Apply Operations to an Array

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ArraysTwo Pointers

You are given a 0-indexed array nums of size n consisting of non-negative integers.

You need to apply n - 1 operations to this array where, in the i^th operation (0-indexed), you will apply the following on the i^th element of nums:

If nums[i] == nums[i + 1], then multiply nums[i] by 2 and set nums[i + 1] to 0. Otherwise, you skip this operation.

After performing all the operations, shift all the 0's to the end of the array.

For example, the array [1,0,2,0,0,1] after shifting all its 0's to the end, is [1,2,1,0,0,0].

Return the resulting array.

Note that the operations are applied sequentially, not all at once.

Example 1:

Input: nums = [1,2,2,1,1,0]
Output: [1,4,2,0,0,0]
Explanation: We do the following operations:
- i = 0: nums[0] and nums[1] are not equal, so we skip this operation.
- i = 1: nums[1] and nums[2] are equal, we multiply nums[1] by 2 and change nums[2] to 0. The array becomes [1,4,0,1,1,0].
- i = 2: nums[2] and nums[3] are not equal, so we skip this operation.
- i = 3: nums[3] and nums[4] are equal, we multiply nums[3] by 2 and change nums[4] to 0. The array becomes [1,4,0,2,0,0].
- i = 4: nums[4] and nums[5] are equal, we multiply nums[4] by 2 and change nums[5] to 0. The array becomes [1,4,0,2,0,0].
After that, we shift the 0's to the end, which gives the array [1,4,2,0,0,0].

Example 2:

Input: nums = [0,1]
Output: [1,0]
Explanation: No operation can be applied, we just shift the 0 to the end.

Constraints:

2 <= nums.length <= 2000
0 <= nums[i] <= 1000

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of values for the integers in the input array? Can the array contain negative numbers, zeros, or floating-point numbers?
What should the function return if the input array is null or empty?
Are we modifying the input array in-place, or should I return a new array with the modified elements?
Could you clarify the order in which operations should be applied if there are multiple possible pairs that meet the condition?
Is there a maximum size for the input array that I should consider?

Brute Force Solution

Approach

The brute force method for this array problem means we'll explore all possible sets of operations on the numbers. We'll try applying the operations in every conceivable way and then check the resulting array to see if it meets the desired conditions.

Here's how the algorithm would work step-by-step:

Start by looking at the first number and the number next to it.
Try applying the operation to these two numbers, and then try not applying the operation.
Next, move to the second number and the number next to it, repeating the process of applying or not applying the operation.
Continue this process for all pairs of adjacent numbers in the entire list.
Each time you've made a decision about applying or not applying operations to all pairs, look at the resulting modified list.
Check if the modified list fulfills the required specifications for the problem.
If it does, store the list, or a measure of how 'good' it is.
Repeat this entire process until all possible combinations of operation applications have been tried.
Finally, compare all the stored lists (or their 'goodness' scores) and select the best one according to the problem's requirements.

Code Implementation

def apply_operations_brute_force(numbers):
    best_array = numbers[:] # Initialize with a copy of the original array
    
    def check_and_update(current_array):
        nonlocal best_array
        # This is a placeholder for problem specific criteria
        # We check if the current_array is 'better' than the best_array
        if not best_array or sum(current_array) > sum(best_array):
            best_array = current_array[:]
    
    def generate_combinations(index, current_array):
        # When index equals length, we have finished going through all pairs
        if index >= len(numbers) - 1:
            check_and_update(current_array)
            return

        # Explore without applying the operation
        generate_combinations(index + 1, current_array)

        # Create a copy to modify for applying operations to adjacent numbers
        temp_array = current_array[:]

        # Apply the operation: multiply if equal, else do nothing (example)
        if temp_array[index] == temp_array[index + 1]:
            temp_array[index] *= 2
            temp_array[index + 1] = 0
        
        # Explore with applying the operation
        generate_combinations(index + 1, temp_array)

    generate_combinations(0, numbers[:])
    return best_array

Big(O) Analysis

Time Complexity

O(2^n) – The brute force approach explores all possible combinations of applying or not applying the operation to adjacent pairs in the array of size n. For each pair, there are two choices: apply the operation or do not apply it. Since there are approximately n such pairs, the total number of combinations is 2 * 2 * ... * 2 (n times), which equals 2^n. Therefore, the algorithm examines an exponential number of possibilities.

Space Complexity

O(2^N) – The algorithm explores all possible combinations of applying or not applying the operation to adjacent number pairs. For an array of size N, there are N-1 pairs. Each decision (apply or don't apply) creates a branching path, resulting in a decision tree with a maximum of 2^(N-1) paths. Each path represents a modified list that needs to be stored temporarily for comparison. Because the size of each modified list is N, the algorithm has to store at least one resulting modified list, or a 'goodness' score derived from it while exploring the rest of the decision tree, which still has to maintain all possible results from each operation. Therefore, in the worst-case, the algorithm has to compare a number of resulting modified lists exponential to the size of the input array N. This leads to an auxiliary space complexity of O(2^N).

Optimal Solution

Approach

The goal is to modify the array as instructed, but do it efficiently. The clever trick is to avoid unnecessary movements and changes by focusing on a two-step process: first perform the instructed calculations, and then shift all the zeros to the end in an optimized manner.

Here's how the algorithm would work step-by-step:

Go through each pair of adjacent numbers in order.
If a pair is identical, multiply the first number in the pair by two, and make the second number zero.
Now, push all the zeros to the end of the sequence by going through the sequence from start to end.
When you find a non-zero number, move it to the next available spot from the start of the sequence if it is not already there. This efficiently places all non-zero numbers at the front.

Code Implementation

def apply_operations(numbers):
    array_length = len(numbers)

    # Perform calculations on adjacent identical numbers
    for index in range(array_length - 1):
        if numbers[index] == numbers[index + 1]:
            numbers[index] *= 2
            numbers[index + 1] = 0

    non_zero_index = 0
    # Shift non-zero elements to the front
    for index in range(array_length):
        if numbers[index] != 0:
            numbers[non_zero_index], numbers[index] = numbers[index], numbers[non_zero_index]

            # Update the index of the next non-zero element
            non_zero_index += 1

    return numbers

Big(O) Analysis

Time Complexity

O(n) – The solution involves two main phases. The first phase iterates through the array once to identify adjacent equal numbers, performing a constant-time operation for each pair, hence O(n). The second phase iterates through the array again, moving non-zero elements to the front. Each element is visited at most once, resulting in another O(n) operation. Therefore, the total time complexity is O(n) + O(n), which simplifies to O(n).

Space Complexity

O(1) – The algorithm operates in place, modifying the input array directly. No auxiliary data structures, such as temporary arrays or hash maps, are created. The operations involve only iterating through the array and swapping elements, which can be done using a constant amount of extra memory, regardless of the input size N.

Edge Cases

Case	How to Handle
Null or empty input array	Return an empty array or throw an IllegalArgumentException as appropriate based on the requirements
Array containing all zeros	The algorithm should handle all zeros correctly, potentially leading to a long sequence of operations
Array containing only one element	Return the array itself, as no operations are applicable
Array with a large number of elements	Ensure the algorithm has linear time complexity and does not exceed memory limits
Array with all identical numbers	Consecutive identical numbers will be doubled and subsequent zeros handled correctly
Array containing negative numbers	The operations should correctly apply regardless of whether the numbers are negative, as only equality checks and multiplications are used.
Array with very large numbers that can potentially cause integer overflow after multiplication	Use a larger integer type (e.g., long) to prevent potential overflow, or implement overflow detection
Array with only a few non-zero elements scattered throughout	The solution should efficiently handle the shifting of zeros to the end of the array, ensuring unnecessary operations are avoided.

Null or empty input array

How to Handle:

Return an empty array or throw an IllegalArgumentException as appropriate based on the requirements

Array containing all zeros

How to Handle:

The algorithm should handle all zeros correctly, potentially leading to a long sequence of operations

Array containing only one element

How to Handle:

Return the array itself, as no operations are applicable

Array with a large number of elements

How to Handle:

Ensure the algorithm has linear time complexity and does not exceed memory limits

Array with all identical numbers

How to Handle:

Consecutive identical numbers will be doubled and subsequent zeros handled correctly

Array containing negative numbers

How to Handle:

The operations should correctly apply regardless of whether the numbers are negative, as only equality checks and multiplications are used.

Array with very large numbers that can potentially cause integer overflow after multiplication

How to Handle:

Use a larger integer type (e.g., long) to prevent potential overflow, or implement overflow detection

Array with only a few non-zero elements scattered throughout

How to Handle:

The solution should efficiently handle the shifting of zeros to the end of the array, ensuring unnecessary operations are avoided.