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**HackerEarth Directory Deletion problem solution,**You are given a directory tree of N directories/folders. Each directory is represented by a particular id which ranges from 1 to N. The id of the root directory is 1, then it has some child directories, those directories may contain some other ones and it goes on. Now you are given a list of directories id's to delete, you need to find the minimum number of directories that need to be deleted so that all the directories in the given list get deleted.## HackerEarth Directory Deletion problem solution.

`#include<bits/stdc++.h>`

#define LL long long int

#define M 1000000007

#define endl "\n"

#define eps 0.00000001

LL pow(LL a,LL b,LL m){ a%=m;LL x=1,y=a;while(b > 0){if(b%2 == 1){x=(x*y);if(x>m) x%=m;}y = (y*y);if(y>m) y%=m;b /= 2;}return x%m;}

LL gcd(LL a,LL b){if(b==0) return a; else return gcd(b,a%b);}

LL gen(LL start,LL end){LL diff = end-start;LL temp = rand()%start;return temp+diff;}

using namespace std;

vector<int> graph[100001];

bool to_delete[100001];

int p[100001];

int ans = 0;

void dfs(int node) {

if(to_delete[node] == 1) {

++ans;

return;

}

for(int i: graph[node]) {

dfs(i);

}

}

int main() {

ios_base::sync_with_stdio(0);

cin.tie(0);

int n;

assert(cin >> n);

assert(n >= 1 && n <= 100000);

for(int i = 1; i <= n; i++) {

assert(cin >> p[i]);

assert(p[i] <= n);

if(p[i] != -1)

graph[p[i]].push_back(i);

}

int k;

assert(cin >> k);

for(int i = 1; i <= k; i++) {

int val;

assert(cin >> val);

assert(val <= n && val >= 1);

to_delete[val] = 1;

}

dfs(1);

cout << ans;

}

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