Today was my first day solving DSA (Data Structures and Algorithms) questions on Leetcode. It was quite challenging, but also very rewarding. I had to put my problem-solving skills to the test and think critically about how to approach each question. I also had to research and learn new algorithms and data structures in order to find the most efficient solution.

As a software engineer, understanding and being proficient in data structures and algorithms is crucial to creating efficient and performant code. Data structures are the building blocks of any program and choosing the right one for a given problem can greatly impact the performance and scalability of the solution. Algorithms, on the other hand, are the set of instructions that are used to manipulate the data structures. Understanding the time and space complexity of different algorithms is important in order to choose the most efficient solution for a given problem.

Solving DSA questions on Leetcode is a great way to improve my skills and understanding of data structures and algorithms. It also helps me to think more critically and creatively when solving problems. I am looking forward to continue solving more questions and becoming more proficient in these areas.

## The Array Question

Given an integer array nums, return an array answer such that answer[i] is equal to the product of all the elements of nums except nums[i].

The product of any prefix or suffix of nums is guaranteed to fit in a 32-bit integer.

You must write an algorithm that runs in O(n) time and without using the division operation.

## Pseudocode:

- Initialize an empty array “output” of the same length as nums
- Initialize a variable “left_product” to 1
- Loop through the nums array from the left to the right a. Multiply the current value of left_product with the current element of the nums array b. Assign the resulting product to the current element of the output array c. Update the value of left_product with the new product
- Initialize a variable “right_product” to 1
- Loop through the nums array from the right to the left a. Multiply the current value of right_product with the current element of the nums array b. Multiply the current element of the output array with the new product of right_product c. Update the value of right_product with the new product
- Return the output array

## Code

```
def product_except_self(nums):
output = [0] * len(nums)
left_product = 1
for i in range(len(nums)):
output[i] = left_product
left_product *= nums[i]
right_product = 1
for i in range(len(nums)-1, -1, -1):
output[i] *= right_product
right_product *= nums[i]
return output
```

## Time and space complexity

The above algorithm uses two loops to iterate through the nums array. In the first loop, it calculates the product of all the elements to the left of the current element and stores it in the corresponding element of the output array. In the second loop, it calculates the product of all the elements to the right of the current element and multiplies it with the value already stored in the output array. By doing this, it avoids using the division operator while still achieving O(n) time complexity.

What is the space and time complexity of the above solution

The time complexity of the above solution is O(n) because it uses two loops to iterate through the input array “nums” of length n. The outer loop takes O(n) time to complete and the inner loop takes O(n) time to complete. Therefore, the overall time complexity is O(n) + O(n) = O(n).

The space complexity of the above solution is O(n) because it uses an additional output array of length n to store the result. The algorithm does not use any other data structures, so the space complexity is determined by the size of this output array. Therefore, the space complexity of the above solution is O(n).

What is the space and time complexity of the above solution

The time complexity of the above solution is O(n) because it uses two loops to iterate through the input array “nums” of length n. The outer loop takes O(n) time to complete and the inner loop takes O(n) time to complete. Therefore, the overall time complexity is O(n) + O(n) = O(n).

The space complexity of the above solution is O(n) because it uses an additional output array of length n to store the result. The algorithm does not use any other data structures, so the space complexity is determined by the size of this output array. Therefore, the space complexity of the above solution is O(n).

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