Skip to content
Programmingoneonone
Programmingoneonone

Learn everything about programming

  • Home
  • CS Subjects
    • Internet of Things (IoT)
    • Digital Communication
    • Human Values
    • Cybersecurity
  • Programming Tutorials
    • C Programming
    • Data structures and Algorithms
    • 100+ Java Programs
    • 100+ C Programs
  • HackerRank Solutions
    • HackerRank Algorithms Solutions
    • HackerRank C problems solutions
    • HackerRank C++ problems solutions
    • HackerRank Java problems solutions
    • HackerRank Python problems solutions
Programmingoneonone
Programmingoneonone

Learn everything about programming

HackerRank Jeanie’s Route problem solution

YASH PAL, 31 July 202430 November 2025

In this HackerRank Jeanie’s Route problem solution Byteland has N cities (numbered from 1 to N) and N – 1 bidirectional road. It is guaranteed that there is a route from any city to any other city.

Jeanie is a postal worker who must deliver K letters to various cities in Byteland. She can start and end her delivery route in any city. Given the destination cities for K letters and the definition of each road in Byteland, find and print the minimum distance Jeanie must travel to deliver all K letters.

HackerRank Jeanie's Route problem solution

Problem solution in Python.

def farthest_node_from_start(nodes,is_goal,subtree,starting_node):
    visited=[False for _ in nodes]
    st=[(0,starting_node)]
    visited[starting_node]=True
    largest = 0,starting_node
    all_dist_sum = 0
    while st:
        dist,u=st.pop()
        for v,d in nodes[u]:
            if not visited[v] and subtree[v]:
                elm=dist+d,v
                all_dist_sum+=d
                if is_goal[v]:
                    largest=max(largest,elm)
                st.append(elm)
                visited[v]=True
    return all_dist_sum,largest

def make_subtree(nodes,is_goal,starting_node):
    subtree=is_goal[:]
    for u,v in dfs(nodes, starting_node):
        subtree[u]=subtree[u] or subtree[v]
    return subtree

def dfs(nodes, starting_node):
    stack = [(starting_node, v) for v,d in nodes[starting_node]]
    edges = list()
    while stack:
        u, v = stack.pop()
        edges.append((u, v))
        stack.extend((v, w) for w,d in nodes[v] if w != u)
    edges.reverse()
    return edges

N, K = list(map(int,input().strip().split()))
goals = list(map(int,input().strip().split()))
is_goal=[False]*(N+1)

for item in goals:
    is_goal[item]=True
    
nodes=[[] for _ in range(N+1)]

for _ in range(N-1):
    u,v,d = list(map(int,input().strip().split()))
    nodes[u].append((v,d))
    nodes[v].append((u,d))

subtree = make_subtree(nodes,is_goal,goals[0])
a,(b,c)=farthest_node_from_start(nodes,is_goal,subtree,goals[0])
Distance,(d,_)=farthest_node_from_start(nodes,is_goal,subtree,c)
print(2*Distance-d)

{“mode”:”full”,”isActive”:false}

Problem solution in Java.

import java.util.Scanner;
import java.util.Queue;
import java.util.ArrayDeque;

class Solution{
    static boolean[] prune(int[][][] adj, boolean[] isLett){
    int n=adj.length;
    int[] degree=new int[n];
    for(int i=0;i<n;++i) degree[i]=adj[i].length;
    boolean[] rem=new boolean[n];
    Queue<Integer> q=new ArrayDeque<>();
    for(int i=0;i<n;++i){
        if(!isLett[i] && degree[i]==1) q.add(i);
    }
    while(!q.isEmpty()){
        int leaf=q.remove();
        rem[leaf]=true;
        for(int[] edge: adj[leaf]){
        int node=edge[0];
        if(isLett[node]) break;
        else if(!rem[node]){
            if(--degree[node] == 1){
            q.add(node);
            break;
            }
        }
        }
    }
    return rem;
    }
    static int[] bfs(int[][][] adj, boolean[] rem, int source){
    int n=adj.length, unvis=-1;
    int[] dist=new int[n];
    for(int i=0;i<n;++i) dist[i]=unvis;
    Queue<Integer> q=new ArrayDeque<>();
    q.add(source);
    dist[source]=0;
    int best=0, total=0;
    while(!q.isEmpty()){
        int x=q.remove();
        for(int[] edge: adj[x]){
        int to=edge[0];
        if(!rem[to] && dist[to]==unvis){
            int weight=edge[1];
            total+=weight;
            q.add(to);
            dist[to]=dist[x]+weight;
            if(dist[to]>dist[best]) best=to;
        }
        }
    }
    int[] result={total,dist[best],best};
    return result;
    }
    static int solve(int[][][] adj, int[] lett){
    boolean[] isLett=new boolean[adj.length];
    for(int i: lett) isLett[i]=true;    
    boolean[] rem=prune(adj,isLett);
    int[] result=bfs(adj,rem,lett[0]);
    int totalWeight=result[0], sink=result[2];
    result=bfs(adj,rem,sink);
    int diameter=result[1];
    return 2*totalWeight-diameter;
    }
    static int[][][] weightedAdjacency(int n, int[] from, int[] to, int[] d){
    int[] count=new int[n];
    for(int f: from) ++count[f];
    for(int t: to) ++count[t];
    int[][][] adj=new int[n][][];
    for(int i=0;i<n;++i) adj[i]=new int[count[i]][];
    for(int i=0;i<from.length;++i){
        adj[from[i]][--count[from[i]]]=new int[]{to[i],d[i]};
        adj[to[i]][--count[to[i]]]=new int[]{from[i],d[i]};
    }
    return adj;
    }
    public static void main(String[] args){
    Scanner sc=new Scanner(System.in);
    int n=sc.nextInt(), k=sc.nextInt();
    int[] lett=new int[k];
    for(int i=0;i<k;++i) lett[i]=sc.nextInt()-1;
    int[] from=new int[n-1], to=new int[n-1], d=new int[n-1];
    for(int i=0;i<n-1;++i){
        from[i]=sc.nextInt()-1;
        to[i]=sc.nextInt()-1;
        d[i]=sc.nextInt();
    }
    sc.close();
    int[][][] adj=weightedAdjacency(n,from,to,d);
    System.out.println(solve(adj,lett));
    }
}

{“mode”:”full”,”isActive”:false}

Problem solution in C++.

#include <bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define MA(x,y) ((x)>(y)?(x):(y))
using namespace std;
const int N=500002;
const int inf=1000000000;
int n,K,s[N],SUM,M;
bool f[N];
vector <int> v[N],d[N];

int input(){
    scanf("%d%d",&n,&K);
    for (int i=0,x;i<K;i++){
        scanf("%d",&x);
        s[x]=1;
        f[x]=1;
    }
    for (int i=1,x,y,z;i<n;i++){
        scanf("%d%d%d",&x,&y,&z);
        v[x].pb(y);
        v[y].pb(x);
        d[x].pb(z);
        d[y].pb(z);
    }
    return 0;
}

int go(int x,int from){
    int D1=-inf;
    int D2=-inf;

    for (int i=0;i<v[x].size();i++)
    if (v[x][i]!=from){
        D1=max(D1,go(v[x][i],x)+d[x][i]);
        if (D1>D2) swap(D1,D2);
        s[x]+=s[v[x][i]];
        if (0<s[v[x][i]] && s[v[x][i]]<K) SUM+=d[x][i];
    }

    if (D1>0) M=max(M,D1+D2);

    if (D2>0 && f[x]) M=max(M,D2);

    if (f[x]) D2=max(D2,0);

    return D2;
}

int sol(){
    go(1,1);
    printf("%dn",SUM*2-M);
    return 0;
}

int main() {
    input();
    sol();
    return 0;
}

{“mode”:”full”,”isActive”:false}

Problem solution in C.

#include <assert.h>
#include <limits.h>
#include <math.h>
#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
typedef struct _node{
  int x;
  int w;
  struct _node *next;
} lnode;
void insert_edge(int x,int y,int w);
void dfs1(int x);
void dfs2(int x,int y,int z);
int max(int x,int y);
int mark[100000]={0},count[100000],trace[100000],l[100000],ul[100000],s[100000],sc[100000],len[100000],ulen[100000],ans=2147483647;
lnode *table[100000]={0};

int main(){
  int N,K,x,y,w,i;
  scanf("%d%d",&N,&K);
  while(K--){
    scanf("%d",&x);
    mark[x-1]=1;
  }
  for(i=0;i<N-1;i++){
    scanf("%d%d%d",&x,&y,&w);
    insert_edge(x-1,y-1,w);
  }
  memset(trace,0,sizeof(trace));
  dfs1(0);
  memset(trace,0,sizeof(trace));
  dfs2(0,-1,-1);
  printf("%d",ans);
  return 0;
}
void insert_edge(int x,int y,int w){
  lnode *t=malloc(sizeof(lnode));
  t->x=y;
  t->w=w;
  t->next=table[x];
  table[x]=t;
  t=malloc(sizeof(lnode));
  t->x=x;
  t->w=w;
  t->next=table[y];
  table[y]=t;
  return;
}
void dfs1(int x){
  lnode *p;
  trace[x]=1;
  if(mark[x])
    count[x]=1;
  else
    count[x]=0;
  l[x]=s[x]=sc[x]=len[x]=0;
  for(p=table[x];p;p=p->next)
    if(!trace[p->x]){
      dfs1(p->x);
      count[x]+=count[p->x];
      if(count[p->x]){
        sc[x]++;
        len[x]+=len[p->x]+2*p->w;
        if(l[p->x]+p->w>l[x]){
          s[x]=l[x];
          l[x]=l[p->x]+p->w;
        }
        else if(l[p->x]+p->w>s[x])
          s[x]=l[p->x]+p->w;
      }
    }
  return;
}
void dfs2(int x,int y,int z){
  lnode *p;
  int rl,rs;
  trace[x]=1;
  if(y==-1)
    ulen[x]=ul[x]=0;
  else{
    if(count[0]-count[x]){
      ulen[x]=len[y]+ulen[y]-len[x];
      if(!count[x])
        ulen[x]+=2*z;
      if(l[y]==z+l[x])
        ul[x]=z+max(s[y],ul[y]);
      else
        ul[x]=z+max(l[y],ul[y]);
    }
    else
      ulen[x]=ul[x]=0;
  }
    rl=l[x];
    rs=s[x];
    if(ul[x]>rl){
      rs=rl;
      rl=ul[x];
    }
    else if(ul[x]>rs)
      rs=ul[x];
    if(len[x]+ulen[x]-rl-rs<ans)
      ans=len[x]+ulen[x]-rl-rs;
  for(p=table[x];p;p=p->next)
    if(!trace[p->x])
      dfs2(p->x,x,p->w);
  return;
}
int max(int x,int y){
  return (x>y)?x:y;
}

{“mode”:”full”,”isActive”:false}

Algorithms coding problems solutions AlgorithmsHackerRank

Post navigation

Previous post
Next post

Are you a student and stuck with your career or worried about real-time things, and don't know how to manage your learning phase? Which profession to choose? and how to learn new things according to your goal, and land a dream job. Then this might help to you.

Hi My name is YASH PAL, founder of this Blog and a Senior Software engineer with 5+ years of Industry experience. I personally helped 40+ students to make a clear goal in their professional lives. Just book a one-on-one personal call with me for 30 minutes for 300 Rupees. Ask all your doubts and questions related to your career to set a clear roadmap for your professional life.

Book session - https://wa.me/qr/JQ2LAS7AASE2M1

Pages

  • About US
  • Contact US
  • Privacy Policy

Follow US

  • YouTube
  • LinkedIn
  • Facebook
  • Pinterest
  • Instagram
©2026 Programmingoneonone | WordPress Theme by SuperbThemes