Question

class HashMap private: HashEntry table ; public: HashMap() table-new HashEntry * [TABLE_SIZE]; for (int 1-0; 1< TABLE SIZE: i++) table [i] = NULL; Hash Function int HashFunc(int key) return key % TABLE-SIZE; Insert Element at a key void Insert(int key, int value) int hashHashFunc(key); while (table [hash] != NULL && table [hash]->key key ) != hash-HashFunc(hash 1); if (table [hash] != NULL) delete table [hash]; table [hash] - new HashEntry(key, value);
Identify the letters and anything associated with it. It is a hash map so please feel free to point out the other important parts beside the arrows. I apologize for the order of the pictures

Answer 1

a. Here, you are creating a 2D array of HashEntry. Because you want to access these elements again we are doing this dynamically and follow up the updates done to Hash Map. It will be in the following format:

table[0] --------> table[0][0] table[0][1]........table[0][TABLE_SIZE-1]

table[1]---------> table[1][0].......................table[1][TABLE_SIZE-1]

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table[TABLE_SIZE-1]---------->table[TABLE_SIZE-1][0]............table[TABLE_SIZE-1][TABLE_SIZE-1]

b. Here, in hashing we try to allocate the elements using modulo function. Since the size of the table is 128, we try to obtain the value at value%128. Hence the %TABLE_SIZE(given TABLE_SIZE=128).

c. The if condition above will check if for element presence at the table. If already an element is present and that element is not equal to the newly inserted element then we do the looping until you find an empty space along the table[hash] series and add the element once found an empty space. If the pointer is NULL then it means no element was added under this series. So create a new space and add the element.

1. Here, we assign the key and values obtained from the parameters to the variables created in the class. Hence, we use 'this' to differentiate between the class variables from the parameter variables..'this' is used for class variables.

2. is explained in a.

3. Here, we are creating rows for the table dynamically. In the for loop we are assigning these pointer to NULL because the point to nothing initially i.e.:

table[0]->>>NULL

table[1]->>>NULL and so on..

I. Here, we are returning -1 because, we didn't find the element in the hash entry. This is found after the while loop where we are searching for the element

II. Since keys are unique, we will remove the elements by key parameter. Hence, we send the key as the parameter.

III. Here, we remove the element at the position table[hash](this is the column value) at the corresponding modulo row.