Final Array State After K Multiplication Operations I

Easy

Microsoft

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ArraysGreedy AlgorithmsStacks

You are given an integer array nums, an integer k, and an integer multiplier.

You need to perform k operations on nums. In each operation:

Find the minimum value x in nums. If there are multiple occurrences of the minimum value, select the one that appears first.
Replace the selected minimum value x with x * multiplier.

Return an integer array denoting the final state of nums after performing all k operations.

Example 1:

Input: nums = [2,1,3,5,6], k = 5, multiplier = 2

Output: [8,4,6,5,6]

Explanation:

Operation	Result
After operation 1	[2, 2, 3, 5, 6]
After operation 2	[4, 2, 3, 5, 6]
After operation 3	[4, 4, 3, 5, 6]
After operation 4	[4, 4, 6, 5, 6]
After operation 5	[8, 4, 6, 5, 6]

Example 2:

Input: nums = [1,2], k = 3, multiplier = 4

Output: [16,8]

Explanation:

Operation	Result
After operation 1	[4, 2]
After operation 2	[4, 8]
After operation 3	[16, 8]

Constraints:

1 <= nums.length <= 100
1 <= nums[i] <= 100
1 <= k <= 10
1 <= multiplier <= 5

Final Array State After K Multiplication Operations I

Easy

Microsoft

0 views

Topics:

ArraysGreedy AlgorithmsStacks

You are given an integer array nums, an integer k, and an integer multiplier.

You need to perform k operations on nums. In each operation:

Find the minimum value x in nums. If there are multiple occurrences of the minimum value, select the one that appears first.
Replace the selected minimum value x with x * multiplier.

Return an integer array denoting the final state of nums after performing all k operations.

Example 1:

Input: nums = [2,1,3,5,6], k = 5, multiplier = 2

Output: [8,4,6,5,6]

Explanation:

Operation	Result
After operation 1	[2, 2, 3, 5, 6]
After operation 2	[4, 2, 3, 5, 6]
After operation 3	[4, 4, 3, 5, 6]
After operation 4	[4, 4, 6, 5, 6]
After operation 5	[8, 4, 6, 5, 6]

Example 2:

Input: nums = [1,2], k = 3, multiplier = 4

Output: [16,8]

Explanation:

Operation	Result
After operation 1	[4, 2]
After operation 2	[4, 8]
After operation 3	[16, 8]

Constraints:

1 <= nums.length <= 100
1 <= nums[i] <= 100
1 <= k <= 10
1 <= multiplier <= 5

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 are the possible ranges for the integers within the input array, and what is the range for K?
Can the input array be empty or null? What should I return in such cases?
Are the multiplication operations applied sequentially? For example, if K = 2, do I multiply the first two elements, then the next two, or is there some other logic?
Is K guaranteed to be less than or equal to the number of elements in the array?
If the length of the array is not perfectly divisible by K, how should I handle the remaining elements?

Brute Force Solution

Approach

The brute force method for this problem involves trying every single combination of multiplication operations. We essentially simulate each operation one by one on a copy of the original list of numbers to see the final result, and we do this for every possible combination within the given operation limit.

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

Start with the original list of numbers.
Consider the first possible multiplication operation from all possible pairs.
Make a copy of the original list and apply the multiplication operation to the numbers on the copied list.
Now, consider the next possible multiplication operation and apply it to the list that was just modified.
Keep repeating this process of considering an operation, making a copy, and applying it until you have used all the allowed number of multiplication operations, 'K'.
Remember that you could also perform fewer than 'K' operations and consider this also as a possible solution.
Store the final list of numbers that results after performing each series of multiplication operations.
Once you have tried every possible combination of multiplication operations, compare all the stored final lists.
Finally, select the stored final list that matches the question's constraints (e.g. the one with the smallest or largest number).

Code Implementation

def final_array_state_after_k_multiplication_operations_i(numbers, max_operations):
    possible_final_arrays = []
    number_of_elements = len(numbers)

    def generate_combinations(current_array, operations_used):
        possible_final_arrays.append(current_array[:])

        if operations_used == max_operations:
            return

        # Try all possible pairs for multiplication
        for first_index in range(number_of_elements):
            for second_index in range(first_index + 1, number_of_elements):
                # Create a copy to avoid modifying the original
                new_array = current_array[:]

                # Apply the multiplication operation
                new_array[first_index] = new_array[first_index] * new_array[second_index]

                generate_combinations(new_array, operations_used + 1)

    generate_combinations(numbers, 0)

    return possible_final_arrays

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Big(O) Analysis

Time Complexity

O(n^k) – The algorithm explores all possible combinations of up to k multiplication operations on n numbers. In the worst case, we consider every possible pair of numbers for multiplication in each of the k operations. The number of ways to choose a pair is approximately n^2. Since we repeat this up to k times, the total number of combinations is proportional to (n^2)^k/2! or n^(2k). We also need to copy the array of size n after each set of operations. In the worst case, the complexity will be proportional to the number of ways to choose up to k pairs of numbers, which is O(n^(2k)). Since n is larger than 2, O(n^(2k)) dominates O(n). Therefore the runtime is O(n^(2k)).

Space Complexity

O(N * 2^((N*(N-1)/2)*K)) – The algorithm makes a copy of the original list of N numbers for each possible combination of multiplication operations. In the worst-case, there are N*(N-1)/2 possible pairs of numbers to multiply, and we can perform up to K operations. The algorithm explores all possible combinations of these operations, potentially leading to 2^((N*(N-1)/2)*K) branches. Each branch requires storing a copy of the list of size N. Therefore, the auxiliary space complexity is O(N * 2^((N*(N-1)/2)*K)).

Optimal Solution

Approach

The key is to realize that multiplying by even numbers results in even numbers, and multiplying by odd numbers might change the parity. We want to strategically use our multiplication operations to make all numbers even, if possible. Our main goal is to minimize the number of odd numbers left at the end.

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

First, check if we have enough operations to make all the numbers even. If we have more operations than odd numbers in our starting list, we can always make them all even.
If we can make everything even, simply multiply each odd number by an even number. The array now only contains even numbers and we are done.
If we don't have enough operations to make everything even, we must choose which numbers to change. We do nothing, since our goal is to return the original array.

Code Implementation

def final_array(numbers, operations):
    odd_count = 0
    for number in numbers:
        if number % 2 != 0:
            odd_count += 1

    # If operations are enough to make everything even
    if operations >= odd_count:
        for index in range(len(numbers)):
            if numbers[index] % 2 != 0:
                # Change odd number to even
                numbers[index] *= 2

    # If operations are not enough, we do nothing.
    # Just return the original array.
    return numbers

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the array of size n once to count the odd numbers. Then, it checks if k is greater than or equal to the count of odd numbers which is O(1). Finally, if it can make all numbers even it iterates through the array again which is O(n). Therefore, the dominant factor is the iteration over the array, making the time complexity O(n).

Space Complexity

O(1) – The provided algorithm operates in place, modifying the input array directly. It only uses a few integer variables to store the number of operations (K) and potentially loop counters, which require constant space regardless of the input array's size (N, where N is the number of elements in the array). No auxiliary data structures like lists, hash maps, or recursion are employed that would scale with the input. Thus, the auxiliary space complexity is constant.

Edge Cases

Case	How to Handle
Null or undefined input array	Return an empty array or throw an IllegalArgumentException if the input array is null or undefined to avoid NullPointerExceptions.
K is zero	Return the original array as no operations are performed when K is zero.
K is larger than half the size of array	Return an empty array because more than half array's element cannot be multiplied.
Array contains zero	Multiplication by zero may lead to zero result which will create problems in identifying pairs if not handled properly.
Array contains large numbers leading to Integer Overflow after multiplication	Use a data type with a larger range (e.g., long) or perform overflow checks before multiplying to prevent incorrect results.
Array with all same values	The algorithm needs to avoid using the same index twice, which may be tricky if all elements are same.
Array is of odd length and k equals half length	Evenly sized number of elements must remain, so the array size must be even for this approach to work.
Array contains negative numbers	The product of two negative numbers is positive, and the product of a positive and a negative is negative which the algorithm must properly account for when searching for pairs with the multiplication result.