Product of Array Except Self

The problem requires computing an output array where each element is the product of all numbers in the input array except the one at the current index. This must be done in O(n) time complexity without using division and preferably with O(1) extra space.

A solution to this problem can be done by utilizing the concept of prefix and suffix products. Here’s how to approach the problem:

Calculate Prefix Products: For each element, calculate the product of all the numbers before it (to the left of it). We can store these products in an output array.
Calculate Suffix Products: For each element, calculate the product of all the numbers after it (to the right of it). This can be done using a single variable, multiplying it with each element in a reverse loop.
Calculate Final Result: Multiply the prefix products and suffix products for each element to get the final result.

Here is the code:

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class Solution:
    def productExceptSelf(self, nums: List[int]) -> List[int]:
        n = len(nums)
        answer = [0] * n

        # Calculate prefix products
        answer[0] = 1
        for i in range(1, n):
            answer[i] = nums[i - 1] * answer[i - 1]

        # Calculate suffix products and multiply with prefix products
        suffix_product = 1
        for i in range(n - 1, -1, -1):
            answer[i] *= suffix_product
            suffix_product *= nums[i]

        return answer

Explanation:

Prefix products are computed by iterating through the array from left to right, using the previous value in the output array.
Suffix products are computed by iterating from right to left, multiplying a temporary variable by each element.
Finally, each element in the answer array is multiplied by the corresponding suffix product.

This algorithm runs in O(n) time complexity, as required, and O(1) extra space complexity, not counting the output array, as the follow-up question asks.

[1,2,3,4]

Numbers: 1 2 3 4 prefix: 1 12 123 suffix: 234 3*4 4
Let’s fill the empty with 1:

Numbers: 1 2 3 4 prefix: 1 1 12 123 suffix: 234 3*4 4 1

[1, 1, 2, 6] [24, 12, 4, 1]

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# @param {Integer[]} nums
# @return {Integer[]}

def product_except_self(nums)
  prefix_accumulator = 1
  prefixes = [1]
  
  for i in (0..nums.size-2)
    prefix_accumulator *= nums[i]  
    prefixes[i+1] = prefix_accumulator
  end
  prefixes
  
  suffix_accmulator = 1
  suffixes = []
  
  (nums.size-1).downto(0).each do |i|
    suffixes[i] = suffix_accmulator
    suffix_accmulator *= nums[i]
  end 

  for i in (0..nums.size-1)
    nums[i] = prefixes[i] * suffixes[i]
  end
  nums
end

Identifying Problem Isomorphism

“Product of Array Except Self” can be approximately mapped to the “Running Sum of 1D Array”.

Reasoning:

“Product of Array Except Self” requires you to calculate a product of all elements except the current one. Similarly, the “Running Sum of 1D Array” requires you to calculate the sum of all previous elements for the current one. The concept of utilizing the result of a computation from all other elements to calculate the result for the current one is the common pattern here.

Here’s the approximate mapping: For each number in the input array for the “Product of Array Except Self” problem, there corresponds the same position number in the input array for the “Running Sum of 1D Array” problem.

The two problems are using different operations: one uses multiplication while the other uses addition. The mapping works in terms of the structure of the problem but not the actual operations performed.

So, while the two problems do not map exactly in terms of their operations, they do map approximately in terms of their structure: both require you to traverse the array, perform an operation based on the other elements, and store the result.

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class Solution:
    def productExceptSelf(self, nums: List[int]) -> List[int]:
        length=len(nums)
        sol=[1]*length
        pre = 1
        post = 1
        for i in range(length):
            sol[i] *= pre
            pre = pre*nums[i]
            sol[length-i-1] *= post
            post = post*nums[length-i-1]
        return(sol)

Problem Classification

The problem is from the domain of Arrays and Mathematics.

The ‘What’ components of the problem are:

An integer array nums.
Each element in the output array is the product of all elements in nums except for the corresponding element in nums.
The algorithm must run in O(n) time complexity.
No division operation can be used.

The problem can be further classified as a product calculation problem with array traversal and element manipulation. It requires computation of array elements, excluding the current index, while maintaining linear time complexity. Constraints also point towards optimization in terms of space complexity. The mention of the prefix or suffix product guarantees to fit in a 32-bit integer indicates there will be handling of large numbers, hence, efficient computation methods need to be devised.

It falls under array manipulation problems involving mathematical computations with optimization constraints on time and space complexity.

Visual Model of the Problem

“Product of Array Except Self” asks you to create a new array where each element at index i is the product of all numbers in the original array except the one at i.

One way to visualize this problem is to think about a row of buckets, each containing a number from the array. For each bucket, you need to multiply the contents of all the other buckets together.

Another visualization could be that you have an array of numbers on a number line. For each number, you’re stretching or compressing the number line such that the number itself stays in place (becomes 1), while all other numbers get multiplied by the original number.

Alternatively, imagine an array as a series of gears, where each gear’s size corresponds to an element in the array. Now, for each gear, you need to calculate the combined effect of all other gears (i.e., the product of all other numbers), but without the influence of the current gear.

In a mathematical way, you can visualize a grid where every cell (i, j) represents the product of the i-th element of the array by the j-th element. The diagonal from top-left to bottom-right would represent the square of each number. The result array would correspond to the product of all elements in each row or column, excluding the diagonal cell.

These are not standard names for these visualization methods, as they are just descriptive methods to help understand the problem. However, we could assign some informal names based on the metaphor used in each explanation:

Bucket Method: This visualization uses the metaphor of a row of buckets, each containing a number from the array. The product at each index is akin to multiplying the contents of all other buckets, excluding the current one.
Number Line Transformation: This method visualizes the problem as a number line where each point is transformed by the product of all other points, while it itself remains unchanged.
Gear System: This metaphor visualizes each number as a gear in a system. The resulting array is calculated as the combined effect of all gears excluding the current one.
Grid Multiplication: In this visualization, the array is seen as a grid where each cell represents the product of two numbers. The resultant array is obtained by summing each row or column, excluding the diagonal.

These names are not standard or universally accepted. The goal of visualization is to help you better understand the problem at hand, so it’s more important that the visualization makes sense to you.

Problem Restatement

Let’s simplify the problem:

You’re given a list of integers, nums. For each element in the list, you need to calculate the product of all other elements in the list and save it in a new list. The order in the new list should be maintained as per the original list.

Now, there are some conditions you need to follow while solving this problem:

You need to finish the calculation in a single pass over the list, which is a fancy way of saying your solution should be linear in terms of time complexity - O(n).
You are not allowed to use division to simplify the calculation.
The computation should be such that the product doesn’t exceed the limit of a 32-bit integer.

There’s also an additional challenge to solve this problem using O(1) extra space, that is, without using any additional data structures (like lists or arrays) that grow with the size of the input. The resulting list doesn’t count towards this space limit.

Terminology

Here are some terms and concepts that are essential for this problem:

Array/List: This is the primary data structure used in this problem. An array or list is a collection of elements identified by index or key. Here, the array nums stores the input integers, and another array is expected as output, with each element being the product of all numbers except the corresponding one in the input array.
Product: This refers to the result of multiplication. In this problem, we’re looking to find the product of all numbers except the one at a particular index.
Time Complexity: This is a computational concept that describes the computational efficiency of an algorithm as the size of the input increases. Here, the problem specifies an O(n) time complexity, which means that the time taken by the solution should increase linearly with the size of the input array.
Space Complexity: Similar to time complexity, space complexity refers to the amount of memory used by an algorithm as a function of the size of the input. The problem’s follow-up question asks for a solution with O(1) space complexity, which means the memory used should be constant and not increase with the size of the input array.
32-bit Integer: This is a data type that can hold integer values within a certain range. It’s important here because the problem statement specifies that the product of any prefix or suffix of nums is guaranteed to fit in a 32-bit integer. This means that we have to ensure our product calculations don’t result in overflow (exceeding the maximum limit) or underflow (going below the minimum limit).
Prefix/Suffix Products: These are terms used in the approach to solve this problem. The prefix product at any index is the product of all numbers before that index, and the suffix product at any index is the product of all numbers after that index. By calculating these, we can obtain the product of all numbers except the one at a particular index without using division.

Problem Simplification and Explanation

You are given a list of numbers, and for each number in that list, you have to find the product of all other numbers. The challenge here is to do this without using division and in linear time complexity.

Let’s take a simpler example to understand it better. Suppose the list is [2, 3, 4]. The task is to replace each number in the list with the product of all other numbers. So, the product for the first number 2 would be 3*4=12, for the second number 3 it would be 2*4=8, and for the last number 4 it would be 2*3=6. Hence, the result would be [12, 8, 6].

Now, let’s use a metaphor to make it more intuitive. Suppose you have three friends, Alice, Bob, and Charlie. They are playing a game where they all show some number of fingers. Alice shows 2, Bob shows 3, and Charlie shows 4 fingers. Now, they want to know for each person, how many total fingers the other two are showing. Alice would see Bob’s 3 and Charlie’s 4 fingers, hence 34=12. Bob would see Alice’s 2 and Charlie’s 4 fingers, so 24=8. Finally, Charlie would see Alice’s 2 and Bob’s 3 fingers, so 2*3=6.

But, there’s a catch: They can only multiply numbers (no addition or division), and they need to find the total for each person as quickly as possible. How can they do it? They can do it by computing and storing prefix and suffix products as mentioned in the Terminology section above, which will allow them to quickly compute the total for each person (analogous to the product of all other numbers in the array).

The interaction of these key concepts is the crux of this problem, and devising a way to implement them efficiently will be the goal as we move forward.

Constraints

Let’s analyze the given constraints:

The number of elements in the array: The array’s length is between 2 and 10^5. The upper limit suggests that a brute force solution of calculating the product for each element individually would not be feasible. A brute force approach would have a time complexity of O(n^2), which would be too large for n = 10^5. Therefore, we are asked to provide a solution with O(n) complexity, implying that we should only be looping through the array a constant number of times.
The range of the elements in the array: Each element in the array can range from -30 to 30. This means that we can’t ignore the possibility of negative numbers and zeros in our array. The presence of a zero has a significant effect because the product of any number with zero is zero. If there is more than one zero in the array, the product of all other numbers will always be zero. For negative numbers, if there is an even number of negative numbers, their product will be positive; if there’s an odd number of negatives, the product will be negative.
Product of any prefix or suffix of nums is guaranteed to fit in a 32-bit integer: This tells us that the intermediate calculations won’t exceed the 32-bit integer limit, so we don’t need to worry about integer overflow during our calculations.
Follow-up question asking for O(1) extra space complexity: This indicates that there might be a way to solve the problem without allocating new space proportional to the size of the input. This hints at the possibility of using the result array to store some intermediate results or a single variable to hold temporary values during the computation.

Recognizing and utilizing these constraints and characteristics will help us develop a more efficient solution to the problem.

Thought Process

Here is the step-by-step thought process and code for this problem:

Cue from problem statement: Since the array can contain negative numbers, 0, and positive numbers, the product of elements can be negative, 0, or positive. We need to handle all these cases. The problem specifically mentions that we can’t use division and we need an O(n) solution.
Approach: We can solve this problem by using the idea of a prefix and suffix product. The prefix product at a position i in the array would be the product of all numbers before i, and the suffix product would be the product of all numbers after i. If we multiply the prefix product and suffix product at position i, it gives us the product of all numbers except the number at i.
To get O(n) time complexity, we compute the prefix and suffix products in two separate loops, and then in a third loop, we multiply the prefix and suffix products to get the final result.
Handling zeros: If the array contains more than one 0, then the product of all other elements except 0 is always 0. If the array contains one 0, then for all positions except where 0 is present, the product is 0 and at the position where 0 is present, the product is the product of all other elements.
Insight: We don’t actually need separate arrays to store the prefix and suffix products. We can directly compute the final product array to save space. This also addresses the follow-up question about solving the problem in O(1) space.

Here is the Python code implementing the above approach:

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def productExceptSelf(nums):
    n = len(nums)
    # Initialize answer array
    answer = [1] * n

    # Calculate prefix product
    prefix_product = 1
    for i in range(n):
        answer[i] *= prefix_product
        prefix_product *= nums[i]

    # Calculate suffix product
    suffix_product = 1
    for i in range(n-1, -1, -1):
        answer[i] *= suffix_product
        suffix_product *= nums[i]

    return answer

The time complexity of this approach is O(n) as we are going over the array three times. The space complexity is O(1) (not counting the space required for output), which fulfills the requirements of the follow-up question. The constraints of the problem dictate that integer overflow is not a concern in this problem, so we don’t need to worry about it here.

Solution Approach and Analysis

The approach to solve this problem can be broken down into the following steps:

Initialization: Start by initializing an array, answer, with the same length as nums and fill it with 1s. This array will ultimately contain our result.
Prefix Product Calculation: Traverse the nums array from left to right (i.e., prefix). For each index i, calculate the product of all the numbers before i (exclusive) and multiply it with the corresponding element in answer.
Metaphorically, consider this process as preparing a recipe. Up to each step (or ingredient), you calculate the ongoing result (or dish) that you’ve prepared so far.
Suffix Product Calculation: Traverse the nums array from right to left (i.e., suffix). For each index i, calculate the product of all the numbers after i (exclusive) and multiply it with the corresponding element in answer.
In terms of our recipe metaphor, this is like adding finishing touches. You again go through each step (or ingredient), but this time in reverse, enhancing the previously prepared dish based on what comes later.
Result: After these steps, answer holds the product of all elements except for the element at each index, which is exactly what we want.

If the size of the nums array (n) increases, our solution will take proportionally longer to compute, as our algorithm’s time complexity is O(n). If there are multiple zeros in nums, all elements in the answer will be zero. If there’s exactly one zero, all elements will be zero, except the one at the zero’s position in nums, which will be the product of all non-zero elements.

Let’s see this in action using nums = [1,2,3,4]:

After initialization: answer = [1,1,1,1]
After prefix product calculation: answer = [1,1,2,6]
After suffix product calculation: answer = [24,12,8,6], which is our final result.

So, answer[i] is indeed the product of all numbers in nums except nums[i].

Let’s break down our solution into a stepwise refinement:

Step 1 - Initialization:
- Create a new array, answer, of the same size as nums.
- Initialize all elements in answer to 1.
Step 2 - Prefix product calculation:
- Initialize a variable, left_product, to 1. This variable will hold the product of all elements to the left of the current element in nums.
- Iterate over nums from left to right. For each i, multiply answer[i] with left_product and update left_product by multiplying it with nums[i].
Step 3 - Suffix product calculation:
- Initialize a variable, right_product, to 1. This variable will hold the product of all elements to the right of the current element in nums.
- Iterate over nums from right to left. For each i, multiply answer[i] with right_product and update right_product by multiplying it with nums[i].

After these steps, answer will contain the product of all elements except for the one at each index, which is our final solution.

In terms of repeatable patterns or parts of the problem that can be solved independently, Steps 2 and 3 can be considered independent calculations. Step 2 calculates the prefix product while Step 3 calculates the suffix product. They are independent in the sense that the result of one doesn’t influence the other, but they both contribute to the final result.

The repeatable pattern here is the product calculation. In both prefix and suffix calculation steps, we’re essentially doing the same operation: multiplying the current accumulated product with the corresponding element in the array. This pattern of cumulative product calculation is at the core of this problem’s solution.

Identification of Applicable Theoretical Concepts

The mathematical concept at play here is the principle of multiplication, specifically the fact that multiplication is both associative and commutative. This means that the order in which you multiply numbers doesn’t change the result.

Because of these properties, we can split the total multiplication into two parts: the multiplication of all numbers to the left of a given index (prefix product) and all numbers to the right of that index (suffix product). The final answer for each index is then the product of the prefix and suffix products.

The algorithmic concept here is prefix and suffix products, a common pattern in array manipulation problems where a running product (or sum) is maintained while iterating over the array. We utilize two passes over the array, one for calculating the prefix product and the other for the suffix product. The result is a product of all elements except the one at each index.

This problem also involves the concept of space optimization. In the follow-up, it’s required to solve the problem with constant extra space. We could use the output array for the intermediate computations, allowing us to adhere to the space constraint.

The concepts of prefix and suffix products and space optimization are prevalent in computer science and are common in many array and string manipulation problems. Understanding these principles is critical in designing efficient algorithms.

From Brute Force to Optimal Solution

Let’s start with a brute force solution for this problem. A brute force approach would be to compute the product of all elements for each index in the array except the current one. Here’s a Python code snippet for this:

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def productExceptSelf(nums):
    n = len(nums)
    output = []
    for i in range(n):
        product = 1
        for j in range(n):
            if i != j:
                product *= nums[j]
        output.append(product)
    return output

This approach works, but it’s not efficient. It has a time complexity of O(n^2) because for each element in the array, we’re going through the array again to calculate the product. If the array has many elements, this solution could become very slow.

Now let’s think about optimizing this.

We can see that for each element in the array, we’re calculating the product of all other elements. We are doing a lot of repeated calculations since the product for each element involves many of the same multiplication operations. This suggests that we could use a dynamic programming approach to store some of these intermediate results and avoid doing the same calculations over and over again.

The optimization comes from the realization that the product of all elements except the one at index i is just the product of all numbers before i and all numbers after i. This way, we can calculate the product of all numbers before i in one pass (prefix product), and the product of all numbers after i in a second pass (suffix product). Multiplying these two values will give us our result for each index.

Here’s the optimized code:

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def productExceptSelf(nums):
    n = len(nums)
    prefix, suffix = [0]*n, [0]*n
    prefix[0], suffix[-1] = nums[0], nums[-1]
    
    for i in range(1, n):
        prefix[i] = prefix[i-1] * nums[i]
    for i in range(n-2, -1, -1):
        suffix[i] = suffix[i+1] * nums[i]
    
    output = [0]*n
    output[0], output[-1] = suffix[1], prefix[-2]
    for i in range(1, n-1):
        output[i] = prefix[i-1] * suffix[i+1]
    return output

This solution calculates the prefix and suffix products in two separate passes over the array, and then combines these to form the result. This solution has a time complexity of O(n), and uses O(n) extra space for the prefix and suffix arrays.

The problem also has a follow-up part that asks us to solve it with O(1) extra space. We can achieve this by using the output array to calculate and store the prefix product, and then use a single variable to keep track of the suffix product as we iterate backwards through the array to calculate the final result:

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def productExceptSelf(nums):
    n = len(nums)
    output = [1]*n
    for i in range(1, n):
        output[i] = output[i-1] * nums[i-1]
    suffix = 1
    for i in range(n-1, -1, -1):
        output[i] *= suffix
        suffix *= nums[i]
    return output

In this final optimized solution, the time complexity remains O(n), but we have improved the space complexity to O(1), which is a significant optimization over our original brute force approach.

So you found the bottleneck in the performance of the brute force solution and addressed it to come up with the optimal solution?

Yes, that’s correct! The brute force solution had a performance bottleneck because it was recalculating the same product values multiple times. By identifying this issue, we were able to design a more efficient solution.

In our optimized solution, we essentially split the product of all other numbers into two parts: the product of numbers before the current one (prefix product) and the product of numbers after it (suffix product). By calculating these values in two passes over the array, we avoided unnecessary recalculations and significantly improved the time complexity from O(n^2) to O(n).

Additionally, by optimizing space usage further, we achieved a space complexity of O(1), if we don’t count the output array as extra space. We did this by reusing the output array to store intermediate prefix products, and a single variable to keep track of the suffix product, which eliminated the need for additional arrays.

This is a typical example of how we can optimize algorithms: by identifying and eliminating redundant calculations, and by using space more efficiently.

Coding Constructs

Sure, let’s analyze the provided Python code which solves the problem of getting the product of all elements except the current one in an array.

High-level problem-solving strategies or techniques: This code uses the idea of prefix and suffix products. It first calculates the product of all elements to the left (prefix) of each element in the array and then multiplies it with the product of all elements to the right (suffix) of each element.
Explaining the purpose of this code to a non-programmer: This code takes a list of numbers and for each number in the list, it calculates the product of every other number in the list. For example, if the list was [2,3,4], it would produce a new list where the first number is the product of 3 and 4 (12), the second number is the product of 2 and 4 (8), and the third number is the product of 2 and 3 (6).
Logical elements or constructs used in this code: The code uses a few fundamental programming constructs:
- Arrays for storing input and output.
- Loops for iterating over arrays.
- Variables for temporary storage of data (suffix).
Algorithmic approach in plain English:
- Create an output array and initialize all elements as 1.
- In the first loop, calculate the product of all numbers before the current number and store it in the output array.
- Initialize a variable (suffix) as 1 which stores the product of all numbers after the current number.
- In the second loop, moving backwards, multiply the previously calculated product with the product of all numbers after the current number and store it in the output array.
Key steps or operations this code is performing:
- Initialization of the output array.
- Calculation of prefix products and storing in output.
- Calculation of suffix products on the fly and updating the output array.
Algorithmic patterns or strategies: This code utilizes the concept of prefix and suffix products for efficient calculation. It reduces the time complexity by avoiding repeated computation of the same products. This is a common pattern used in array processing problems, especially when a property (in this case product) of all other elements affects each element.

Language Agnostic Coding Drills

Targeted Drills in Python

Q&A

Thinking Process

While addressing a particular problem, my initial inclination was to employ a brute force approach. The idea was to use two pointers, one starting from i-1 and moving to 1, the other from i+1 and advancing to the length of the ’nums’ array. At each step, the elements at the pointer positions would be multiplied, and the result appended to the output array. However, having encountered and solved questions related to Prefix Sum and Postfix Sum previously, I saw an opportunity to leverage a similar concept in this case, which could potentially make the solution more efficient.

The central approach I chose is based on the idea that the prefix product of arr[i] (that is, the product of all elements from arr[0] to arr[i-1]) and the postfix product of arr[i] (the product of all elements from arr[i+1] to arr[n-1]) together form the total product except self at index i. Therefore, the task simplifies to calculating the prefix product * postfix product for each element of the array.

While typical implementations of the prefix-postfix concept suggest a solution involving two separate traversals of the array, I was convinced that the solution could be further optimized by a single traversal.

Here’s the plan of action:

I initialize an output solution array, the same size as the input array, filled with 1’s.
Next, I create two variables to hold the prefix and postfix product so far.
I traverse the input array.
For each ‘i’, before updating the prefix and postfix values, I multiply the current solution array value at index ‘i’ with the prefix product of the previous ‘i-1’ elements.
Similarly, I calculate the postfix product value for ’n-i-1’ where ’n’ is the length of the input array at each iteration.
Much like in step 4, before calculating the postfix for the ‘i’th value, I multiply the solution_array[n-i-1] with the postfix product value, which is the product of elements from input[i+1] to input[n-1].

By following this approach, we effectively eliminate the need for a second traversal of the array while achieving the same results, thereby improving the efficiency of the solution.

Practice Phase Template

1. Identify and Breakdown Concepts:

To create a “Practice Phase” template, we first identify the key concepts and the needed programming skills for this problem. For this specific problem, the following concepts are involved:

a. Array or List manipulation.

b. Variables and expressions.

c. Looping constructs (for loop).

d. Array indexing and index manipulation.

e. In-place modification of an array.

2. Coding Drills:

We next transform these concepts into standalone, language-agnostic coding drills, as follows:

a. Create an array or list, and perform basic operations (add, delete, modify, access) on it.

b. Define and manipulate variables, and use them in mathematical expressions.

c. Write a basic for loop that iterates over an array or list.

d. Access elements of an array using indices, including reverse indexing.

e. Modify the content of an array in-place.

3. Practice Objectives:

The final step is to provide clear practice objectives, focusing on integrating the learned skills into a cohesive solution. These objectives should be directly aligned with the identified concepts and drills.

a. Use a list to store the results of an operation, with the operation being some kind of computation that involves the current and previous values from the same list.

b. Use a for loop to iterate through the list, using the loop variable as the list’s index.

c. Use in-place modification to change the values of the list while iterating through it.

d. Utilize two separate loops simultaneously to process the list in two different directions (forward and reverse).

These three steps form the “Practice Phase” template, which serves as a guide for effectively practicing and learning the skills needed to solve the problem.

Study Plan Template for Python

Let’s create Python-based drills for each of the practice objectives identified above:

Drill 1 - Create an array or list, and perform basic operations

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# Creating an array
array = [1, 2, 3, 4, 5]

# Adding an element
array.append(6)
print(array)  # [1, 2, 3, 4, 5, 6]

# Removing an element
array.remove(3)
print(array)  # [1, 2, 4, 5, 6]

# Modifying an element
array[1] = 7
print(array)  # [1, 7, 4, 5, 6]

# Accessing an element
print(array[2])  # 4

Drill 2 - Define and manipulate variables, and use them in mathematical expressions

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# Define a variable
a = 5

# Manipulate variable
a *= 2
print(a)  # 10

# Mathematical expression
b = a + 3
print(b)  # 13

Drill 3 - Write a basic for loop that iterates over an array or list

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# Define an array
array = [1, 2, 3, 4, 5]

# For loop
for element in array:
    print(element)

Drill 4 - Access elements of an array using indices, including reverse indexing

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# Define an array
array = [1, 2, 3, 4, 5]

# Forward indexing
print(array[1])  # 2

# Reverse indexing
print(array[-1])  # 5

Drill 5 - Modify the content of an array in-place

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# Define an array
array = [1, 2, 3, 4, 5]

# Modify in-place
for i in range(len(array)):
    array[i] *= 2

print(array)  # [2, 4, 6, 8, 10]

These drills will help you practice the key coding concepts used in the solution. Practice these drills separately and then try to combine them to form a solution for the given problem.

Practice Drills

The practice drills designed for the “productExceptSelf” function in Python do cover all the six steps in the action plan.

The first drill, which is about initializing an array of a specific size, mirrors the first step in the action plan.
The second drill, about creating two variables and using them within a function, corresponds to the second step in the action plan.
The third drill corresponds to the third step in the action plan, where we iterate over an array.
The fourth drill mirrors the fourth step, where you’re asked to manipulate the array elements based on the prefix product concept.
The fifth drill corresponds to the fifth step, where the postfix product value is calculated and stored.
The sixth drill, which asks you to adjust the values in your solution array based on the postfix product, corresponds to the last step in the action plan.

Thus, the practice drills are designed to give you hands-on experience with all the steps required to implement the “productExceptSelf” function as outlined in the action plan.

Previous Problems

Prefix Sum

A simple coding drill for practicing the concept of postfix sums. The goal of this drill is to calculate the postfix sum for every element in a list. The postfix sum of an element at position i in a list arr is the sum of all elements from position i+1 to the end of the list.

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# Given a list of integers:
nums = [4, 2, 1, 3, 5]

# Your task is to write a function to calculate the postfix sum for each element:
def postfix_sum(nums):
    result = [0] * len(nums)
    # TODO: Fill in your code here
    return result

# When you call your function with the provided list, it should return: [11, 9, 7, 5, 0]
print(postfix_sum(nums))

Fill in the missing code to complete the function. You need to implement the logic to calculate the postfix sum for each element in the list. As a hint, you might find it helpful to iterate over the list in reverse order.

Certainly, here is a simple exercise for the concept of a Postfix Sum. In this exercise, the objective is to compute an array where each element is the sum of all the elements that come after it in the input array.

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# Here's a list of integers:
nums = [1, 2, 3, 4, 5]

# Write a function that calculates the postfix sum for each element:
def compute_postfix_sum(nums):
    postfix_sum = [0] * len(nums)
    # TODO: Fill in your code here
    return postfix_sum

# Calling your function with the above list should return: [14, 14, 12, 9, 5, 0]
print(compute_postfix_sum(nums))

Now, it’s your task to fill in the missing code in the function. The function should iterate over the list in reverse order, and for each element, it should calculate the sum of all elements that come after it in the input array. In other words, for an element at index i, the function should sum all elements at indices greater than i.

Postfix Sum

Absolutely, here is a simple exercise for the concept of a Prefix Sum. In this exercise, you will create an array where each element is the sum of all the elements that come before it in the input array.

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# Here's a list of integers:
nums = [1, 2, 3, 4, 5]

# Write a function that calculates the prefix sum for each element:
def compute_prefix_sum(nums):
    prefix_sum = [0] * len(nums)
    # TODO: Fill in your code here
    return prefix_sum

# Calling your function with the above list should return: [0, 1, 3, 6, 10, 15]
print(compute_prefix_sum(nums))

Now, you need to fill in the missing code in the function. The function should iterate over the list from left to right, and for each element, it should calculate the sum of all elements that come before it in the input array. In other words, for an element at index i, the function should sum all elements at indices less than i.

Drills

We can break down the concept of calculating a Postfix Sum into simpler drills:

Reverse Iterating over a list: To calculate the Postfix Sum, one must know how to iterate over a list in reverse.

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# Write a function that takes a list of numbers and prints each number with its index, in reverse order
def print_elements_with_indices_reverse(nums):
    # TODO: Fill in your code here
    pass

# Your function should print the index and value for each element in the list, in reverse order
print_elements_with_indices_reverse([1, 2, 3, 4, 5])  # Expected: 4 5, 3 4, 2 3, 1 2, 0 1

Summing elements of a list from a given index to the end: This is a precursor to calculating the postfix sum of a list.

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# Write a function that takes a list of numbers and an index, and returns the sum of all elements from that index to the end
def sum_from_index(nums, index):
    # TODO: Fill in your code here
    pass

# Your function should return the sum of all elements from the given index to the end of the list
print(sum_from_index([1, 2, 3, 4, 5], 2))  # Expected: 12

These foundational exercises will guide you towards understanding how to compute a Postfix Sum.

Let’s break it down into some foundational programming concepts for Postfix Sum.

Creating an empty list: This is a fundamental concept that you need to know to solve a wide range of problems in Python.

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# Write a function that creates an empty list of size n
def create_empty_list(n):
    # TODO: Fill in your code here
    pass

# Your function should return a list with n elements all initialized to 0
print(create_empty_list(5))  # Expected: [0, 0, 0, 0, 0]

Iterating over a list with indices: Iterating over a list is one of the most common tasks in Python programming. It’s important to understand how to iterate over a list and access elements using indices.

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# Write a function that takes a list of numbers and prints each number with its index
def print_elements_with_indices(nums):
    # TODO: Fill in your code here
    pass

# Your function should print the index and value for each element in the list
print_elements_with_indices([1, 2, 3, 4, 5])  # Expected: 0 1, 1 2, 2 3, 3 4, 4 5

Summing elements of a list up to a given index: This is a precursor to calculating the prefix sum of a list.

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# Write a function that takes a list of numbers and an index, and returns the sum of all elements up to that index
def sum_up_to_index(nums, index):
    # TODO: Fill in your code here
    pass

# Your function should return the sum of all elements up to the given index
print(sum_up_to_index([1, 2, 3, 4, 5], 3))  # Expected: 10

After practicing these drills, calculating the prefix sum should become much more manageable.