Jobs

Interviews

Discussions

Number of Ways to Stay in the Same Place After Some Steps

Hard

Asked by:

20 views

Topics:

Dynamic Programming

You have a pointer at index 0 in an array of size arrLen. At each step, you can move 1 position to the left, 1 position to the right in the array, or stay in the same place (The pointer should not be placed outside the array at any time).

Given two integers steps and arrLen, return the number of ways such that your pointer is still at index 0 after exactly steps steps. Since the answer may be too large, return it modulo 10⁹ + 7.

Example 1:

Input: steps = 3, arrLen = 2
Output: 4
Explanation: There are 4 differents ways to stay at index 0 after 3 steps.
Right, Left, Stay
Stay, Right, Left
Right, Stay, Left
Stay, Stay, Stay

Example 2:

Input: steps = 2, arrLen = 4
Output: 2
Explanation: There are 2 differents ways to stay at index 0 after 2 steps
Right, Left
Stay, Stay

Example 3:

Input: steps = 4, arrLen = 2
Output: 8

Constraints:

1 <= steps <= 500
1 <= arrLen <= 10⁶

Try coding it in LeetCode

Number of Ways to Stay in the Same Place After Some Steps

Hard

Asked by:

20 views

Topics:

Dynamic Programming

Example 1:

Input: steps = 3, arrLen = 2
Output: 4
Explanation: There are 4 differents ways to stay at index 0 after 3 steps.
Right, Left, Stay
Stay, Right, Left
Right, Stay, Left
Stay, Stay, Stay

Example 2:

Input: steps = 2, arrLen = 4
Output: 2
Explanation: There are 2 differents ways to stay at index 0 after 2 steps
Right, Left
Stay, Stay

Example 3:

Input: steps = 4, arrLen = 2
Output: 8

Constraints:

1 <= steps <= 500
1 <= arrLen <= 10⁶

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 are the constraints on the number of steps and the length of the array? Specifically, what are the maximum possible values?
Can the number of steps be zero? If so, what should the return value be?
Is the array guaranteed to have at least one element?
Since the result can be very large, what should I return the result modulo?
Are the array indices 0-based or 1-based?

Brute Force Solution

Approach

The brute force method involves trying every single combination of moves possible. We explore each path until we either reach the end or exceed the allowed number of steps. At the end, we count how many paths lead us back to the starting point.

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

Start at the initial position.
Consider all possible moves: stay, move left, or move right.
For each move, repeat the process: consider all possible moves from the new position.
Keep repeating this until the number of steps taken equals the allowed number of steps.
If, after taking all the steps, you are back at the initial position, count that as one successful way.
Do this for every possible combination of moves from the very beginning.
The total count of all the successful ways is the final answer.

Code Implementation

def number_of_ways_brute_force(steps, array_length):
    number_of_successful_ways = 0

    def explore_path(current_position, steps_remaining):
        nonlocal number_of_successful_ways

        # Base case: No steps remaining
        if steps_remaining == 0:
            if current_position == 0:
                # Increment if we are back at the starting position.
                number_of_successful_ways += 1
            return

        # Explore the 'stay' option
        explore_path(current_position, steps_remaining - 1)

        # Explore moving left, if possible
        if current_position > 0:
            # Ensure we don't move past the start of the array.
            explore_path(current_position - 1, steps_remaining - 1)

        # Explore moving right, if possible
        if current_position < array_length - 1:
            # Ensure we don't move past the end of the array
            explore_path(current_position + 1, steps_remaining - 1)

    explore_path(0, steps)
    return number_of_successful_ways

Big(O) Analysis

Time Complexity

O(3^steps) – The brute force approach explores all possible combinations of moves at each step. At each step, we have three choices: move left, move right, or stay. Therefore, for 'steps' number of steps, we have 3 options at each step. Thus, we potentially explore 3 * 3 * ... * 3 (steps times) different paths. Hence, the time complexity is O(3^steps), where 'steps' represents the number of steps allowed.

Space Complexity

O(steps) – The brute force approach described uses recursion to explore all possible move combinations. Each recursive call corresponds to a step taken. The maximum depth of the recursion is equal to the number of steps allowed. Each call requires allocating space on the call stack, therefore the auxiliary space is proportional to the maximum recursion depth which is the number of steps. This gives a space complexity of O(steps).

Optimal Solution

Approach

The most efficient way to solve this is by breaking down the problem into smaller, overlapping subproblems. We use a technique to remember the answers to these subproblems, so we don't have to recalculate them repeatedly, which avoids unnecessary work.

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

Imagine you're standing on a number line, and you can only move left, right, or stay put.
Figure out how many ways you can reach a specific spot after a certain number of moves. Consider each possible previous position: from the left, from the right, or from staying in the same spot.
Notice that the number of ways to reach a spot depends only on the number of steps taken so far and your current position.
Create a table to store the results of how many ways to reach a certain position after a specific number of steps.
Start filling in the table, beginning with the base case: there's one way to be at the starting position at the beginning.
Then, use the previous results to calculate the next values. For example, to find the number of ways to reach a spot after N steps, add the number of ways you could have reached the positions immediately to the left, right, and the current spot after N-1 steps.
Remember that you cannot move past the edges of the spots you are allowed to move on, so don't try to move left when you are at the start or move right when you are at the end.
After filling the table, the answer you're looking for is the number of ways to be at the starting position after the given number of steps, which will be stored in your table.

Code Implementation

def number_of_ways_to_stay
(steps, array_length):
    modulo = 10**9 + 7
    max_column = min(steps, array_length - 1)
    # dp_table[i][j] stores number of ways to reach position j after i steps.
    dp_table = [[0] * (max_column + 1) for _ in range(steps + 1)]

    dp_table[0][0] = 1

    for current_step in range(1, steps + 1):
        for current_position in range(max_column + 1):
            # Staying put is always an option.
            dp_table[current_step][current_position] = dp_table[current_step - 1][current_position]

            # Moving left is possible if we're not at the beginning.
            if current_position > 0:
                dp_table[current_step][current_position] =
                (dp_table[current_step][current_position] +
                 dp_table[current_step - 1][current_position - 1]) % modulo

            # Moving right is possible if within bounds.
            if current_position < max_column:
                dp_table[current_step][current_position] =
                (dp_table[current_step][current_position] +
                 dp_table[current_step - 1][current_position + 1]) % modulo

    # Result is stored at dp_table[steps][0] - ways to be at position 0 after 'steps' steps.
    return dp_table[steps][0]

Big(O) Analysis

Time Complexity

O(steps * arrLen) – The described dynamic programming solution builds a table where the dimensions are dependent on the number of steps and the array length. The table has 'steps + 1' rows and 'min(steps/2 + 1, arrLen)' columns, where 'steps' is the number of steps allowed, and 'arrLen' is the length of the array. Each cell in the table is computed in constant time, by summing up to three adjacent values from the previous row. Therefore, the overall time complexity is determined by the size of the table, which is O(steps * min(steps, arrLen)). Since min(steps,arrLen) can be approximated to arrLen (in the worst case when steps > arrLen) or steps (when arrLen > steps), the complexity simplifies to O(steps * arrLen).

Space Complexity

O(steps * arrLen) – The provided solution utilizes dynamic programming with a table to store intermediate results. This table represents the number of ways to reach a certain position after a specific number of steps. The table has dimensions (steps + 1) x (arrLen), where steps is the number of steps allowed and arrLen is the length of the array. Therefore, the space required to store this table is proportional to the product of steps and arrLen. This auxiliary space dominates the overall space complexity, overshadowing any other constant space usage.

Edge Cases

Case	How to Handle
steps is zero	If steps is zero, the only way to stay in the same place is if arrayLen is at least 1; return 1 if arrayLen >= 1, otherwise return 0.
arrayLen is 1	If arrayLen is 1, the only possible move is to stay in place; return 1 if steps is even, 0 otherwise (since you must take steps in pairs).
steps exceeds arrayLen - 1	The maximum distance you can move from the start is `steps`, so cap arrayLen used in DP calculations at `min(steps + 1, arrayLen)`.
Large steps value leading to integer overflow in DP table	Use modulo arithmetic to prevent integer overflow during DP calculations using the provided modulo value.
steps and arrayLen are both very large, potentially exceeding memory limits for DP table	The DP table should be sized at `steps + 1` x `min(steps + 1, arrayLen)`, which is sufficient and prevents excessive memory usage.
No possible way to return to the starting position	The algorithm inherently covers cases where no valid solution exists; it will return 0 if no path returns to the starting position.
steps is odd and arrayLen is 1	When arrayLen is 1 and steps is odd it is impossible to stay in the same place, so return 0.
Steps is a very large number close to the maximum integer limit.	Ensure intermediate calculations within the DP table do not exceed the maximum integer value even with modulo applied, as repeated additions might still lead to issues.

steps is zero

How to Handle:

If steps is zero, the only way to stay in the same place is if arrayLen is at least 1; return 1 if arrayLen >= 1, otherwise return 0.

arrayLen is 1

How to Handle:

If arrayLen is 1, the only possible move is to stay in place; return 1 if steps is even, 0 otherwise (since you must take steps in pairs).

steps exceeds arrayLen - 1

How to Handle:

The maximum distance you can move from the start is `steps`, so cap arrayLen used in DP calculations at `min(steps + 1, arrayLen)`.

Large steps value leading to integer overflow in DP table

How to Handle:

Use modulo arithmetic to prevent integer overflow during DP calculations using the provided modulo value.

steps and arrayLen are both very large, potentially exceeding memory limits for DP table

How to Handle:

The DP table should be sized at `steps + 1` x `min(steps + 1, arrayLen)`, which is sufficient and prevents excessive memory usage.

No possible way to return to the starting position

How to Handle:

The algorithm inherently covers cases where no valid solution exists; it will return 0 if no path returns to the starting position.

steps is odd and arrayLen is 1

How to Handle:

When arrayLen is 1 and steps is odd it is impossible to stay in the same place, so return 0.

Steps is a very large number close to the maximum integer limit.

How to Handle:

Ensure intermediate calculations within the DP table do not exceed the maximum integer value even with modulo applied, as repeated additions might still lead to issues.

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 constraints on the number of steps and the length of the array? Specifically, what are the maximum possible values?
Can the number of steps be zero? If so, what should the return value be?
Is the array guaranteed to have at least one element?
Since the result can be very large, what should I return the result modulo?
Are the array indices 0-based or 1-based?

Brute Force Solution

Approach

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

Start at the initial position.
Consider all possible moves: stay, move left, or move right.
For each move, repeat the process: consider all possible moves from the new position.
Keep repeating this until the number of steps taken equals the allowed number of steps.
If, after taking all the steps, you are back at the initial position, count that as one successful way.
Do this for every possible combination of moves from the very beginning.
The total count of all the successful ways is the final answer.

Code Implementation

def number_of_ways_brute_force(steps, array_length):
    number_of_successful_ways = 0

    def explore_path(current_position, steps_remaining):
        nonlocal number_of_successful_ways

        # Base case: No steps remaining
        if steps_remaining == 0:
            if current_position == 0:
                # Increment if we are back at the starting position.
                number_of_successful_ways += 1
            return

        # Explore the 'stay' option
        explore_path(current_position, steps_remaining - 1)

        # Explore moving left, if possible
        if current_position > 0:
            # Ensure we don't move past the start of the array.
            explore_path(current_position - 1, steps_remaining - 1)

        # Explore moving right, if possible
        if current_position < array_length - 1:
            # Ensure we don't move past the end of the array
            explore_path(current_position + 1, steps_remaining - 1)

    explore_path(0, steps)
    return number_of_successful_ways

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Imagine you're standing on a number line, and you can only move left, right, or stay put.
Figure out how many ways you can reach a specific spot after a certain number of moves. Consider each possible previous position: from the left, from the right, or from staying in the same spot.
Notice that the number of ways to reach a spot depends only on the number of steps taken so far and your current position.
Create a table to store the results of how many ways to reach a certain position after a specific number of steps.
Start filling in the table, beginning with the base case: there's one way to be at the starting position at the beginning.
Then, use the previous results to calculate the next values. For example, to find the number of ways to reach a spot after N steps, add the number of ways you could have reached the positions immediately to the left, right, and the current spot after N-1 steps.
Remember that you cannot move past the edges of the spots you are allowed to move on, so don't try to move left when you are at the start or move right when you are at the end.
After filling the table, the answer you're looking for is the number of ways to be at the starting position after the given number of steps, which will be stored in your table.

Code Implementation

def number_of_ways_to_stay
(steps, array_length):
    modulo = 10**9 + 7
    max_column = min(steps, array_length - 1)
    # dp_table[i][j] stores number of ways to reach position j after i steps.
    dp_table = [[0] * (max_column + 1) for _ in range(steps + 1)]

    dp_table[0][0] = 1

    for current_step in range(1, steps + 1):
        for current_position in range(max_column + 1):
            # Staying put is always an option.
            dp_table[current_step][current_position] = dp_table[current_step - 1][current_position]

            # Moving left is possible if we're not at the beginning.
            if current_position > 0:
                dp_table[current_step][current_position] =
                (dp_table[current_step][current_position] +
                 dp_table[current_step - 1][current_position - 1]) % modulo

            # Moving right is possible if within bounds.
            if current_position < max_column:
                dp_table[current_step][current_position] =
                (dp_table[current_step][current_position] +
                 dp_table[current_step - 1][current_position + 1]) % modulo

    # Result is stored at dp_table[steps][0] - ways to be at position 0 after 'steps' steps.
    return dp_table[steps][0]

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
steps is zero	If steps is zero, the only way to stay in the same place is if arrayLen is at least 1; return 1 if arrayLen >= 1, otherwise return 0.
arrayLen is 1	If arrayLen is 1, the only possible move is to stay in place; return 1 if steps is even, 0 otherwise (since you must take steps in pairs).
steps exceeds arrayLen - 1	The maximum distance you can move from the start is `steps`, so cap arrayLen used in DP calculations at `min(steps + 1, arrayLen)`.
Large steps value leading to integer overflow in DP table	Use modulo arithmetic to prevent integer overflow during DP calculations using the provided modulo value.
steps and arrayLen are both very large, potentially exceeding memory limits for DP table	The DP table should be sized at `steps + 1` x `min(steps + 1, arrayLen)`, which is sufficient and prevents excessive memory usage.
No possible way to return to the starting position	The algorithm inherently covers cases where no valid solution exists; it will return 0 if no path returns to the starting position.
steps is odd and arrayLen is 1	When arrayLen is 1 and steps is odd it is impossible to stay in the same place, so return 0.
Steps is a very large number close to the maximum integer limit.	Ensure intermediate calculations within the DP table do not exceed the maximum integer value even with modulo applied, as repeated additions might still lead to issues.

steps is zero

How to Handle:

If steps is zero, the only way to stay in the same place is if arrayLen is at least 1; return 1 if arrayLen >= 1, otherwise return 0.

arrayLen is 1

How to Handle:

If arrayLen is 1, the only possible move is to stay in place; return 1 if steps is even, 0 otherwise (since you must take steps in pairs).

steps exceeds arrayLen - 1

How to Handle:

The maximum distance you can move from the start is `steps`, so cap arrayLen used in DP calculations at `min(steps + 1, arrayLen)`.

Large steps value leading to integer overflow in DP table

How to Handle:

Use modulo arithmetic to prevent integer overflow during DP calculations using the provided modulo value.

steps and arrayLen are both very large, potentially exceeding memory limits for DP table

How to Handle:

The DP table should be sized at `steps + 1` x `min(steps + 1, arrayLen)`, which is sufficient and prevents excessive memory usage.

No possible way to return to the starting position

How to Handle:

The algorithm inherently covers cases where no valid solution exists; it will return 0 if no path returns to the starting position.

steps is odd and arrayLen is 1

How to Handle:

When arrayLen is 1 and steps is odd it is impossible to stay in the same place, so return 0.

Steps is a very large number close to the maximum integer limit.

How to Handle:

Ensure intermediate calculations within the DP table do not exceed the maximum integer value even with modulo applied, as repeated additions might still lead to issues.