Hello coders, today we are going to solve **Dynamic Array HackerRank Solution 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

*shelf.*

**xth***2 x y*: Print the number of pages in the

*book on the*

**yth***shelf.*

**xth***3 x*: Print the number of books on the

**shelf.**

*xth*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

*, the number of requests.*

**total_number_of_queries**Each of the following

*lines contains a request in one of the three specified formats.*

**total_number_of_queries****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
shelf at*xth*index.**yth** **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

**2nd**shelf is 78.

– 5 The number of books on the

**shelf is 2.**

*0th***Solution – Dynamic Array in C**

**C**

#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); int total_number_of_queries; scanf("%d", &total_number_of_queries); total_number_of_books = (int *)malloc(sizeof(int)*total_number_of_shelves); total_number_of_pages = (int **)malloc(sizeof(int *)*total_number_of_shelves); for(int i = 0; i<total_number_of_shelves; i++){ *(total_number_of_books + i) = 0; } while(total_number_of_queries--){ int type_of_query; scanf("%d", &type_of_query); if(type_of_query == 1){ int x, y; scanf("%d %d", &x, &y); int booksInShelf = *(total_number_of_books + x); *(total_number_of_pages + x) = (int*)realloc(*(total_number_of_pages+x),sizeof(int)*(booksInShelf+1)); *(*(total_number_of_pages+x)+booksInShelf) = y; *(total_number_of_books + x) += 1; } 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 **Hacker Rank** but the Solution is Provided by **CodingBroz**. This tutorial is only for **Educational** and **Learning** Purpose.