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Reach a Number

#653 Most AskedMedium

Reach a Number #653 Most Asked

Medium

Bloomberg LP

6 views

Topics:

Greedy AlgorithmsArrays

You are standing at position 0 on an infinite number line. There is a destination at position target.

You can make some number of moves numMoves so that:

On each move, you can either go left or right.
During the i^th move (starting from i == 1 to i == numMoves), you take i steps in the chosen direction.

Given the integer target, return the minimum number of moves required (i.e., the minimum numMoves) to reach the destination.

Example 1:

Input: target = 2
Output: 3
Explanation:
On the 1^st move, we step from 0 to 1 (1 step).
On the 2^nd move, we step from 1 to -1 (2 steps).
On the 3^rd move, we step from -1 to 2 (3 steps).

Example 2:

Input: target = 3
Output: 2
Explanation:
On the 1^st move, we step from 0 to 1 (1 step).
On the 2^nd move, we step from 1 to 3 (2 steps).

Constraints:

-10⁹ <= target <= 10⁹
target != 0

Try coding it in LeetCode

Reach a Number

#653 Most AskedMedium

Reach a Number #653 Most Asked

Medium

Bloomberg LP

6 views

Topics:

Greedy AlgorithmsArrays

You are standing at position 0 on an infinite number line. There is a destination at position target.

You can make some number of moves numMoves so that:

On each move, you can either go left or right.
During the i^th move (starting from i == 1 to i == numMoves), you take i steps in the chosen direction.

Given the integer target, return the minimum number of moves required (i.e., the minimum numMoves) to reach the destination.

Example 1:

Input: target = 2
Output: 3
Explanation:
On the 1^st move, we step from 0 to 1 (1 step).
On the 2^nd move, we step from 1 to -1 (2 steps).
On the 3^rd move, we step from -1 to 2 (3 steps).

Example 2:

Input: target = 3
Output: 2
Explanation:
On the 1^st move, we step from 0 to 1 (1 step).
On the 2^nd move, we step from 1 to 3 (2 steps).

Constraints:

-10⁹ <= target <= 10⁹
target != 0

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:

Is the target number guaranteed to be non-negative?
Can the initial position be considered as step 0?
If there are multiple ways to reach the target, should I return the minimum number of steps?
What is the maximum value of the target number, or are there any constraints on its magnitude?
If the target cannot be reached, what should I return?

Brute Force Solution

Approach

The brute force method explores every possible way to reach the target number. We can think of each move as either adding or subtracting, and the approach tries all combinations of adding and subtracting up to a certain number of steps to see if any result in the target.

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

Start from zero.
Try adding 1, and then subtracting 1. These are your first two possibilities.
For each of those possibilities, try adding 2 and subtracting 2. Now you have four possibilities.
Keep doing this, adding and subtracting the next number in the sequence to each of your existing possibilities.
With each addition or subtraction, check if you've reached the target number.
If you reach the target, you've found one way to do it. If you continue this process for a certain number of steps, you'll eventually explore all possibilities within that range.
After trying all combinations up to a certain limit of steps, check the smallest number of steps it took to reach the target number.
If no combination of adds and subtracts reached the target within the step limit, the problem might not be solvable with this method within the explored range.

Code Implementation

def reach_number_brute_force(target):
    max_steps = 15 # Define an arbitrary limit for steps
    queue = [(0, 0)] # (current_number, steps_taken)
    
    for step_count in range(1, max_steps + 1):
        new_queue = []
        for current_number, steps_taken in queue:
            # Try adding the current step number
            next_number_add = current_number + step_count

            # Check if the target is reached
            if next_number_add == target:
                return steps_taken + 1
            new_queue.append((next_number_add, steps_taken + 1))

            # Try subtracting the current step number
            next_number_subtract = current_number - step_count

            # Check if the target is reached
            if next_number_subtract == target:
                return steps_taken + 1
            new_queue.append((next_number_subtract, steps_taken + 1))

        queue = new_queue

    return -1 # Target not reached within max_steps

Big(O) Analysis

Time Complexity

O(2^n) – The brute force approach explores all possible combinations of adding and subtracting numbers from 1 up to n. At each step i, we have two choices: add i or subtract i. This creates a binary tree of possibilities. After n steps, we have explored 2^n possible paths. Therefore, the time complexity is O(2^n) because we potentially visit each of these 2^n nodes in the search tree.

Space Complexity

O(2^N) – The brute force approach explores all possible combinations of adding and subtracting numbers from 1 up to N, where N is the number of steps. In each step, the number of possibilities doubles (add or subtract), leading to a branching factor of 2. Therefore, the maximum number of possibilities to store at any given time is proportional to 2 raised to the power of N. Consequently, the space complexity becomes O(2^N) as we need to store these intermediate results during the exploration. Each of these intermediate results represent an auxiliary data structure to contribute to the space complexity.

Optimal Solution

Approach

The goal is to reach a specific number by adding or subtracting consecutive numbers starting from 1. Instead of trying every possible combination of pluses and minuses, we figure out the closest sum and then adjust only what's necessary. This significantly reduces the calculations.

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

First, calculate the sum of consecutive numbers starting from 1 until the sum is equal to or greater than the target number.
If the sum is exactly equal to the target, we're done. The number of steps we took to reach the sum is the answer.
If the sum is greater than the target, find the difference between the sum and the target.
If the difference is an even number, then we can change the sign of a number in the sequence to make the sum equal to the target. The number of steps taken so far is the answer.
If the difference is an odd number, we need to keep adding numbers until the difference becomes even. Each time we add a number, we are essentially adding a new step to the total number of steps required.
Repeat the process in the previous step until the difference between the current sum and the target becomes an even number. The total number of steps is then the answer.

Code Implementation

def reach_a_number(target):    target = abs(target)
    step_count = 0
    current_sum = 0

    while current_sum < target:
        step_count += 1
        current_sum += step_count

    # If the sum equals the target, we're done.
    if current_sum == target:
        return step_count

    difference = current_sum - target

    # If the difference is even, we can flip a sign.
    if difference % 2 == 0:
        return step_count

    # Need to find when the difference becomes even.
    while difference % 2 != 0:
        step_count += 1
        current_sum += step_count
        difference = current_sum - target

    return step_count

Big(O) Analysis

Time Complexity

O(sqrt(target)) – The dominant operation is incrementing the sum until it either equals or exceeds the target. The sum is calculated as 1 + 2 + 3 + ... + n, which is approximately n*(n+1)/2. To find the 'n' where the sum exceeds the target, we solve n*(n+1)/2 >= target. This means n is proportional to the square root of the target. The while loop in step 5 iterates at most a few times after n is found to ensure the difference between the sum and the target is even, so that does not change the overall complexity. Therefore the time complexity is O(sqrt(target)).

Space Complexity

O(1) – The algorithm described uses a constant amount of extra space. It calculates the sum iteratively and stores intermediate values such as the current sum, the difference, and the number of steps. These variables occupy a fixed amount of memory, irrespective of the target number. Therefore, the auxiliary space complexity is O(1).

Edge Cases

Case	How to Handle
Target is 0	The algorithm should correctly identify the steps needed to reach 0, possibly involving moving back and forth.
Target is a large positive number	The algorithm should scale efficiently without integer overflow issues by potentially using long data type or an iterative approach.
Target is a large negative number	The algorithm should correctly handle negative target values, potentially by considering the absolute value of the target.
Integer overflow during step calculation	Use a long data type for step and sum calculations to prevent integer overflow issues.
No solution possible within reasonable bounds	Include a condition to check if the sum is eventually greater than or equal to the target to avoid infinite loop.
Steps needed result in stack overflow with recursive solution	Convert the solution to an iterative approach instead of recursion to avoid stack overflow errors.
Target requires a very large number of steps	Ensure the while loop or iterative process terminates when the solution is found or a defined limit is exceeded.
The target is negative, but the sum is converging towards positive infinity and vice versa.	Handle the case where the sum and target have opposite signs; negate sum to continue to approach target.

Target is 0

How to Handle:

The algorithm should correctly identify the steps needed to reach 0, possibly involving moving back and forth.

Target is a large positive number

How to Handle:

The algorithm should scale efficiently without integer overflow issues by potentially using long data type or an iterative approach.

Target is a large negative number

How to Handle:

The algorithm should correctly handle negative target values, potentially by considering the absolute value of the target.

Integer overflow during step calculation

How to Handle:

Use a long data type for step and sum calculations to prevent integer overflow issues.

No solution possible within reasonable bounds

How to Handle:

Include a condition to check if the sum is eventually greater than or equal to the target to avoid infinite loop.

Steps needed result in stack overflow with recursive solution

How to Handle:

Convert the solution to an iterative approach instead of recursion to avoid stack overflow errors.

Target requires a very large number of steps

How to Handle:

Ensure the while loop or iterative process terminates when the solution is found or a defined limit is exceeded.

The target is negative, but the sum is converging towards positive infinity and vice versa.

How to Handle:

Handle the case where the sum and target have opposite signs; negate sum to continue to approach target.

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:

Is the target number guaranteed to be non-negative?
Can the initial position be considered as step 0?
If there are multiple ways to reach the target, should I return the minimum number of steps?
What is the maximum value of the target number, or are there any constraints on its magnitude?
If the target cannot be reached, what should I return?

Brute Force Solution

Approach

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

Start from zero.
Try adding 1, and then subtracting 1. These are your first two possibilities.
For each of those possibilities, try adding 2 and subtracting 2. Now you have four possibilities.
Keep doing this, adding and subtracting the next number in the sequence to each of your existing possibilities.
With each addition or subtraction, check if you've reached the target number.
If you reach the target, you've found one way to do it. If you continue this process for a certain number of steps, you'll eventually explore all possibilities within that range.
After trying all combinations up to a certain limit of steps, check the smallest number of steps it took to reach the target number.
If no combination of adds and subtracts reached the target within the step limit, the problem might not be solvable with this method within the explored range.

Code Implementation

def reach_number_brute_force(target):
    max_steps = 15 # Define an arbitrary limit for steps
    queue = [(0, 0)] # (current_number, steps_taken)
    
    for step_count in range(1, max_steps + 1):
        new_queue = []
        for current_number, steps_taken in queue:
            # Try adding the current step number
            next_number_add = current_number + step_count

            # Check if the target is reached
            if next_number_add == target:
                return steps_taken + 1
            new_queue.append((next_number_add, steps_taken + 1))

            # Try subtracting the current step number
            next_number_subtract = current_number - step_count

            # Check if the target is reached
            if next_number_subtract == target:
                return steps_taken + 1
            new_queue.append((next_number_subtract, steps_taken + 1))

        queue = new_queue

    return -1 # Target not reached within max_steps

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

First, calculate the sum of consecutive numbers starting from 1 until the sum is equal to or greater than the target number.
If the sum is exactly equal to the target, we're done. The number of steps we took to reach the sum is the answer.
If the sum is greater than the target, find the difference between the sum and the target.
If the difference is an even number, then we can change the sign of a number in the sequence to make the sum equal to the target. The number of steps taken so far is the answer.
If the difference is an odd number, we need to keep adding numbers until the difference becomes even. Each time we add a number, we are essentially adding a new step to the total number of steps required.
Repeat the process in the previous step until the difference between the current sum and the target becomes an even number. The total number of steps is then the answer.

Code Implementation

def reach_a_number(target):    target = abs(target)
    step_count = 0
    current_sum = 0

    while current_sum < target:
        step_count += 1
        current_sum += step_count

    # If the sum equals the target, we're done.
    if current_sum == target:
        return step_count

    difference = current_sum - target

    # If the difference is even, we can flip a sign.
    if difference % 2 == 0:
        return step_count

    # Need to find when the difference becomes even.
    while difference % 2 != 0:
        step_count += 1
        current_sum += step_count
        difference = current_sum - target

    return step_count

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Target is 0	The algorithm should correctly identify the steps needed to reach 0, possibly involving moving back and forth.
Target is a large positive number	The algorithm should scale efficiently without integer overflow issues by potentially using long data type or an iterative approach.
Target is a large negative number	The algorithm should correctly handle negative target values, potentially by considering the absolute value of the target.
Integer overflow during step calculation	Use a long data type for step and sum calculations to prevent integer overflow issues.
No solution possible within reasonable bounds	Include a condition to check if the sum is eventually greater than or equal to the target to avoid infinite loop.
Steps needed result in stack overflow with recursive solution	Convert the solution to an iterative approach instead of recursion to avoid stack overflow errors.
Target requires a very large number of steps	Ensure the while loop or iterative process terminates when the solution is found or a defined limit is exceeded.
The target is negative, but the sum is converging towards positive infinity and vice versa.	Handle the case where the sum and target have opposite signs; negate sum to continue to approach target.

Target is 0

How to Handle:

The algorithm should correctly identify the steps needed to reach 0, possibly involving moving back and forth.

Target is a large positive number

How to Handle:

The algorithm should scale efficiently without integer overflow issues by potentially using long data type or an iterative approach.

Target is a large negative number

How to Handle:

The algorithm should correctly handle negative target values, potentially by considering the absolute value of the target.

Integer overflow during step calculation

How to Handle:

Use a long data type for step and sum calculations to prevent integer overflow issues.

No solution possible within reasonable bounds

How to Handle:

Include a condition to check if the sum is eventually greater than or equal to the target to avoid infinite loop.

Steps needed result in stack overflow with recursive solution

How to Handle:

Convert the solution to an iterative approach instead of recursion to avoid stack overflow errors.

Target requires a very large number of steps

How to Handle:

Ensure the while loop or iterative process terminates when the solution is found or a defined limit is exceeded.

The target is negative, but the sum is converging towards positive infinity and vice versa.

How to Handle:

Handle the case where the sum and target have opposite signs; negate sum to continue to approach target.

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