Hello Programmers, In this post, you will know how to solve the Dynamic Array in C HackerRank Solution. This problem is a part of the HackerRank C Programming Series.
One more thing to add, don’t directly look for the solutions, first try to solve the problems of Hackerrank by yourself. If you find any difficulty after trying several times, then you can look for solutions.
Dynamic Array in C
Problem
Snow Howler is the librarian at the central library of the city of HuskyLand. He must handle requests which come in the following forms:
1 x y : Insert a book with y pages at the end of the xth shelf.
2 x y : Print the number of pages in the yth book on the xth shelf.
3 x : Print the number of books on the xth shelf.
Snow Howler has got an assistant, Oshie, provided by the Department of Education. Although inexperienced, Oshie can handle all of the queries of types 2 and 3.
Help Snow Howler deal with all the queries of type 1.
Oshie has used two arrays:
int* total_number_of_books; /* * This stores the total number of books on each shelf. */ int** total_number_of_pages; /* * This stores the total number of pages in each book of each shelf. * The rows represent the shelves and the columns represent the books. */
Input Format :
The first line contains an integer total_number_of_shelves, the number of shelves in the library. The second line contains an integer total_number_of_queries, the number of requests.
Each of the following total_number_of_queries lines contains a request in one of the three specified formats.
Constraints :
- 1<= total_number_of_shelves <= 10^5
- 1<= total_number_of_queries <= 10^5
- For each query of the second type, it is guaranteed that a book is present on the xth shelf at yth index.
- 0 <= x < total_number_of_shelves
- Both the shelves and the books are numbered starting from 0.
- Maximum number of books per shelf <= 1100
Output Format :
Write the logic for the requests of type 1. The logic for requests of types 2 and 3 are provided.
Sample Input 0 :
5 5 1 0 15 1 0 20 1 2 78 2 2 0 3 0
Sample Output 0 :
78 2
Explanation 0 :
There are 5 shelves and 5 requests, or queries.
– 1 Place a 15 page book at the end of shelf 0.
– 2 Place a 20 page book at the end of shelf 0.
– 3 Place a 78 page book at the end of shelf 2.
– 4 The number of pages in the 0th book on the 2th shelf is 78.
– 5 The number of books on the 0th shelf is 2.
Dynamic Array in C HackerRank Solution
#include <stdio.h>
#include <stdlib.h>
/*
* This stores the total number of books in each shelf.
*/
int* total_number_of_books;
/*
* This stores the total number of pages in each book of each shelf.
* The rows represent the shelves and the columns represent the books.
*/
int** total_number_of_pages;
int main()
{
int total_number_of_shelves;
scanf("%d", &total_number_of_shelves);
total_number_of_books = calloc(total_number_of_shelves, sizeof(int));
int total_number_of_queries;
scanf("%d", &total_number_of_queries);
total_number_of_pages = malloc(total_number_of_shelves * sizeof(int *));
for (int i = 0; i < total_number_of_shelves; i++)
{
total_number_of_pages[i] = calloc(1100, sizeof(int));
}
while (total_number_of_queries--)
{
int type_of_query;
scanf("%d", &type_of_query);
if (type_of_query == 1)
{
/*
* Process the query of first type here.
*/
int shelf, pages;
scanf("%d %d", &shelf, &pages);
total_number_of_books[shelf]++;
int *book = total_number_of_pages[shelf];
while (*book != 0)
book++;
*book = pages;
}
else if (type_of_query == 2)
{
int x, y;
scanf("%d %d", &x, &y);
printf("%d\n", *(*(total_number_of_pages + x) + y));
}
else
{
int x;
scanf("%d", &x);
printf("%d\n", *(total_number_of_books + x));
}
}
if (total_number_of_books)
{
free(total_number_of_books);
}
for (int i = 0; i < total_number_of_shelves; i++)
{
if (*(total_number_of_pages + i))
{
free(*(total_number_of_pages + i));
}
}
if (total_number_of_pages)
{
free(total_number_of_pages);
}
return 0;
}Disclaimer: The above Problem (Dynamic Array in C) is generated by Hackerrank but the Solution is Provided by BrokenProgrammers. This tutorial is only for Educational and Learning purposes.
