Maximize the number of uncolored vertices appearing along the path from root vertex and the colored vertices

Given a tree with N vertices numbered 1 through N with vertex 1 as root vertex and N – 1 edges. We have to color exactly k number of vertices and count the number of uncolored vertices between root vertex and every colored vertex. We have to include the root vertex in the count if it is not colored. The task to maximize the number of uncolored vertices occurring between the path from root vertex and the colored vertices. 

Examples:

Input :

           1
         / |  \
       /   |    \
     2     3      4
          / \      \
         /   \      \
        5     6      7

k = 4 
Output : 7
Explanation:
If we color vertex 2, 5, 6 and 7, 
the number of uncolored vertices between the path
from root to colored vertices is maximum which is 7.

Input :

          1
         / \
        /   \
       2     3
      /
     /
    4

k = 1
Output : 2
 

Approach:
To solve the above-mentioned problem we observe that if a vertex is chosen to be uncolored then its parent must be chosen to be uncolored. Then we can calculate how many uncolored vertices we will get if we choose a certain path to the colored vertex. Simply calculate the difference between the number of vertices between root to each vertex and the number of vertices that occur below the current vertex. Take the largest k of all the difference and calculate the sum. Use nth_element stl to get an O(n) solution.

Below is the implementation of the above approach: 

C++

// C++ program to Maximize the number
// of uncolored vertices occurring between
// the path from root vertex and the colored vertices
#include <bits/stdc++.h>
using namespace std;

// Comparator function
bool cmp(int a, int b)
{
    return a > b;
}

class graph
{
    vector<vector<int> > g;
    vector<int> depth;
    vector<int> subtree;
    int* diff;

public:
    // Constructor
    graph(int n)
    {
        g = vector<vector<int> >(n + 1);

        depth = vector<int>(n + 1);

        subtree = vector<int>(n + 1);

        diff = new int[n + 1];
    }

    // Function to push edges
    void push(int a, int b)
    {
        g[a].push_back(b);

        g[b].push_back(a);
    }

    // function for dfs traversal
    int dfs(int v, int p)
    {

        // Store depth of vertices
        depth[v] = depth[p] + 1;

        subtree[v] = 1;

        for (auto i : g[v]) {
            if (i == p)
                continue;

            // Calculate number of vertices
            // in subtree of all vertices
            subtree[v] += dfs(i, v);
        }

        // Computing the difference
        diff[v] = depth[v] - subtree[v];

        return subtree[v];
    }

    // Function that print maximum number of
    // uncolored vertices occur between root vertex
    // and all colored vertices
    void solution(int n, int k)
    {

        // Computing first k largest difference
        nth_element(diff + 1, diff + k, diff + n + 1, cmp);

        int sum = 0;

        for (int i = 1; i <= k; i++) {
            sum += diff[i];
        }

        // Print the result
        cout << sum << "\n";
    }
};

// Driver Code
int main()
{

    int N = 7;
    int k = 4;

    // initialise graph
    graph g(N);

    g.push(1, 2);
    g.push(1, 3);
    g.push(1, 4);
    g.push(3, 5);
    g.push(3, 6);
    g.push(4, 7);

    g.dfs(1, 0);

    g.solution(N, k);

    return 0;
}

Python3

# Python3 program to Maximize the number
# of uncolored vertices occurring between
# the path from root vertex and the colored vertices
g = []
depth = []
subtree = []
diff = []

# Constructor
def graph(n):
    global g, depth, subtree, diff
    g = [[] for i in range(n + 1)]
    depth = [0]*(n + 1)
    subtree = [0]*(n + 1)
    diff = [0]*(n + 1)

# Function to push edges
def push(a, b):
    global g
    g[a].append(b)
    g[b].append(a)

# function for dfs traversal
def dfs(v, p):
    global depth, subtree, g, diff

    # Store depth of vertices
    depth[v] = depth[p] + 1
    subtree[v] = 1
    for i in g[v]:
        if (i == p):
            continue

        # Calculate number of vertices
        # in subtree of all vertices
        subtree[v] += dfs(i, v)

    # Computing the difference
    diff[v] = depth[v] - subtree[v]
    return subtree[v]

# Function that print maximum number of
# uncolored vertices occur between root vertex
# and all colored vertices
def solution(n, k):
    global diff

    # Computing first k largest difference
    diff = sorted(diff)[::-1]

    # nth_element(diff + 1, diff + k, diff + n + 1, cmp)
    sum = 2
    for i in range(1, k + 1):
        sum += diff[i]

    # Print the result
    print (sum)

# Driver Code
if __name__ == '__main__':
    N = 7
    k = 4

    # initialise graph
    graph(N)
    push(1, 2)
    push(1, 3)
    push(1, 4)
    push(3, 5)
    push(3, 6)
    push(4, 7)
    dfs(1, 0)
    solution(N, k)

# This code is contributed by mohit kumar 29.

Javascript

<script>
// Javascript program to Maximize the number
// of uncolored vertices occurring between
// the path from root vertex and the colored vertices

let g = [];
let depth = [];
let subtree = [];
let diff = [];

function graph(n)
{
    g = new Array(n + 1);

        depth = Array(n + 1);

        subtree = Array(n + 1);

        diff = new Array(n + 1);

       for(let i = 0; i < (n + 1); i++)
    {
        g[i] = [];
        depth[i] = 0;
        subtree[i] = 0;
        diff[i] = 0;
    }
}

//Function to push edges
function push(a, b)
{
    g[a].push(b);

        g[b].push(a);
}
// function for dfs traversal
function  dfs(v, p)
{
    // Store depth of vertices
        depth[v] = depth[p] + 1;

        subtree[v] = 1;

        for (let i=0;i< g[v].length;i++) {
            if (g[v][i] == p)
                continue;

            // Calculate number of vertices
            // in subtree of all vertices
            subtree[v] += dfs(g[v][i], v);
        }

        // Computing the difference
        diff[v] = depth[v] - subtree[v];

        return subtree[v];
}

// Function that print maximum number of
// uncolored vertices occur between root vertex
// and all colored vertices
function solution(n, k)
{
    diff.sort(function(a,b){return b-a;});

    let sum = 2,i;
    for(i = 1; i < k + 1; i++)
    {
        sum += diff[i];
    }

    document.write(sum);

}

// Driver Code
let N = 7,
    k = 4;
graph(N)
push(1, 2)
push(1, 3)
push(1, 4)
push(3, 5)
push(3, 6)
push(4, 7)
dfs(1, 0)
solution(N, k)

// This code is contributed by patel2127
</script>

Java

// C3 program to Maximize the number
// of uncolored vertices occurring between
// the path from root vertex and the colored vertices
import java.util.*;

class Graph {
    List<List<Integer> > g;
    int[] depth;
    int[] subtree;
    int[] diff;

    public Graph(int n)
    {
        g = new ArrayList<>(n + 1);
        for (int i = 0; i <= n; i++) {
            g.add(new ArrayList<>());
        }
        depth = new int[n + 1];
        subtree = new int[n + 1];
        diff = new int[n + 1];
    }
 // Function to push edges
    public void push(int a, int b)
    {
        g.get(a).add(b);
        g.get(b).add(a);
    }
  // function for dfs traversal
    public int dfs(int v, int p)
    { // Store depth of vertices
        depth[v] = depth[p] + 1;
        subtree[v] = 1;
        for (int i : g.get(v)) {
            if (i == p)
                continue;
            // Calculate number of vertices
            // in subtree of all vertices
            subtree[v] += dfs(i, v);
        }
        // Computing the difference
        diff[v] = depth[v] - subtree[v];
        return subtree[v];
    }
 // Function that print maximum number of
    // uncolored vertices occur between root vertex
    // and all colored vertices
    public void solution(int n, int k)
    {
        Arrays.sort(diff, 1, n + 1);
        int sum = 0;
        for (int i = n; i >= n - k + 1; i--) {
            sum += diff[i];
        }
      //Print the result
        System.out.println(sum);
    }
}
//Driver code
public class Main {
    public static void main(String[] args)
    {
        int N = 7;
        int k = 4;
    // initialise graph
        Graph g = new Graph(N);
        g.push(1, 2);
        g.push(1, 3);
        g.push(1, 4);
        g.push(3, 5);
        g.push(3, 6);
        g.push(4, 7);

        g.dfs(1, 0);

        g.solution(N, k);
    }
}

C#

// C# program to Maximize the number
// of uncolored vertices occurring between
// the path from root vertex and the colored vertices
using System;
using System.Collections.Generic;

class Graph {
    List<List<int> > g;
    int[] depth;
    int[] subtree;
    int[] diff;

    public Graph(int n)
    {
        g = new List<List<int> >(n + 1);
        for (int i = 0; i <= n; i++) {
            g.Add(new List<int>());
        }
        depth = new int[n + 1];
        subtree = new int[n + 1];
        diff = new int[n + 1];
    }
    // Function to push edges
    public void Push(int a, int b)
    {
        g[a].Add(b);
        g[b].Add(a);
    }
    // function for dfs traversal
    public int Dfs(int v, int p)
    {
        // Store depth of vertices
        depth[v] = depth[p] + 1;
        subtree[v] = 1;
        foreach(int i in g[v])
        {
            if (i == p)
                continue;
            // Calculate number of vertices
            // in subtree of all vertices
            subtree[v] += Dfs(i, v);
        }
        diff[v] = depth[v] - subtree[v];
        return subtree[v];
    }
    // Function that print maximum number of
    // uncolored vertices occur between root vertex
    // and all colored vertices
    public void Solution(int n, int k)
    {
        if (k > n) {
            Console.WriteLine(
                "Error: k should be less than or equal to n.");
            return;
        }

        Array.Sort(diff, 1, n);
        int sum = 2;
        for (int i = n - 1; i >=n - k; i--) {
            sum += diff[i];
        }
        // Print the result
        Console.WriteLine(sum);
    }
}

class MainClass {
    // Driver code
    public static void Main(string[] args)
    {
        int N = 7;
        int k = 4;

        Graph g = new Graph(N);
        g.Push(1, 2);
        g.Push(1, 3);
        g.Push(1, 4);
        g.Push(3, 5);
        g.Push(3, 6);
        g.Push(4, 7);

        g.Dfs(1, 0);
        g.Solution(N, k);
    }
}

Output:

7

Time Complexity: O(N), where N is the number of vertices.
Auxiliary Space: O(N), for using extra vectors of size N.