Maximum difference of count of black and white vertices in a path containing vertex V

Given a Tree with N vertices and N – 1 edges where the vertices are numbered from 0 to N – 1, and a vertex V present in the tree. It is given that each vertex in the tree has a colour assigned to it which is either white or black and the respective colours of the vertices is represented by an array arr[]. The task is to find the maximum difference between the number of white coloured vertices and the number of black coloured vertices from any possible subtree from the given tree that contains the given vertex V.

Examples:

Input: V = 0,
arr[] = {'b', 'w', 'w', 'w', 'b',
         'b', 'b', 'b', 'w'}
Tree:
           0 b
         /   \
       /       \
     1 w        2 w
    /          / \
   /          /   \
  5 b      w 3     4 b
  |          |     |
  |          |     |
  7 b      b 6     8 w
Output: 2
Explanation:
We can take the subtree
containing the vertex 0 
which contains vertices
0, 1, 2, 3 such that 
the difference between
the number of white 
and the number of black vertices
is maximum which is equal to 2.

Input:
V = 2,
arr[] = {'b', 'b', 'w', 'b'}
Tree:
        0 b
     /  |  \
    /   |   \
   1    2    3
   b    w     b
Output: 1 

Approach: The idea is to use the concept of dynamic programming to solve this problem.

  • Firstly, make a vector for colour array and for white colour, push 1 and for black colour, push -1.
  • Make an array dp[] to calculate the maximum possible difference between the number of white and black vertices in some subtree containing the vertex V.
  • Now, traverse through the tree using depth first search traversal and update the values in dp[] array.
  • Finally, the minimum value present in the dp[] array is the required answer.

Below is the implementation of the above approach:

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// C++ program to find maximum
// difference between count of
// black and white vertices in
// a path containing vertex V
  
#include <bits/stdc++.h>
using namespace std;
  
// Defining the tree class
class tree {
    vector<int> dp;
    vector<vector<int> > g;
    vector<int> c;
  
public:
    // Constructor
    tree(int n)
    {
        dp = vector<int>(n);
        g = vector<vector<int> >(n);
        c = vector<int>(n);
    }
  
    // Function for adding edges
    void addEdge(int u, int v)
    {
        g[u].push_back(v);
        g[v].push_back(u);
    }
  
    // Function to perform DFS
    // on the given tree
    void dfs(int v, int p = -1)
    {
        dp[v] = c[v];
  
        for (auto i : g[v]) {
            if (i == p)
                continue;
  
            dfs(i, v);
  
            // Returning calculated maximum
            // difference between white
            // and black for current vertex
            dp[v] += max(0, dp[i]);
        }
    }
  
    // Function that prints the
    // maximum difference between
    // white and black vertices
    void maximumDifference(int v,
                           char color[],
                           int n)
    {
        for (int i = 0; i < n; i++) {
  
            // Condition for white vertex
            if (color[i] == 'w')
                c[i] = 1;
  
            // Condition for black vertex
            else
                c[i] = -1;
        }
  
        // Calling dfs function for vertex v
        dfs(v);
  
        // Printing maximum difference between
        // white and black vertices
        cout << dp[v] << "\n";
    }
};
  
// Driver code
int main()
{
    tree t(9);
  
    t.addEdge(0, 1);
    t.addEdge(0, 2);
    t.addEdge(2, 3);
    t.addEdge(2, 4);
    t.addEdge(1, 5);
    t.addEdge(3, 6);
    t.addEdge(5, 7);
    t.addEdge(4, 8);
  
    int V = 0;
  
    char color[] = { 'b', 'w', 'w',
                     'w', 'b', 'b',
                     'b', 'b', 'w' };
  
    t.maximumDifference(V, color, 9);
  
    return 0;
}

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Output:

2

Time Complexity: O(N), where N is the number of vertices in the tree.

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