Post Transition in C | HackerRank Solution

Hello coders, today we are going to solve Post Transition HackerRank Solution in C.

Post Transition


We live in a big country. This country has towns_count towns in it. Each town has some post offices in which packages are stored and transferred.

Post offices have different inner structure. Specifically, each of them has some limitations on the packages it can store – their weight should be between min_weight and max_weight inclusively, where min_weight and max_weight are fixed for each office.

Packages are stored in some order in the office queue. That means, that they are processed using this order when sending and receiving.

Sometimes two post offices, even in different towns, may organize the following transaction: the first one sends all its packages to the second one. The second one accepts the packages that satisfy the weight condition for the second office and rejects all other ones. These rejected packages return to the first office back and are stored in the same order they were stored before they were sent. The accepted packages move to the tail of the second office’s queue in the same order they were stored in the first office.

You should process several queries in your program. You’ll be provided with structures packagepost_office and town. in order to complete this task, you should fill the following functions:

 print_all_packages – given the town t, print all packages in this town. They should be printed as follows:


where 01 etc are the numbers of post offices and id0id1… are the ids of packages from the 0th post office in the order of its queue,id2id3 are from the 1st one etc. There should be one ‘\t’ symbol before post office numbers and two ‘\t’ symbols before the ids.

send_all_acceptable_packages – given the towns source and target and post office indices source_office_index and target_office_index, manage the transaction described above between the post office #source_office_index in town  and the post office #target_office_index in town target.

town_with_most_packages – given all towns, find the one with the most number of packages in all post offices altogether. If there are several of them, find the first one from the collection towns.

 find_town – given all towns and a string name, find the town with the name name. It’s guaranteed that the town exists.

Input Format

First line of the input contains a single integer towns_counttowns_count blocks follow, each describing a town.

Every town block contains several lines. On the first line there is a string  town_name – the name of the town. On the second line there is an integer offices_count – the number of the offices in the town. offices_count blocks follow then, each describing an office.

Every office block also contains several lines. On the first line there are three integers separated by single spaces: packages_count (the number of packages in the office), min_weight and max_weight (described above). package_count blocks follow, each describing a package.

Every package block contains exactly two lines. On the first line there is a string id which is an id of the package. On the second line there is an integer weight which is the weight of the package.

Then, there is a single integer queries on the line which is the number of queries. queries blocks follow, each describing a query.

Every query block contains several lines. On the first line there is an integer 12 or 3. If this integer is 1, on the second line there is a string town_name – the name of town for which all packages should be printed. If this integer is 2, on the second line there are string source_name, integer source_office_index, string target_name and integer target_office_index separated by single spaces. That means transactions between post office #source_office_index in the town source_name and post office #target_office_index in the town target_name should be processed.

If the integer is 3, no lines follow and the town with the most number of packages should be found.


  • All integer are between 0 and 10
  • towns_count > 0, offices_count > 0 .
  • All strings have length <= 5
  • All towns’ names have only uppercase english letters and are unique.
  • All packages’ ids have only lowercase english letters and are unique.
  • For each post office, min_weight <= max_weight.
  • All queries are valid, that means, towns with the given names always exist, post offices with the given indices always exist.

Output Format

For queries of type 1, print all packages in the format provided in the problem statement. For queries of type 3, print “Town with the most number of packages is townname” on a separate line.

Sample Input 0

2 1 3
a 2
b 3
1 2 4
c 2
4 1 4
d 1
e 2
f 3
h 4
2 B 0 A 1
1 A
1 B

Sample Output 0

Town with the most number of packages is B
Town with the most number of packages is A

Explanation 0

Before all queries, town B has 4 packages in total, town A has 3. But after transaction all packages from B’s 0th post office go to the 1st post office of A, except package d because it’s too light.

Solution – Post Transition in C


#include <stdio.h>
#include <stdlib.h>

struct package
   char* id;
   int weight;

typedef struct package package;

struct post_office
   int min_weight;
   int max_weight;
   package* packages;
   int packages_count;

typedef struct post_office post_office;

struct town
   char* name;
   post_office* offices;
   int offices_count;

typedef struct town town;

void print_all_packages(town t)
    for (int i = 0; i < t.offices_count; i++)
        printf("\t%d:\n", i);
        for (int j = 0; j < t.offices[i].packages_count; j++)
            printf("\t\t%s\n", t.offices[i].packages[j].id);

void send_all_acceptable_packages(town* source, int source_office_index, town* target, int target_office_index)
    int n = 0;
    for (int i = 0; i < source->offices[source_office_index].packages_count; i++)
        if (source->offices[source_office_index].packages[i].weight >= target->offices[target_office_index].min_weight &&
            source->offices[source_office_index].packages[i].weight <= target->offices[target_office_index].max_weight)
    package* newPackages = malloc(sizeof(package)*(n + target->offices[target_office_index].packages_count));
    package* oldPackages = malloc(sizeof(package)*(source->offices[source_office_index].packages_count - n));
    for (int i = 0; i < target->offices[target_office_index].packages_count; i++)
        newPackages[i] = target->offices[target_office_index].packages[i];
    n = target->offices[target_office_index].packages_count;
    int m = 0;
    for (int i = 0; i < source->offices[source_office_index].packages_count; i++)
        if (source->offices[source_office_index].packages[i].weight >= target->offices[target_office_index].min_weight &&
            source->offices[source_office_index].packages[i].weight <= target->offices[target_office_index].max_weight)
            newPackages[n] = source->offices[source_office_index].packages[i];
            oldPackages[m] = source->offices[source_office_index].packages[i];
    target->offices[target_office_index].packages_count = n;
    target->offices[target_office_index].packages = newPackages;
    source->offices[source_office_index].packages_count = m;
    source->offices[source_office_index].packages = oldPackages;

int number_of_packages(town t)
    int ans = 0;
    for (int i = 0; i < t.offices_count; i++)
        ans += t.offices[i].packages_count;
    return ans;

town town_with_most_packages(town* towns, int towns_count)
    int ans;
    int max_packages = -1;
    for (int i = 0; i < towns_count; i++)
        if (number_of_packages(towns[i]) > max_packages)
            max_packages = number_of_packages(towns[i]);
            ans = i;
    return towns[ans];

town* find_town(town* towns, int towns_count, char* name)
    for (int i = 0; i < towns_count; i++)
        if (!strcmp(towns[i].name, name))
            return &(towns[i]);
    return &towns[0];

int main()
   int towns_count;
   scanf("%d", &towns_count);
   town* towns = malloc(sizeof(town)*towns_count);
   for (int i = 0; i < towns_count; i++) {
      towns[i].name = malloc(sizeof(char) * MAX_STRING_LENGTH);
      scanf("%s", towns[i].name);
      scanf("%d", &towns[i].offices_count);
      towns[i].offices = malloc(sizeof(post_office)*towns[i].offices_count);
      for (int j = 0; j < towns[i].offices_count; j++) {
         scanf("%d%d%d", &towns[i].offices[j].packages_count, &towns[i].offices[j].min_weight, &towns[i].offices[j].max_weight);
         towns[i].offices[j].packages = malloc(sizeof(package)*towns[i].offices[j].packages_count);
         for (int k = 0; k < towns[i].offices[j].packages_count; k++) {
            towns[i].offices[j].packages[k].id = malloc(sizeof(char) * MAX_STRING_LENGTH);
            scanf("%s", towns[i].offices[j].packages[k].id);
            scanf("%d", &towns[i].offices[j].packages[k].weight);
   int queries;
   scanf("%d", &queries);
   char town_name[MAX_STRING_LENGTH];
   while (queries--) {
      int type;
      scanf("%d", &type);
      switch (type) {
      case 1:
         scanf("%s", town_name);
         town* t = find_town(towns, towns_count, town_name);
      case 2:
         scanf("%s", town_name);
         town* source = find_town(towns, towns_count, town_name);
         int source_index;
         scanf("%d", &source_index);
         scanf("%s", town_name);
         town* target = find_town(towns, towns_count, town_name);
         int target_index;
         scanf("%d", &target_index);
         send_all_acceptable_packages(source, source_index, target, target_index);
      case 3:
         printf("Town with the most number of packages is %s\n", town_with_most_packages(towns, towns_count).name);
   return 0;

Disclaimer: The above Problem (Post Transition) is generated by Hacker Rank but the Solution is Provided by CodingBroz. This tutorial is only for Educational and Learning Purpose.

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