# Count nodes within K-distance from all nodes in a set

Given an undirected tree with some marked nodes and a positive number K. We need to print the count of all such nodes which have distance from all marked nodes less than K that means every node whose distance from all marked nodes is less than K, should be counted in the result.

**Examples:**

In above tree we can see that node with index 0, 2, 3, 5, 6, 7 have distances less than 3 from all the marked nodes. so answer will be 6

We can solve this problem using breadth first search. Main thing to observe in this problem is that if we find two marked nodes which are at largest distance from each other considering all pairs of marked nodes then if a node is at a distance less than K from both of these two nodes then it will be at a distance less than K from all the marked nodes because these two nodes represents the extreme limit of all marked nodes, if a node lies in this limit then it will be at a distance less than K from all marked nodes otherwise not.

As in above example, node-1 and node-4 are most distant marked node so nodes which are at distance less than 3 from these two nodes will also be at distance less than 3 from node 2 also. Now first distant marked node we can get by doing a bfs from any random node, second distant marked node we can get by doing another bfs from marked node we just found from the first bfs and in this bfs we can also found distance of all nodes from first distant marked node and to find distance of all nodes from second distant marked node we will do one more bfs, so after doing these three bfs we can get distance of all nodes from two extreme marked nodes which can be compared with K to know which nodes fall in K-distance range from all marked nodes.

## C++

`// C++ program to count nodes inside K distance ` `// range from marked nodes ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Utility bfs method to fill distance vector and returns ` `// most distant marked node from node u ` `int` `bfsWithDistance(vector<` `int` `> g[], ` `bool` `mark[], ` `int` `u, ` ` ` `vector<` `int` `>& dis) ` `{ ` ` ` `int` `lastMarked; ` ` ` `queue<` `int` `> q; ` ` ` ` ` `// push node u in queue and initialize its distance as 0 ` ` ` `q.push(u); ` ` ` `dis[u] = 0; ` ` ` ` ` `// loop untill all nodes are processed ` ` ` `while` `(!q.empty()) ` ` ` `{ ` ` ` `u = q.front(); q.pop(); ` ` ` `// if node is marked, update lastMarked variable ` ` ` `if` `(mark[u]) ` ` ` `lastMarked = u; ` ` ` ` ` `// loop over all neighbors of u and update their ` ` ` `// distance before pushing in queue ` ` ` `for` `(` `int` `i = 0; i < g[u].size(); i++) ` ` ` `{ ` ` ` `int` `v = g[u][i]; ` ` ` ` ` `// if not given value already ` ` ` `if` `(dis[v] == -1) ` ` ` `{ ` ` ` `dis[v] = dis[u] + 1; ` ` ` `q.push(v); ` ` ` `} ` ` ` `} ` ` ` `} ` ` ` `// return last updated marked value ` ` ` `return` `lastMarked; ` `} ` ` ` `// method returns count of nodes which are in K-distance ` `// range from marked nodes ` `int` `nodesKDistanceFromMarked(` `int` `edges[][2], ` `int` `V, ` ` ` `int` `marked[], ` `int` `N, ` `int` `K) ` `{ ` ` ` `// vertices in a tree are one more than number of edges ` ` ` `V = V + 1; ` ` ` `vector<` `int` `> g[V]; ` ` ` ` ` `// fill vector for graph ` ` ` `int` `u, v; ` ` ` `for` `(` `int` `i = 0; i < (V - 1); i++) ` ` ` `{ ` ` ` `u = edges[i][0]; ` ` ` `v = edges[i][1]; ` ` ` ` ` `g[u].push_back(v); ` ` ` `g[v].push_back(u); ` ` ` `} ` ` ` ` ` `// fill boolean array mark from marked array ` ` ` `bool` `mark[V] = {` `false` `}; ` ` ` `for` `(` `int` `i = 0; i < N; i++) ` ` ` `mark[marked[i]] = ` `true` `; ` ` ` ` ` `// vectors to store distances ` ` ` `vector<` `int` `> tmp(V, -1), dl(V, -1), dr(V, -1); ` ` ` ` ` `// first bfs(from any random node) to get one ` ` ` `// distant marked node ` ` ` `u = bfsWithDistance(g, mark, 0, tmp); ` ` ` ` ` `/* second bfs to get other distant marked node ` ` ` `and also dl is filled with distances from first ` ` ` `chosen marked node */` ` ` `u = bfsWithDistance(g, mark, u, dl); ` ` ` ` ` `// third bfs to fill dr by distances from second ` ` ` `// chosen marked node ` ` ` `bfsWithDistance(g, mark, u, dr); ` ` ` ` ` `int` `res = 0; ` ` ` `// loop over all nodes ` ` ` `for` `(` `int` `i = 0; i < V; i++) ` ` ` `{ ` ` ` `// increase res by 1, if current node has distance ` ` ` `// less than K from both extreme nodes ` ` ` `if` `(dl[i] <= K && dr[i] <= K) ` ` ` `res++; ` ` ` `} ` ` ` `return` `res; ` `} ` ` ` `// Driver code to test above methods ` `int` `main() ` `{ ` ` ` `int` `edges[][2] = ` ` ` `{ ` ` ` `{1, 0}, {0, 3}, {0, 8}, {2, 3}, ` ` ` `{3, 5}, {3, 6}, {3, 7}, {4, 5}, ` ` ` `{5, 9} ` ` ` `}; ` ` ` `int` `V = ` `sizeof` `(edges) / ` `sizeof` `(edges[0]); ` ` ` ` ` `int` `marked[] = {1, 2, 4}; ` ` ` `int` `N = ` `sizeof` `(marked) / ` `sizeof` `(marked[0]); ` ` ` ` ` `int` `K = 3; ` ` ` `cout << nodesKDistanceFromMarked(edges, V, marked, N, K); ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

## Python3

# Python3 program to count nodes inside

# K distance range from marked nodes

import queue

# Utility bfs method to fill distance

# vector and returns most distant

# marked node from node u

def bfsWithDistance(g, mark, u, dis):

lastMarked = 0

q = queue.Queue()

# push node u in queue and initialize

# its distance as 0

q.put(u)

dis[u] = 0

# loop untill all nodes are processed

while (not q.empty()):

u = q.get()

# if node is marked, update

# lastMarked variable

if (mark[u]):

lastMarked = u

# loop over all neighbors of u and

# update their distance before

# pushing in queue

for i in range(len(g[u])):

v = g[u][i]

# if not given value already

if (dis[v] == -1):

dis[v] = dis[u] + 1

q.put(v)

# return last updated marked value

return lastMarked

# method returns count of nodes which

# are in K-distance range from marked nodes

def nodesKDistanceFromMarked(edges, V, marked, N, K):

# vertices in a tree are one

# more than number of edges

V = V + 1

g = [[] for i in range(V)]

# fill vector for graph

u, v = 0, 0

for i in range(V – 1):

u = edges[i][0]

v = edges[i][1]

g[u].append(v)

g[v].append(u)

# fill boolean array mark from

# marked array

mark = [False] * V

for i in range(N):

mark[marked[i]] = True

# vectors to store distances

tmp = [-1] * V

dl = [-1] * V

dr = [-1] * V

# first bfs(from any random node)

# to get one distant marked node

u = bfsWithDistance(g, mark, 0, tmp)

# second bfs to get other distant

# marked node and also dl is filled

# with distances from first chosen

# marked node

u = bfsWithDistance(g, mark, u, dl)

# third bfs to fill dr by distances

# from second chosen marked node

bfsWithDistance(g, mark, u, dr)

res = 0

# loop over all nodes

for i in range(V):

# increase res by 1, if current node

# has distance less than K from both

# extreme nodes

if (dl[i] <= K and dr[i] <= K):
res += 1
return res
# Driver Code
if __name__ == '__main__':
edges = [[1, 0], [0, 3], [0, 8],
[2, 3], [3, 5], [3, 6],
[3, 7], [4, 5], [5, 9]]
V = len(edges)
marked = [1, 2, 4]
N = len(marked)
K = 3
print(nodesKDistanceFromMarked(edges, V,
marked, N, K))
# This code is contributed by PranchalK
[tabbyending]
**Output:**

6

This article is contributed by **Utkarsh Trivedi**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

## Recommended Posts:

- Count the nodes of the tree which make a pangram when concatenated with the sub-tree nodes
- Count the nodes in the given tree whose weight is even
- Count the number of non-reachable nodes
- Count the number of nodes at a given level in a tree using DFS
- Count the number of nodes at given level in a tree using BFS.
- Count the nodes in the given tree whose weight is even parity
- Count the nodes in the given tree whose sum of digits of weight is odd
- Count the nodes in the given tree whose weight is prime
- Count the nodes in the given tree whose weight is a power of two
- Count the nodes of the given tree whose weight has X as a factor
- Count the nodes whose weight is a perfect square
- Count the nodes of the given tree whose weighted string is a palindrome
- Count the nodes of a tree whose weighted string does not contain any duplicate characters
- Count the nodes of a tree whose weighted string is an anagram of the given string
- XOR of all the nodes in the sub-tree of the given node