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Find the Maximum Achievable Number

Easy

Find the Maximum Achievable Number

Easy

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 `num` and `t`? Can they be negative, zero, or very large?
What happens if `t` is initially zero? Should I return `num` in that case?
Are we looking for the absolute maximum achievable number, or is there a limit to the number of moves I can make?
Is there a specific data type required for the return value (e.g., int, long)?
Are there any constraints on the number of moves, or can I perform as many moves as needed as long as `t` doesn't go below zero?

Brute Force Solution

Approach

The brute force approach for this problem involves checking all possible transformations of the given number. We'll look at every single possibility to find the biggest result that fits the rules. This is like trying every door until you find the one that opens.

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

Imagine we can perform the allowed operation a certain number of times.
We need to figure out what that 'certain number' could be.
Then, for every possible number of operations (from zero up to that 'certain number'), we try it out.
Each time we try, we perform the operation one step at a time, picking one way to apply the operation and getting a result.
Then, we try a different way to apply the operation and get a different result.
We keep doing this, trying all the different sequences of operations we can.
We keep track of the largest number we managed to achieve during all of these attempts.
Finally, after trying out all the combinations, the largest number we saw is our answer.

Code Implementation

def find_maximum_achievable_number_brute_force(number, increment):
    maximum_achieved_number = number

    # Operations can happen up to increment times as described
    for number_of_operations in range(increment + 1):
        current_number = number

        # Try a simple sequence where we just add and subtract the increment
        for _ in range(number_of_operations):
            current_number = current_number + increment
            current_number = current_number - increment

        maximum_achieved_number = max(maximum_achieved_number, current_number)

        current_number = number

        for _ in range(number_of_operations):
            current_number = current_number - increment
            current_number = current_number + increment

        maximum_achieved_number = max(maximum_achieved_number, current_number)

        current_number = number

        # Key decision point: Explore adding the total possible increment.
        current_number = current_number + increment * number_of_operations
        maximum_achieved_number = max(maximum_achieved_number, current_number)

        current_number = number

        # Key decision point: Explore subtracting the total possible increment.
        current_number = current_number - increment * number_of_operations
        maximum_achieved_number = max(maximum_achieved_number, current_number)

        # This loop simulates trying a few different addition/subtraction sequences. 
        for i in range(number_of_operations):
            current_number = number
            temp_increment = increment

            for j in range(number_of_operations):
                if j % 2 == 0:
                    current_number += temp_increment
                else:
                    current_number -= temp_increment
            maximum_achieved_number = max(maximum_achieved_number, current_number)

    # Key decision point: Return the largest number achieved.
    return maximum_achieved_number

Big(O) Analysis

Time Complexity

O(infinite) – The proposed brute force approach involves checking every possible transformation. The number of operations allowed is theoretically unlimited, meaning we can repeatedly add k to num and subtract k from num resulting in an endless loop of potential transformations. Since there is no bound on the number of operations and no constraint on the values we reach, the algorithm will never terminate. Therefore the time complexity is practically infinite, as the algorithm will run indefinitely exploring an unbounded search space.

Space Complexity

O(K^N) – The brute force approach described involves exploring all possible combinations of operations. We try applying the operation a certain number of times (up to K times) where K is the maximum allowed number of operations. For each operation, we explore different ways to apply it. The algorithm keeps track of the largest number seen, and it generates many intermediate results during its operation. The total space required to hold intermediate results and to explore the combinations can grow up to O(K^N) where N could potentially represent the number of times we choose to apply the operation. Therefore, the space complexity is exponential.

Optimal Solution

Approach

The problem asks us to maximize a number based on a simple operation. Instead of trying a bunch of different options, we can directly calculate the maximum achievable number. This relies on understanding the relationship between the input numbers and the desired result.

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

Recognize that the key to maximizing the number is to perform the given operation in a specific direction.
Notice that the maximum achievable number can be obtained by simply adding the difference between the given goal and the given number to the goal.
Calculate the difference between the given number and goal, and then add the difference to the goal, which can be simplified to some basic arithmetic.

Code Implementation

def find_maximum_achievable_number(given_number, given_goal):
    # Calculate the difference between the number and the goal
    difference_between_numbers = given_goal - given_number

    # The maximum number is the goal plus the difference
    maximum_achievable_number = given_goal + difference_between_numbers

    # This is the optimized calculation, equivalent to goal + (goal - number).
    # It directly computes the maximum reachable number.
    return maximum_achievable_number

Big(O) Analysis

Time Complexity

O(1) – The solution involves a single arithmetic calculation: goal + (goal - num). This operation consists of subtraction and addition, both of which take constant time. Therefore, the time complexity is independent of the input size and remains constant. The Big O notation is O(1).

Space Complexity

O(1) – The solution involves a direct calculation using the input numbers and does not create any auxiliary data structures like arrays, lists, or hash maps. Only a few constant space variables are used to store the input values and the result of the arithmetic operations. Therefore, the space complexity is independent of the input size and remains constant. This means the space used does not grow with the input, leading to O(1) space complexity.

Edge Cases

Case	How to Handle
num and t are both zero	The maximum achievable number is 0 since any move will decrease t and increase num equally, and the formula num + 2*t handles this correctly.
t is negative	The algorithm still works correctly as it effectively decreases num while increasing the absolute value of t, handled correctly by num + 2*t.
num is a large positive number and t is a large positive number leading to integer overflow	Use a language that can handle larger integer ranges (like long long in C++, or Python's arbitrary precision integers) or implement overflow detection/prevention and handle the result accordingly.
num is a large negative number and t is a large positive number	The formula num + 2*t handles negative numbers correctly, calculating the correct achievable number.
t is zero	No moves can be made, so the maximum achievable number is num, which is what num + 2*t evaluates to when t is 0.
t is extremely large causing many iterations and potential performance issues in iterative solutions.	The direct calculation num + 2*t provides constant time complexity, avoiding iterative processes.
num is very small and t is large	num + 2*t correctly calculates a larger number by adding twice the value of t to num.
Both num and t are extremely large positive numbers that could cause calculation overflow even with large number support.	Implement overflow detection during the multiplication or addition and return a special error value or handle the overflow accordingly based on problem constraints.

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 `num` and `t`? Can they be negative, zero, or very large?
What happens if `t` is initially zero? Should I return `num` in that case?
Are we looking for the absolute maximum achievable number, or is there a limit to the number of moves I can make?
Is there a specific data type required for the return value (e.g., int, long)?
Are there any constraints on the number of moves, or can I perform as many moves as needed as long as `t` doesn't go below zero?

Brute Force Solution

Approach

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

Imagine we can perform the allowed operation a certain number of times.
We need to figure out what that 'certain number' could be.
Then, for every possible number of operations (from zero up to that 'certain number'), we try it out.
Each time we try, we perform the operation one step at a time, picking one way to apply the operation and getting a result.
Then, we try a different way to apply the operation and get a different result.
We keep doing this, trying all the different sequences of operations we can.
We keep track of the largest number we managed to achieve during all of these attempts.
Finally, after trying out all the combinations, the largest number we saw is our answer.

Code Implementation

def find_maximum_achievable_number_brute_force(number, increment):
    maximum_achieved_number = number

    # Operations can happen up to increment times as described
    for number_of_operations in range(increment + 1):
        current_number = number

        # Try a simple sequence where we just add and subtract the increment
        for _ in range(number_of_operations):
            current_number = current_number + increment
            current_number = current_number - increment

        maximum_achieved_number = max(maximum_achieved_number, current_number)

        current_number = number

        for _ in range(number_of_operations):
            current_number = current_number - increment
            current_number = current_number + increment

        maximum_achieved_number = max(maximum_achieved_number, current_number)

        current_number = number

        # Key decision point: Explore adding the total possible increment.
        current_number = current_number + increment * number_of_operations
        maximum_achieved_number = max(maximum_achieved_number, current_number)

        current_number = number

        # Key decision point: Explore subtracting the total possible increment.
        current_number = current_number - increment * number_of_operations
        maximum_achieved_number = max(maximum_achieved_number, current_number)

        # This loop simulates trying a few different addition/subtraction sequences. 
        for i in range(number_of_operations):
            current_number = number
            temp_increment = increment

            for j in range(number_of_operations):
                if j % 2 == 0:
                    current_number += temp_increment
                else:
                    current_number -= temp_increment
            maximum_achieved_number = max(maximum_achieved_number, current_number)

    # Key decision point: Return the largest number achieved.
    return maximum_achieved_number

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Recognize that the key to maximizing the number is to perform the given operation in a specific direction.
Notice that the maximum achievable number can be obtained by simply adding the difference between the given goal and the given number to the goal.
Calculate the difference between the given number and goal, and then add the difference to the goal, which can be simplified to some basic arithmetic.

Code Implementation

def find_maximum_achievable_number(given_number, given_goal):
    # Calculate the difference between the number and the goal
    difference_between_numbers = given_goal - given_number

    # The maximum number is the goal plus the difference
    maximum_achievable_number = given_goal + difference_between_numbers

    # This is the optimized calculation, equivalent to goal + (goal - number).
    # It directly computes the maximum reachable number.
    return maximum_achievable_number

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
num and t are both zero	The maximum achievable number is 0 since any move will decrease t and increase num equally, and the formula num + 2*t handles this correctly.
t is negative	The algorithm still works correctly as it effectively decreases num while increasing the absolute value of t, handled correctly by num + 2*t.
num is a large positive number and t is a large positive number leading to integer overflow	Use a language that can handle larger integer ranges (like long long in C++, or Python's arbitrary precision integers) or implement overflow detection/prevention and handle the result accordingly.
num is a large negative number and t is a large positive number	The formula num + 2*t handles negative numbers correctly, calculating the correct achievable number.
t is zero	No moves can be made, so the maximum achievable number is num, which is what num + 2*t evaluates to when t is 0.
t is extremely large causing many iterations and potential performance issues in iterative solutions.	The direct calculation num + 2*t provides constant time complexity, avoiding iterative processes.
num is very small and t is large	num + 2*t correctly calculates a larger number by adding twice the value of t to num.
Both num and t are extremely large positive numbers that could cause calculation overflow even with large number support.	Implement overflow detection during the multiplication or addition and return a special error value or handle the overflow accordingly based on problem constraints.