A tree with N nodes and N-1 edges is given with 2 different colours for its nodes.
Find the sub-tree with minimum colour difference i.e. abs(1-colour nodes – 2-colour nodes) is minimum.
Example:
Input :
Edges : 1 2
1 3
2 4
3 5
Colours : 1 1 2 2 1 [1-based indexing where
index denotes the node]
Output : 2
Explanation : The sub-tree {1-2} and {1-2-3-5}
have color difference of 2. Sub-tree {1-2} has two
1-colour nodes and zero 2-colour nodes. So, color
difference is 2. Sub-tree {1-2-3-5} has three 1-colour
nodes and one 2-colour nodes. So color diff = 2.
Method 1 : The problem can be solved by checking every possible sub-tree from every node of the tree. This will take exponential time as we will check for sub-trees from every node.
Method 2 : (Efficient) If we observe, we are solving a portion of the tree several times. This produces recurring sub-problems. We can use Dynamic Programming approach to get the minimum color difference in one traversal. To make things simpler, we can have color values as 1 and -1. Now, if we have a sub-tree with both colored nodes equal, our sum of colors will be 0. To get the minimum difference, we should have maximum negative sum or maximum positive sum.
- Case 1 When we need to have a sub-tree with maximum sum : We take a node if its value > 0, i.e. sum(parent) += max(0, sum(child))
- Case 2 When we need to have a sub-tree with minimum sum(or max negative sum) : We take a node if its value < 0, i.e. sum(parent) += min(0, sum(child))
To get the minimum sum, we can interchange the colors of nodes, i.e. -1 becomes 1 and vice-versa.
Below is the implementation :
C++
// CPP code to find the sub-tree with minimum color// difference in a 2-coloured tree#include <bits/stdc++.h>using namespace std;// Tree traversal to compute minimum differencevoid dfs(int node, int parent, vector<int> tree[], int colour[], int answer[]){ // Initial min difference is the color of node answer[node] = colour[node]; // Traversing its children for (auto u : tree[node]) { // Not traversing the parent if (u == parent) continue; dfs(u, node, tree, colour, answer); // If the child is adding positively to // difference, we include it in the answer // Otherwise, we leave the sub-tree and // include 0 (nothing) in the answer answer[node] += max(answer[u], 0); }}int maxDiff(vector<int> tree[], int colour[], int N){ int answer[N + 1]; memset(answer, 0, sizeof(answer)); // DFS for colour difference : 1colour - 2colour dfs(1, 0, tree, colour, answer); // Minimum colour difference is maximum answer value int high = 0; for (int i = 1; i <= N; i++) { high = max(high, answer[i]); // Clearing the current value // to check for colour2 as well answer[i] = 0; } // Interchanging the colours for (int i = 1; i <= N; i++) { if (colour[i] == -1) colour[i] = 1; else colour[i] = -1; } // DFS for colour difference : 2colour - 1colour dfs(1, 0, tree, colour, answer); // Checking if colour2 makes the minimum colour // difference for (int i = 1; i < N; i++) high = max(high, answer[i]); return high;}// Driver codeint main(){ // Nodes int N = 5; // Adjacency list representation vector<int> tree[N + 1]; // Edges tree[1].push_back(2); tree[2].push_back(1); tree[1].push_back(3); tree[3].push_back(1); tree[2].push_back(4); tree[4].push_back(2); tree[3].push_back(5); tree[5].push_back(3); // Index represent the colour of that node // There is no Node 0, so we start from // index 1 to N int colour[] = { 0, 1, 1, -1, -1, 1 }; // Printing the result cout << maxDiff(tree, colour, N); return 0;} |
Java
// Java code to find the sub-tree // with minimum color difference // in a 2-coloured treeimport java.util.*;class GFG{// Tree traversal to compute minimum differencestatic void dfs(int node, int parent, Vector<Integer> tree[], int colour[], int answer[]){ // Initial min difference is // the color of node answer[node] = colour[node]; // Traversing its children for (Integer u : tree[node]) { // Not traversing the parent if (u == parent) continue; dfs(u, node, tree, colour, answer); // If the child is adding positively to // difference, we include it in the answer // Otherwise, we leave the sub-tree and // include 0 (nothing) in the answer answer[node] += Math.max(answer[u], 0); }}static int maxDiff(Vector<Integer> tree[], int colour[], int N){ int []answer = new int[N + 1]; // DFS for colour difference : 1colour - 2colour dfs(1, 0, tree, colour, answer); // Minimum colour difference is // maximum answer value int high = 0; for (int i = 1; i <= N; i++) { high = Math.max(high, answer[i]); // Clearing the current value // to check for colour2 as well answer[i] = 0; } // Interchanging the colours for (int i = 1; i <= N; i++) { if (colour[i] == -1) colour[i] = 1; else colour[i] = -1; } // DFS for colour difference : 2colour - 1colour dfs(1, 0, tree, colour, answer); // Checking if colour2 makes the // minimum colour difference for (int i = 1; i < N; i++) high = Math.max(high, answer[i]); return high;}// Driver codepublic static void main(String []args){ // Nodes int N = 5; // Adjacency list representation Vector<Integer> tree[] = new Vector[N + 1]; for(int i = 0; i < N + 1; i++) tree[i] = new Vector<Integer>(); // Edges tree[1].add(2); tree[2].add(1); tree[1].add(3); tree[3].add(1); tree[2].add(4); tree[4].add(2); tree[3].add(5); tree[5].add(3); // Index represent the colour of that node // There is no Node 0, so we start from // index 1 to N int colour[] = { 0, 1, 1, -1, -1, 1 }; // Printing the result System.out.println(maxDiff(tree, colour, N));}}// This code is contributed by 29AjayKumar |
Python3
# Python3 code to find the sub-tree# with minimum color difference# in a 2-coloured tree# Tree traversal to compute minimum differencedef dfs(node, parent, tree, colour, answer): # Initial min difference is # the color of node answer[node] = colour[node] # Traversing its children for u in tree[node]: # Not traversing the parent if (u == parent): continue dfs(u, node, tree, colour, answer) # If the child is Adding positively to # difference, we include it in the answer # Otherwise, we leave the sub-tree and # include 0 (nothing) in the answer answer[node] += max(answer[u], 0)def maxDiff(tree, colour, N): answer = [0 for _ in range(N+1)] # DFS for colour difference : 1colour - 2colour dfs(1, 0, tree, colour, answer) # Minimum colour difference is # maximum answer value high = 0 for i in range(1, N+1): high = max(high, answer[i]) # Clearing the current value # to check for colour2 as well answer[i] = 0 # Interchanging the colours for i in range(1, N+1): if colour[i] == -1: colour[i] = 1 else: colour[i] = -1 # DFS for colour difference : 2colour - 1colour dfs(1, 0, tree, colour, answer) # Checking if colour2 makes the # minimum colour difference for i in range(1, N): high = max(high, answer[i]) return high# Driver code# NodesN = 5# Adjacency list representationtree = [list() for _ in range(N+1)]# Edgestree[1].append(2)tree[2].append(1)tree[1].append(3)tree[3].append(1)tree[2].append(4)tree[4].append(2)tree[3].append(5)tree[5].append(3)# Index represent the colour of that node# There is no Node 0, so we start from# index 1 to Ncolour = [0, 1, 1, -1, -1, 1]# Printing the resultprint(maxDiff(tree, colour, N))# This code is contributed by nitibi9839. |
C#
// C# code to find the sub-tree // with minimum color difference // in a 2-coloured treeusing System;using System.Collections.Generic;class GFG{// Tree traversal to compute minimum differencestatic void dfs(int node, int parent, List<int> []tree, int []colour, int []answer){ // Initial min difference is // the color of node answer[node] = colour[node]; // Traversing its children foreach (int u in tree[node]) { // Not traversing the parent if (u == parent) continue; dfs(u, node, tree, colour, answer); // If the child is Adding positively to // difference, we include it in the answer // Otherwise, we leave the sub-tree and // include 0 (nothing) in the answer answer[node] += Math.Max(answer[u], 0); }}static int maxDiff(List<int> []tree, int []colour, int N){ int []answer = new int[N + 1]; // DFS for colour difference : 1colour - 2colour dfs(1, 0, tree, colour, answer); // Minimum colour difference is // maximum answer value int high = 0; for (int i = 1; i <= N; i++) { high = Math.Max(high, answer[i]); // Clearing the current value // to check for colour2 as well answer[i] = 0; } // Interchanging the colours for (int i = 1; i <= N; i++) { if (colour[i] == -1) colour[i] = 1; else colour[i] = -1; } // DFS for colour difference : 2colour - 1colour dfs(1, 0, tree, colour, answer); // Checking if colour2 makes the // minimum colour difference for (int i = 1; i < N; i++) high = Math.Max(high, answer[i]); return high;}// Driver codepublic static void Main(String []args){ // Nodes int N = 5; // Adjacency list representation List<int> []tree = new List<int>[N + 1]; for(int i = 0; i < N + 1; i++) tree[i] = new List<int>(); // Edges tree[1].Add(2); tree[2].Add(1); tree[1].Add(3); tree[3].Add(1); tree[2].Add(4); tree[4].Add(2); tree[3].Add(5); tree[5].Add(3); // Index represent the colour of that node // There is no Node 0, so we start from // index 1 to N int []colour = { 0, 1, 1, -1, -1, 1 }; // Printing the result Console.WriteLine(maxDiff(tree, colour, N));}}// This code is contributed by Rajput-Ji |
Javascript
<script>// JavaScript code to find the sub-tree // with minimum color difference // in a 2-coloured tree// Tree traversal to compute minimum differencefunction dfs(node, parent, tree, colour, answer){ // Initial min difference is // the color of node answer[node] = colour[node]; // Traversing its children for(var u of tree[node]) { // Not traversing the parent if (u == parent) continue; dfs(u, node, tree, colour, answer); // If the child is Adding positively to // difference, we include it in the answer // Otherwise, we leave the sub-tree and // include 0 (nothing) in the answer answer[node] += Math.max(answer[u], 0); }}function maxDiff(tree, colour, N){ var answer = Array(N+1).fill(0); // DFS for colour difference : 1colour - 2colour dfs(1, 0, tree, colour, answer); // Minimum colour difference is // maximum answer value var high = 0; for (var i = 1; i <= N; i++) { high = Math.max(high, answer[i]); // Clearing the current value // to check for colour2 as well answer[i] = 0; } // Interchanging the colours for (var i = 1; i <= N; i++) { if (colour[i] == -1) colour[i] = 1; else colour[i] = -1; } // DFS for colour difference : 2colour - 1colour dfs(1, 0, tree, colour, answer); // Checking if colour2 makes the // minimum colour difference for (var i = 1; i < N; i++) high = Math.max(high, answer[i]); return high;}// Driver code// Nodesvar N = 5;// Adjacency list representationvar tree = Array.from(Array(N+1), ()=>Array());// Edgestree[1].push(2);tree[2].push(1);tree[1].push(3);tree[3].push(1);tree[2].push(4);tree[4].push(2);tree[3].push(5);tree[5].push(3);// Index represent the colour of that node// There is no Node 0, so we start from // index 1 to Nvar colour = [0, 1, 1, -1, -1, 1];// Printing the resultdocument.write(maxDiff(tree, colour, N));</script> |
Output
2
Time complexity: O(n) where n is the number of nodes in the binary tree.
Auxiliary Space: O(n)
This article is contributed by Rohit Thapliyal. If you like neveropen and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the neveropen main page and help other Geeks.
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
