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.

Examples:

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 :2Explanation :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 C++ implementation :

// 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 difference void 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 code int 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; }

Output:

2

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