(g, curMax, i, g[i][j]); res = max(res, path1 * path2); } } return res; } // Utility function to add an undirected edge (u,v) void addEdge(vector< int > g[], int u, int v) { g[u].push_back(v); g[v].push_back(u); } // Driver code to test above methods int main() { int edges[][2] = {{1, 8}, {2, 6}, {3, 1}, {5, 3}, {7, 8}, {8, 4}, {8, 6} }; int N = sizeof (edges)/ sizeof (edges[0]); // there are N edges, so +1 for nodes and +1 // for 1-based indexing vector< int > g[N + 2]; for ( int i = 0; i < N; i++) addEdge(g, edges[i][0], edges[i][1]); cout << maxProductOfTwoPaths(g, N) << endl; return 0; } |
Java
// Java program to find maximum product // of two non-intersecting paths import java.util.*; class GFG{ static int curMax; // Returns maximum length path in // subtree rooted at u after // removing edge connecting u and v static int dfs(Vector<Integer> g[], int u, int v) { // To find lengths of first and // second maximum in subtrees. // currMax is to store overall // maximum. int max1 = 0 , max2 = 0 , total = 0 ; // Loop through all neighbors of u for ( int i = 0 ; i < g[u].size(); i++) { // If neighbor is v, then skip it if (g[u].get(i) == v) continue ; // Call recursively with current // neighbor as root total = Math.max(total, dfs( g, g[u].get(i), u)); // Get max from one side and update if (curMax > max1) { max2 = max1; max1 = curMax; } else max2 = Math.max(max2, curMax); } // Store total length by adding max // and second max total = Math.max(total, max1 + max2); // Update current max by adding 1, i.e. // current node is included curMax = max1 + 1 ; return total; } // Method returns maximum product of // length of two non-intersecting paths static int maxProductOfTwoPaths(Vector<Integer> g[], int N) { int res = Integer.MIN_VALUE; int path1, path2; // One by one removing all edges and // calling dfs on both subtrees for ( int i = 1 ; i < N + 2 ; i++) { for ( int j = 0 ; j < g[i].size(); j++) { // Calling dfs on subtree rooted at // g[i][j], excluding edge from g[i][j] // to i. curMax = 0 ; path1 = dfs(g, g[i].get(j), i); // Calling dfs on subtree rooted at // i, edge from i to g[i][j] curMax = 0 ; path2 = dfs(g,i, g[i].get(j)); res = Math.max(res, path1 * path2); } } return res; } // Utility function to add an // undirected edge (u,v) static void addEdge(Vector<Integer> g[], int u, int v) { g[u].add(v); g[v].add(u); } // Driver code public static void main(String[] args) { int edges[][] = { { 1 , 8 }, { 2 , 6 }, { 3 , 1 }, { 5 , 3 }, { 7 , 8 }, { 8 , 4 }, { 8 , 6 } }; int N = edges.length; // There are N edges, so +1 for nodes // and +1 for 1-based indexing @SuppressWarnings ( "unchecked" ) Vector<Integer> []g = new Vector[N + 2 ]; for ( int i = 0 ; i < g.length; i++) g[i] = new Vector<Integer>(); for ( int i = 0 ; i < N; i++) addEdge(g, edges[i][ 0 ], edges[i][ 1 ]); System.out.print(maxProductOfTwoPaths(g, N) + "\n" ); } } // This code is contributed by Princi Singh |
Python3
# Python3 program to find maximum product of two # non-intersecting paths # Returns maximum length path in subtree rooted # at u after removing edge connecting u and v def dfs(g, curMax, u, v): # To find lengths of first and second maximum # in subtrees. currMax is to store overall # maximum. max1 = 0 max2 = 0 total = 0 # loop through all neighbors of u for i in range ( len (g[u])): # if neighbor is v, then skip it if (g[u][i] = = v): continue # call recursively with current neighbor as root total = max (total, dfs(g, curMax, g[u][i], u)) # get max from one side and update if (curMax[ 0 ] > max1): max2 = max1 max1 = curMax[ 0 ] else : max2 = max (max2, curMax[ 0 ]) # store total length by adding max # and second max total = max (total, max1 + max2) # update current max by adding 1, i.e. # current node is included curMax[ 0 ] = max1 + 1 return total # method returns maximum product of length of # two non-intersecting paths def maxProductOfTwoPaths(g, N): res = - 999999999999 path1, path2 = None , None # one by one removing all edges and calling # dfs on both subtrees for i in range (N): for j in range ( len (g[i])): # calling dfs on subtree rooted at # g[i][j], excluding edge from g[i][j] # to i. curMax = [ 0 ] path1 = dfs(g, curMax, g[i][j], i) # calling dfs on subtree rooted at # i, edge from i to g[i][j] curMax = [ 0 ] path2 = dfs(g, curMax, i, g[i][j]) res = max (res, path1 * path2) return res # Utility function to add an undirected edge (u,v) def addEdge(g, u, v): g[u].append(v) g[v].append(u) # Driver code if __name__ = = '__main__' : edges = [[ 1 , 8 ], [ 2 , 6 ], [ 3 , 1 ], [ 5 , 3 ], [ 7 , 8 ], [ 8 , 4 ], [ 8 , 6 ]] N = len (edges) # there are N edges, so +1 for nodes and +1 # for 1-based indexing g = [[] for i in range (N + 2 )] for i in range (N): addEdge(g, edges[i][ 0 ], edges[i][ 1 ]) print (maxProductOfTwoPaths(g, N)) # This code is contributed by PranchalK |
C#
// C# program to find maximum product // of two non-intersecting paths using System; using System.Collections.Generic; public class GFG { static int curMax; // Returns maximum length path in // subtree rooted at u after // removing edge connecting u and v static int dfs(List< int > []g, int u, int v) { // To find lengths of first and // second maximum in subtrees. // currMax is to store overall // maximum. int max1 = 0, max2 = 0, total = 0; // Loop through all neighbors of u for ( int i = 0; i < g[u].Count; i++) { // If neighbor is v, then skip it if (g[u][i] == v) continue ; // Call recursively with current // neighbor as root total = Math.Max(total, dfs( g, g[u][i], u)); // Get max from one side and update if (curMax > max1) { max2 = max1; max1 = curMax; } else max2 = Math.Max(max2, curMax); } // Store total length by adding max // and second max total = Math.Max(total, max1 + max2); // Update current max by adding 1, i.e. // current node is included curMax = max1 + 1; return total; } // Method returns maximum product of // length of two non-intersecting paths static int maxProductOfTwoPaths(List< int > []g, int N) { int res = int .MinValue; int path1, path2; // One by one removing all edges and // calling dfs on both subtrees for ( int i = 1; i < N + 2; i++) { for ( int j = 0; j < g[i].Count; j++) { // Calling dfs on subtree rooted at // g[i,j], excluding edge from g[i,j] // to i. curMax = 0; path1 = dfs(g, g[i][j], i); // Calling dfs on subtree rooted at // i, edge from i to g[i,j] curMax = 0; path2 = dfs(g,i, g[i][j]); res = Math.Max(res, path1 * path2); } } return res; } // Utility function to add an // undirected edge (u,v) static void addEdge(List< int > []g, int u, int v) { g[u].Add(v); g[v].Add(u); } // Driver code public static void Main(String[] args) { int [,]edges = { { 1, 8 }, { 2, 6 }, { 3, 1 }, { 5, 3 }, { 7, 8 }, { 8, 4 }, { 8, 6 } }; int N = edges.GetLength(0); // There are N edges, so +1 for nodes // and +1 for 1-based indexing List< int > []g = new List< int >[N + 2]; for ( int i = 0; i < g.Length; i++) g[i] = new List< int >(); for ( int i = 0; i < N; i++) addEdge(g, edges[i,0], edges[i,1]); Console.Write(maxProductOfTwoPaths(g, N) + "\n" ); } } // This code is contributed by aashish1995 |
Javascript
<script> // Javascript program to find maximum product of two // non-intersecting paths let curMax; // Returns maximum length path in // subtree rooted at u after // removing edge connecting u and v function dfs(g, u, v) { // To find lengths of first and // second maximum in subtrees. // currMax is to store overall // maximum. let max1 = 0, max2 = 0, total = 0; // Loop through all neighbors of u for (let i = 0; i < g[u].length; i++) { // If neighbor is v, then skip it if (g[u][i] == v) continue ; // Call recursively with current // neighbor as root total = Math.max(total, dfs( g, g[u][i], u)); // Get max from one side and update if (curMax > max1) { max2 = max1; max1 = curMax; } else max2 = Math.max(max2, curMax); } // Store total length by adding max // and second max total = Math.max(total, max1 + max2); // Update current max by adding 1, i.e. // current node is included curMax = max1 + 1; return total; } // Method returns maximum product of // length of two non-intersecting paths function maxProductOfTwoPaths(g, N) { let res = Number.MIN_VALUE; let path1, path2; // One by one removing all edges and // calling dfs on both subtrees for (let i = 1; i < N + 2; i++) { for (let j = 0; j < g[i].length; j++) { // Calling dfs on subtree rooted at // g[i,j], excluding edge from g[i,j] // to i. curMax = 0; path1 = dfs(g, g[i][j], i); // Calling dfs on subtree rooted at // i, edge from i to g[i,j] curMax = 0; path2 = dfs(g,i, g[i][j]); res = Math.max(res, path1 * path2); } } return res; } // Utility function to add an // undirected edge (u,v) function addEdge(g, u, v) { g[u].push(v); g[v].push(u); } let edges = [ [ 1, 8 ], [ 2, 6 ], [ 3, 1 ], [ 5, 3 ], [ 7, 8 ], [ 8, 4 ], [ 8, 6 ] ]; let N = edges.length; // There are N edges, so +1 for nodes // and +1 for 1-based indexing let g = []; for (let i = 0; i < N + 2; i++) { g.push([]); } for (let i = 0; i < g.length; i++) g[i] = []; for (let i = 0; i < N; i++) addEdge(g, edges[i][0], edges[i][1]); document.write(maxProductOfTwoPaths(g, N) + "</br>" ); // This code is contributed by divyesh072019. </script> |
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