Post transition - Hackerrank Solution in C language

Post transition - Hackerrank Solution in C language

Welcome Guys!!
In this post we will see the Post Transition HackerRank solution in C.
It's one of the hard problems for subtopic Struct and Enums.

The Link to the Problem statement of Post Transition is as follows :

Post Transition - Hacker Rank Solution:

For print_all_packages you should simply traverse every town, office and package in the correct order.
For send_all_acceptable_packages first you should determine the number of packages in two offices after transaction. After you determined it, create two dynamic arrays for the packages for sender office and for recipient one. Fill it with appropriate packages and assign to the field packages of responsive offices. Don't forget to free the old ones!
For town_with_most_packages you should create a helper function number_of_packages, that returns the number of packages in the town. After doing this, simply go through all towns and compare the number of packages in each.

#include <stdio.h>
#include <stdlib.h>
#define MAX_STRING_LENGTH 6

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)
{
    printf("%s:\n", t.name);
    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)
            ++n;
    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];
            ++n;
        }
        else
        {
            oldPackages[m] = source->offices[source_office_index].packages[i];
            ++m;
        }
    target->offices[target_office_index].packages_count = n;
    free(target->offices[target_office_index].packages);
    target->offices[target_office_index].packages = newPackages;
    source->offices[source_office_index].packages_count = m;
    free(source->offices[source_office_index].packages);
    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);
         print_all_packages(*t);
         break;
      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);
         break;
      case 3:
         printf("Town with the most number of packages is %s\n", town_with_most_packages(towns, towns_count).name);
         break;
      }
   }
   return 0;
}
Please read and go through the code and try to analysis the code .
IF you have a doubt feel free to share it in the comment section.
See you in the next post ;)

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