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Keep Multiplying Found Values by Two

Easy

Asked by:

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Topics:

Arrays

You are given an array of integers nums. You are also given an integer original which is the first number that needs to be searched for in nums.

You then do the following steps:

If original is found in nums, multiply it by two (i.e., set original = 2 * original).
Otherwise, stop the process.
Repeat this process with the new number as long as you keep finding the number.

Return the final value of original.

Example 1:

Input: nums = [5,3,6,1,12], original = 3
Output: 24
Explanation: 
- 3 is found in nums. 3 is multiplied by 2 to obtain 6.
- 6 is found in nums. 6 is multiplied by 2 to obtain 12.
- 12 is found in nums. 12 is multiplied by 2 to obtain 24.
- 24 is not found in nums. Thus, 24 is returned.

Example 2:

Input: nums = [2,7,9], original = 4
Output: 4
Explanation:
- 4 is not found in nums. Thus, 4 is returned.

Constraints:

1 <= nums.length <= 1000
1 <= nums[i], original <= 1000

Try coding it in LeetCode

Keep Multiplying Found Values by Two

Easy

Asked by:

9 views

Topics:

Arrays

You are given an array of integers nums. You are also given an integer original which is the first number that needs to be searched for in nums.

You then do the following steps:

If original is found in nums, multiply it by two (i.e., set original = 2 * original).
Otherwise, stop the process.
Repeat this process with the new number as long as you keep finding the number.

Return the final value of original.

Example 1:

Input: nums = [5,3,6,1,12], original = 3
Output: 24
Explanation: 
- 3 is found in nums. 3 is multiplied by 2 to obtain 6.
- 6 is found in nums. 6 is multiplied by 2 to obtain 12.
- 12 is found in nums. 12 is multiplied by 2 to obtain 24.
- 24 is not found in nums. Thus, 24 is returned.

Example 2:

Input: nums = [2,7,9], original = 4
Output: 4
Explanation:
- 4 is not found in nums. Thus, 4 is returned.

Constraints:

1 <= nums.length <= 1000
1 <= nums[i], original <= 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 expected behavior if the initial `original` value is not found in the `nums` array?
Can the input array `nums` contain duplicate values, and if so, how should they be handled?
What are the possible ranges for values in the `nums` array and the initial `original` value? Can they be negative or zero?
Is there a limit to the number of times I should multiply `original` by two if it's consistently found in `nums`?
Is the input array `nums` guaranteed to be sorted?

Brute Force Solution

Approach

We're given a starting number and a collection of numbers. The brute force approach involves going through the numbers one by one and multiplying our starting number by two whenever we find a match.

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

Start with the given starting number.
Look at the first number in the collection.
If it's the same as our current number, multiply our current number by two.
If it's not the same, do nothing.
Move on to the next number in the collection and repeat the comparison and multiplication steps.
Continue this process until you've checked every number in the collection.
The final value of our starting number after all the checks is the result.

Code Implementation

def keep_multiplying_found_values_by_two(numbers_to_check, starting_number):
    current_number = starting_number

    for number_in_collection in numbers_to_check:
        # If a match is found, double current_number
        if number_in_collection == current_number:

            current_number *= 2
   
    return current_number

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the collection of numbers once. The input size, n, represents the number of elements in this collection. For each element, a single comparison operation is performed. Therefore, the number of operations grows linearly with the input size n, resulting in a time complexity of O(n).

Space Complexity

O(1) – The provided solution iterates through the collection of numbers, but it doesn't create any auxiliary data structures to store intermediate results or track visited numbers. Only the starting number, which is modified in place, is being stored. Therefore, the algorithm uses a constant amount of extra space, regardless of the size of the input collection. The space complexity is O(1).

Optimal Solution

Approach

The most efficient way to solve this is to check if each number in the list exists in a set. If it does, we double it and keep looking for the new number in the set until it's no longer found. This avoids having to check every number multiple times.

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

First, make a collection (like a bag) of all the numbers in the original list so we can quickly check if a number is present.
Start with the given starting value.
See if the current number is in our collection of original numbers. If it is not, then we are done.
If the current number is in the collection, double it.
Repeat steps 3 and 4 with the new doubled number until the doubled number is no longer in the collection.
Once we have a number that's not in the collection, that's our final answer.

Code Implementation

def keep_multiplying_found_values_by_two(numbers, original_value):
    number_set = set(numbers)

    current_number = original_value

    # Keep going until we find a number not in the set.
    while True:
        # Check if the current number exists in the original list.
        if current_number not in number_set:
            
            return current_number

        # Double the current number since it was found.
        current_number *= 2

Big(O) Analysis

Time Complexity

O(n) – The initial step of creating a set from the input array nums of size n takes O(n) time. Then, starting with original, we repeatedly check if a doubled value exists in the set. In the worst-case scenario, we might double the value until it exceeds the maximum possible integer value, or until we reach a value not present in the original set. However, each doubling step effectively removes a value from the potential set of numbers we need to consider. Because the set is constructed from n elements, we can perform at most n doubling operations. Therefore, the while loop runs at most n times. The overall time complexity is dominated by creating the set and the potential for n lookups and doublings, leading to O(n).

Space Complexity

O(N) – The described solution uses a collection (like a bag or set) to store all numbers from the original input list. This collection allows for quick checks to see if a number is present. In the worst case, all N numbers from the input list are unique and will be stored in this collection. Therefore, the auxiliary space used by this collection is proportional to the number of elements in the input list, resulting in O(N) space complexity.

Edge Cases

Case	How to Handle
Empty nums array	Return the original value if the array is empty as there are no numbers to multiply.
nums array is null	Throw IllegalArgumentException or return a default value (e.g., original) based on problem specifications.
Large nums array with many duplicates	Use a HashSet to efficiently check for presence and avoid redundant multiplications.
Target value is zero	The loop will terminate immediately because multiplying by two will never result in zero (unless array contains zero and its handled elsewhere).
No number in nums results in multiplication to be another number in nums	The original value is returned since no further multiplication is possible.
Integer overflow during multiplication	Check if the multiplication result exceeds the maximum integer value and return an appropriate error or default value based on problem definition.
nums array contains very large positive integers	The multiplication may lead to integer overflow, check the multiplied value against Integer.MAX_VALUE and respond according to the problem description.
nums array contains only the initial value itself	Since there are no other numbers to derive from the original value, it should return the original value

Empty nums array

How to Handle:

Return the original value if the array is empty as there are no numbers to multiply.

nums array is null

How to Handle:

Throw IllegalArgumentException or return a default value (e.g., original) based on problem specifications.

Large nums array with many duplicates

How to Handle:

Use a HashSet to efficiently check for presence and avoid redundant multiplications.

Target value is zero

How to Handle:

The loop will terminate immediately because multiplying by two will never result in zero (unless array contains zero and its handled elsewhere).

No number in nums results in multiplication to be another number in nums

How to Handle:

The original value is returned since no further multiplication is possible.

Integer overflow during multiplication

How to Handle:

Check if the multiplication result exceeds the maximum integer value and return an appropriate error or default value based on problem definition.

nums array contains very large positive integers

How to Handle:

The multiplication may lead to integer overflow, check the multiplied value against Integer.MAX_VALUE and respond according to the problem description.

nums array contains only the initial value itself

How to Handle:

Since there are no other numbers to derive from the original value, it should return the original value

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 expected behavior if the initial `original` value is not found in the `nums` array?
Can the input array `nums` contain duplicate values, and if so, how should they be handled?
What are the possible ranges for values in the `nums` array and the initial `original` value? Can they be negative or zero?
Is there a limit to the number of times I should multiply `original` by two if it's consistently found in `nums`?
Is the input array `nums` guaranteed to be sorted?

Brute Force Solution

Approach

We're given a starting number and a collection of numbers. The brute force approach involves going through the numbers one by one and multiplying our starting number by two whenever we find a match.

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

Start with the given starting number.
Look at the first number in the collection.
If it's the same as our current number, multiply our current number by two.
If it's not the same, do nothing.
Move on to the next number in the collection and repeat the comparison and multiplication steps.
Continue this process until you've checked every number in the collection.
The final value of our starting number after all the checks is the result.

Code Implementation

def keep_multiplying_found_values_by_two(numbers_to_check, starting_number):
    current_number = starting_number

    for number_in_collection in numbers_to_check:
        # If a match is found, double current_number
        if number_in_collection == current_number:

            current_number *= 2
   
    return current_number

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

First, make a collection (like a bag) of all the numbers in the original list so we can quickly check if a number is present.
Start with the given starting value.
See if the current number is in our collection of original numbers. If it is not, then we are done.
If the current number is in the collection, double it.
Repeat steps 3 and 4 with the new doubled number until the doubled number is no longer in the collection.
Once we have a number that's not in the collection, that's our final answer.

Code Implementation

def keep_multiplying_found_values_by_two(numbers, original_value):
    number_set = set(numbers)

    current_number = original_value

    # Keep going until we find a number not in the set.
    while True:
        # Check if the current number exists in the original list.
        if current_number not in number_set:
            
            return current_number

        # Double the current number since it was found.
        current_number *= 2

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty nums array	Return the original value if the array is empty as there are no numbers to multiply.
nums array is null	Throw IllegalArgumentException or return a default value (e.g., original) based on problem specifications.
Large nums array with many duplicates	Use a HashSet to efficiently check for presence and avoid redundant multiplications.
Target value is zero	The loop will terminate immediately because multiplying by two will never result in zero (unless array contains zero and its handled elsewhere).
No number in nums results in multiplication to be another number in nums	The original value is returned since no further multiplication is possible.
Integer overflow during multiplication	Check if the multiplication result exceeds the maximum integer value and return an appropriate error or default value based on problem definition.
nums array contains very large positive integers	The multiplication may lead to integer overflow, check the multiplied value against Integer.MAX_VALUE and respond according to the problem description.
nums array contains only the initial value itself	Since there are no other numbers to derive from the original value, it should return the original value

Empty nums array

How to Handle:

Return the original value if the array is empty as there are no numbers to multiply.

nums array is null

How to Handle:

Throw IllegalArgumentException or return a default value (e.g., original) based on problem specifications.

Large nums array with many duplicates

How to Handle:

Use a HashSet to efficiently check for presence and avoid redundant multiplications.

Target value is zero

How to Handle:

The loop will terminate immediately because multiplying by two will never result in zero (unless array contains zero and its handled elsewhere).

No number in nums results in multiplication to be another number in nums

How to Handle:

The original value is returned since no further multiplication is possible.

Integer overflow during multiplication

How to Handle:

Check if the multiplication result exceeds the maximum integer value and return an appropriate error or default value based on problem definition.

nums array contains very large positive integers

How to Handle:

The multiplication may lead to integer overflow, check the multiplied value against Integer.MAX_VALUE and respond according to the problem description.

nums array contains only the initial value itself

How to Handle:

Since there are no other numbers to derive from the original value, it should return the original value