# HackerEarth Finding pairs problem solution

In this HackerEarth Finding pairs problem solution You are given a rooted tree with N nodes and node 1 as a root. There is a unique path between any two nodes. Here, d(i,j) is defined as a number of edges in a unique path between nodes i and j.

You have to find the number of pairs (i,j) such that and d(i,j) = d(i,1) - d(j,1).

## HackerEarth Finding pairs problem solution.

`#include<bits/stdc++.h>using namespace std;#define FAST ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);#include <ext/pb_ds/assoc_container.hpp>#include <ext/pb_ds/tree_policy.hpp>using namespace __gnu_pbds;#define ordered_set tree<int, null_type,lessl<int>, rb_tree_tag,tree_order_statistics_node_update>#define ll long long#define int long long#define double long double#define all(a)      (a).begin(),(a).end()#define sz(x)       (int)x.size()#define ff first#define ss second#define mp make_pair#define pb push_back#define endl "\n"#define f(i,l,r) for(int i=l;i<=r;i++)#define rf(i,r,l) for(int i=r;i>=l;i--)#define bp __builtin_popcountll#define inf 1e18const int N=1e5+5;const int M=1e9+7;vector<int> v[N];vector<int> lvl(N,0);void dfs(int x,int p){    for(auto X:v[x])    {        if(X==p) continue;        lvl[X]=lvl[x]+1;        dfs(X,x);    }}void solve(){    int n;    cin>>n;    for(int i=1;i<n;i++)    {        int t1,t2;        cin>>t1>>t2;        v[t1].pb(t2);        v[t2].pb(t1);    }    lvl[1]=1;    dfs(1,0);    int ans=0;    for(int i=1;i<=n;i++)        ans+=lvl[i];    cout<<ans<<endl;}signed main(){    FAST        int t=1;    // cin>>t;    for(int tc=1;tc<=t;tc++)    {        // cout<<"Case #"<<tc<<": ";        solve();    }}`

### Second solution

`#include<bits/stdc++.h>using namespace std;#define FAST ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);#include <ext/pb_ds/assoc_container.hpp>#include <ext/pb_ds/tree_policy.hpp>using namespace __gnu_pbds;#define ordered_set tree<int, null_type,lessl<int>, rb_tree_tag,tree_order_statistics_node_update>#define ll long long#define int long long#define double long double#define all(a)      (a).begin(),(a).end()#define sz(x)       (int)x.size()#define ff first#define ss second#define mp make_pair#define pb push_back#define endl "\n"#define f(i,l,r) for(int i=l;i<=r;i++)#define rf(i,r,l) for(int i=r;i>=l;i--)#define bp __builtin_popcountll#define inf 1e18const int N=1e5+5;const int M=1e9+7;vector<int> v[N];vector<int> lvl(N,0);void dfs(int x,int p){    for(auto X:v[x])    {        if(X==p) continue;        lvl[X]=lvl[x]+1;        dfs(X,x);    }}void solve(){    int n;    cin>>n;    for(int i=1;i<n;i++)    {        int t1,t2;        cin>>t1>>t2;        v[t1].pb(t2);        v[t2].pb(t1);    }    lvl[1]=1;    dfs(1,0);    int ans=0;    for(int i=1;i<=n;i++)        ans+=lvl[i];    cout<<ans<<endl;}signed main(){    FAST        int t=1;    // cin>>t;    for(int tc=1;tc<=t;tc++)    {        // cout<<"Case #"<<tc<<": ";        solve();    }}`