Final Answer:
To create a hash table for efficient O(1) indexing, you can modify the provided code as follows:
```python
def MyHash(L, N):
hash_table = [None] * N
for i in range(N):
hash_table[L[i]] = i
return hash_table
L = [5, 2, 0, 1, 4, 3]
N = len(L)
hash_table = MyHash(L, N)
search_item = 2
index = hash_table[search_item]
print("Index of", search_item, "is:", index)
```
This program creates a hash table using the given list `L`, allowing you to quickly locate an item's index with O(1) time complexity. In the example, the index of the search item '2' is retrieved and printed.
Step-by-step explanation:
The provided code defines a function `MyHash` that takes a list `L` and its length `N`, creating a hash table where the indices correspond to the values in `L`. Each element in the hash table holds the index of the corresponding value. The modified code initializes an empty hash table of size `N` and populates it by iterating through the list `L`. The resulting hash table allows for direct indexing of items.
In the example, the list `L` is [5, 2, 0, 1, 4, 3], and the hash table is created using the `MyHash` function. The index of the search item '2' is then obtained from the hash table, demonstrating the O(1) time complexity for item retrieval. This approach provides a fast and efficient way to find the index of any value in the given list.
In summary, the modified program efficiently creates a hash table and demonstrates O(1) indexing, making it a practical solution for quickly locating items in a list with minimal time complexity.