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Number of Ways to Reach a Position After Exactly k Steps

Medium

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Topics:

Dynamic Programming

You are given two positive integers startPos and endPos. Initially, you are standing at position startPos on an infinite number line. With one step, you can move either one position to the left, or one position to the right.

Given a positive integer k, return the number of different ways to reach the position endPos starting from startPos, such that you perform exactly k steps. Since the answer may be very large, return it modulo 10⁹ + 7.

Two ways are considered different if the order of the steps made is not exactly the same.

Note that the number line includes negative integers.

Example 1:

Input: startPos = 1, endPos = 2, k = 3
Output: 3
Explanation: We can reach position 2 from 1 in exactly 3 steps in three ways:
- 1 -> 2 -> 3 -> 2.
- 1 -> 2 -> 1 -> 2.
- 1 -> 0 -> 1 -> 2.
It can be proven that no other way is possible, so we return 3.

Example 2:

Input: startPos = 2, endPos = 5, k = 10
Output: 0
Explanation: It is impossible to reach position 5 from position 2 in exactly 10 steps.

Constraints:

1 <= startPos, endPos, k <= 1000

Try coding it in LeetCode

Number of Ways to Reach a Position After Exactly k Steps

Medium

Asked by:

19 views

Topics:

Dynamic Programming

Two ways are considered different if the order of the steps made is not exactly the same.

Note that the number line includes negative integers.

Example 1:

Input: startPos = 1, endPos = 2, k = 3
Output: 3
Explanation: We can reach position 2 from 1 in exactly 3 steps in three ways:
- 1 -> 2 -> 3 -> 2.
- 1 -> 2 -> 1 -> 2.
- 1 -> 0 -> 1 -> 2.
It can be proven that no other way is possible, so we return 3.

Example 2:

Input: startPos = 2, endPos = 5, k = 10
Output: 0
Explanation: It is impossible to reach position 5 from position 2 in exactly 10 steps.

Constraints:

1 <= startPos, endPos, k <= 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:

Can the starting position, end position, and the number of steps k be negative?
What are the maximum values for the start position, end position, and k? Are there any constraints on their ranges?
If it's impossible to reach the end position in exactly k steps, what should I return (e.g., 0, -1, or throw an exception)?
Is it possible for the start position and end position to be the same?
Does the problem statement imply that each step must be of size 1 (moving only one unit left or right)? Or, can the step size vary?

Brute Force Solution

Approach

We're figuring out how many ways to reach a target spot by taking a specific number of steps, where each step can be either forward or backward. The brute force method tries out absolutely every single path of steps we could take.

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

Start at the beginning position.
For each step we are allowed to take, consider moving one position forward and one position backward. Each of these is a different potential path.
Continue making choices for each of the remaining steps. Each time we make a choice, we are exploring a new possible route.
After taking the allowed number of steps, check if we arrived at the target position.
If we did reach the target position, count that path as a valid way to get there.
Repeat these steps, exploring all possible combinations of forward and backward movements until we've exhausted every possibility.
Finally, add up all the valid paths that landed us at the target position after the correct number of steps. That total is our answer.

Code Implementation

def number_of_ways_brute_force(
    start_position, end_position, number_of_steps
):
    number_of_ways = 0

    def explore_paths(current_position, steps_remaining):
        nonlocal number_of_ways

        # If we've taken all the steps, check if we're at the end
        if steps_remaining == 0:
            if current_position == end_position:
                number_of_ways += 1
            return

        # Explore moving forward
        explore_paths(
            current_position + 1, steps_remaining - 1
        )

        # Explore moving backward
        explore_paths(
            current_position - 1, steps_remaining - 1
        )

    explore_paths(start_position, number_of_steps)
    return number_of_ways

Big(O) Analysis

Time Complexity

O(2^k) – The brute force solution explores every possible combination of forward and backward steps. For each of the k steps, we have two choices: move forward or move backward. This creates a binary tree of possibilities where each level represents a step. Therefore, the total number of paths explored is 2 multiplied by itself k times, resulting in 2^k. The time complexity is directly proportional to the number of paths explored, making it O(2^k).

Space Complexity

O(k) – The brute force approach explores all possible paths using recursion. Each step taken adds a new frame to the call stack, representing a choice between moving forward or backward. With 'k' steps, the maximum depth of the recursion, and therefore the maximum number of stack frames, will be 'k'. Each stack frame stores the current position and remaining steps, contributing to auxiliary space.

Optimal Solution

Approach

The goal is to find how many ways you can reach a target position from a starting position in exactly a certain number of steps, where each step can be either forward or backward. The efficient approach avoids trying every single path and instead uses a smart way to calculate the possible paths using repeated subproblems.

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

Notice that the number of steps to the left and right determine the final position. The number of steps is fixed, so if we know how many steps to the right we take, we automatically know how many steps to the left we take.
Recognize that not every combination of left and right steps is possible. For example, if the target is very far away and the number of steps is small, there is no way to reach the target.
Imagine building up the number of ways to reach a position using smaller step counts. Each position can be reached either from one step to the left or from one step to the right in one fewer step.
Create a table that stores how many ways to reach each position using each possible step count. The table is filled starting from a step count of zero all the way up to the given step count.
For each step count and each position, the number of ways to reach that position is equal to the sum of the ways to reach the positions one step away in one fewer step.
Return the number of ways to reach the target position using the given step count. That value is stored in the table.

Code Implementation

def number_of_ways_to_reach_a_position_after_exactly_k_steps(start_position, end_position, number_of_steps):
    distance = abs(end_position - start_position)

    # If distance exceeds steps, target is unreachable.
    if distance > number_of_steps:
        return 0

    # Optimization: If difference is odd/even.
    if (number_of_steps - distance) % 2 != 0:
        return 0

    max_position = 500  # Define a reasonable bound.
    dp_table = [[0] * (2 * max_position + 1) for _ in range(number_of_steps + 1)]

    # Initialize the starting position with one way (no steps taken).
    dp_table[0][start_position + max_position] = 1

    for step_count in range(1, number_of_steps + 1):
        for current_position in range(-max_position, max_position + 1):
            index = current_position + max_position

            # Consider moving one step to the left.
            if current_position - 1 >= -max_position:
                dp_table[step_count][index] += dp_table[step_count - 1][index - 1]

            # Consider moving one step to the right.
            if current_position + 1 <= max_position:
                dp_table[step_count][index] += dp_table[step_count - 1][index + 1]

    # Result is the number of ways to reach end position after k steps.
    return dp_table[number_of_steps][end_position + max_position]

Big(O) Analysis

Time Complexity

O(k * (k + abs(target - startPos))) – The algorithm utilizes dynamic programming to compute the number of ways to reach the target. It constructs a table (implicitly or explicitly) where rows represent the number of steps from 0 to k, and columns represent positions centered around the starting position. The width of this table depends on k (the number of steps) and the distance between the start and target positions since we only need to consider positions reachable within k steps. Therefore, the time complexity is determined by iterating through this table, resulting in O(k * (k + abs(target - startPos))) where k represents the number of steps and abs(target - startPos) is the absolute difference between the target and start position.

Space Complexity

O(k * range) – The solution utilizes a table (dynamic programming array) to store the number of ways to reach each position using each possible step count. The number of rows in the table is determined by the number of steps, k. The number of columns depends on the possible range of positions, which is influenced by k (since you can move at most k steps away from the starting position) and the difference between the start and target position. Therefore the space needed is proportional to k multiplied by the range of positions reachable within k steps. The space complexity is O(k * range) where range represents the span of reachable positions.

Edge Cases

Case	How to Handle
k = 0, start != end	Return 0, as no steps are allowed and the start and end positions differ.
k = 0, start == end	Return 1, as the starting position already matches the ending position with no moves.
Large k causing potential integer overflow in calculations.	Use modulo operation to prevent integer overflow if the number of ways can be very large.
k is smaller than the absolute difference between start and end.	Return 0, as not enough steps exist to reach the target.
Large absolute difference between start and end requiring many steps.	Check for maximum recursion depth allowed to avoid stack overflow if using a recursive solution.
start and end are very large positive numbers, and k is also large.	Use dynamic programming to optimize for space and avoid recalculating previously computed results.
start and end are very large negative numbers, and k is also large.	Ensure calculations handle negative indices gracefully, possibly by offsetting to positive values.
k is odd, but the difference between start and end is even (or vice versa)	Return 0, because it's impossible to reach the target using only allowed steps.

k = 0, start != end

How to Handle:

Return 0, as no steps are allowed and the start and end positions differ.

k = 0, start == end

How to Handle:

Return 1, as the starting position already matches the ending position with no moves.

Large k causing potential integer overflow in calculations.

How to Handle:

Use modulo operation to prevent integer overflow if the number of ways can be very large.

k is smaller than the absolute difference between start and end.

How to Handle:

Return 0, as not enough steps exist to reach the target.

Large absolute difference between start and end requiring many steps.

How to Handle:

Check for maximum recursion depth allowed to avoid stack overflow if using a recursive solution.

start and end are very large positive numbers, and k is also large.

How to Handle:

Use dynamic programming to optimize for space and avoid recalculating previously computed results.

start and end are very large negative numbers, and k is also large.

How to Handle:

Ensure calculations handle negative indices gracefully, possibly by offsetting to positive values.

k is odd, but the difference between start and end is even (or vice versa)

How to Handle:

Return 0, because it's impossible to reach the target using only allowed steps.

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:

Can the starting position, end position, and the number of steps k be negative?
What are the maximum values for the start position, end position, and k? Are there any constraints on their ranges?
If it's impossible to reach the end position in exactly k steps, what should I return (e.g., 0, -1, or throw an exception)?
Is it possible for the start position and end position to be the same?
Does the problem statement imply that each step must be of size 1 (moving only one unit left or right)? Or, can the step size vary?

Brute Force Solution

Approach

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

Start at the beginning position.
For each step we are allowed to take, consider moving one position forward and one position backward. Each of these is a different potential path.
Continue making choices for each of the remaining steps. Each time we make a choice, we are exploring a new possible route.
After taking the allowed number of steps, check if we arrived at the target position.
If we did reach the target position, count that path as a valid way to get there.
Repeat these steps, exploring all possible combinations of forward and backward movements until we've exhausted every possibility.
Finally, add up all the valid paths that landed us at the target position after the correct number of steps. That total is our answer.

Code Implementation

def number_of_ways_brute_force(
    start_position, end_position, number_of_steps
):
    number_of_ways = 0

    def explore_paths(current_position, steps_remaining):
        nonlocal number_of_ways

        # If we've taken all the steps, check if we're at the end
        if steps_remaining == 0:
            if current_position == end_position:
                number_of_ways += 1
            return

        # Explore moving forward
        explore_paths(
            current_position + 1, steps_remaining - 1
        )

        # Explore moving backward
        explore_paths(
            current_position - 1, steps_remaining - 1
        )

    explore_paths(start_position, number_of_steps)
    return number_of_ways

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Notice that the number of steps to the left and right determine the final position. The number of steps is fixed, so if we know how many steps to the right we take, we automatically know how many steps to the left we take.
Recognize that not every combination of left and right steps is possible. For example, if the target is very far away and the number of steps is small, there is no way to reach the target.
Imagine building up the number of ways to reach a position using smaller step counts. Each position can be reached either from one step to the left or from one step to the right in one fewer step.
Create a table that stores how many ways to reach each position using each possible step count. The table is filled starting from a step count of zero all the way up to the given step count.
For each step count and each position, the number of ways to reach that position is equal to the sum of the ways to reach the positions one step away in one fewer step.
Return the number of ways to reach the target position using the given step count. That value is stored in the table.

Code Implementation

def number_of_ways_to_reach_a_position_after_exactly_k_steps(start_position, end_position, number_of_steps):
    distance = abs(end_position - start_position)

    # If distance exceeds steps, target is unreachable.
    if distance > number_of_steps:
        return 0

    # Optimization: If difference is odd/even.
    if (number_of_steps - distance) % 2 != 0:
        return 0

    max_position = 500  # Define a reasonable bound.
    dp_table = [[0] * (2 * max_position + 1) for _ in range(number_of_steps + 1)]

    # Initialize the starting position with one way (no steps taken).
    dp_table[0][start_position + max_position] = 1

    for step_count in range(1, number_of_steps + 1):
        for current_position in range(-max_position, max_position + 1):
            index = current_position + max_position

            # Consider moving one step to the left.
            if current_position - 1 >= -max_position:
                dp_table[step_count][index] += dp_table[step_count - 1][index - 1]

            # Consider moving one step to the right.
            if current_position + 1 <= max_position:
                dp_table[step_count][index] += dp_table[step_count - 1][index + 1]

    # Result is the number of ways to reach end position after k steps.
    return dp_table[number_of_steps][end_position + max_position]

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
k = 0, start != end	Return 0, as no steps are allowed and the start and end positions differ.
k = 0, start == end	Return 1, as the starting position already matches the ending position with no moves.
Large k causing potential integer overflow in calculations.	Use modulo operation to prevent integer overflow if the number of ways can be very large.
k is smaller than the absolute difference between start and end.	Return 0, as not enough steps exist to reach the target.
Large absolute difference between start and end requiring many steps.	Check for maximum recursion depth allowed to avoid stack overflow if using a recursive solution.
start and end are very large positive numbers, and k is also large.	Use dynamic programming to optimize for space and avoid recalculating previously computed results.
start and end are very large negative numbers, and k is also large.	Ensure calculations handle negative indices gracefully, possibly by offsetting to positive values.
k is odd, but the difference between start and end is even (or vice versa)	Return 0, because it's impossible to reach the target using only allowed steps.

k = 0, start != end

How to Handle:

Return 0, as no steps are allowed and the start and end positions differ.

k = 0, start == end

How to Handle:

Return 1, as the starting position already matches the ending position with no moves.

Large k causing potential integer overflow in calculations.

How to Handle:

Use modulo operation to prevent integer overflow if the number of ways can be very large.

k is smaller than the absolute difference between start and end.

How to Handle:

Return 0, as not enough steps exist to reach the target.

Large absolute difference between start and end requiring many steps.

How to Handle:

Check for maximum recursion depth allowed to avoid stack overflow if using a recursive solution.

start and end are very large positive numbers, and k is also large.

How to Handle:

Use dynamic programming to optimize for space and avoid recalculating previously computed results.

start and end are very large negative numbers, and k is also large.

How to Handle:

Ensure calculations handle negative indices gracefully, possibly by offsetting to positive values.

k is odd, but the difference between start and end is even (or vice versa)

How to Handle:

Return 0, because it's impossible to reach the target using only allowed steps.