The problem "Number Of Ways To Stay In The Same Place After Some Steps" on LeetCode can be summarized as follows:

You start at position 0 on a number line and you have to take n steps, either left or right. You have a probability p of taking a step to the right and (1-p) probability of taking a step to the left. You want to find the number of ways you can end up back at position 0 after taking n steps.

To solve this problem, we can use dynamic programming. Let f(i, j) denote the number of ways to end up at position i after j steps. We are interested in finding f(0, n).

The base cases for this problem are:

f(0, 0) = 1 (the number of ways to stay at position 0 after 0 steps is 1) f(i, 0) = 0 (you cannot end up at any position other than 0 after 0 steps)

The recursive formula for this problem is:

f(i, j) = p * f(i-1, j-1) + (1-p) * f(i+1, j-1)

This formula says that to end up at position i after j steps, either you took a step to the right from position i-1 (with probability p) and ended up at position i, or you took a step to the left from position i+1 (with probability 1-p) and ended up at position i.

Using this recursive formula, we can fill out a 2D array of size (n+1) x (n+1) where the row corresponds to the position and the column corresponds to the number of steps taken. Then, the answer to the problem is f(0, n).

Here's the complete Python code to solve this problem:

def numWays(n: int, p: float) -> int: # Initialize the 2D array dp = [[0] * (n+1) for _ in range(n+1)] dp[0][0] = 1

# Fill out the 2D array
for j in range(1, n+1):
    for i in range(-j, j+1):
        if i == -j:
            dp[i][j] = p * dp[i+1][j-1]
        elif i == j:
            dp[i][j] = (1-p) * dp[i-1][j-1]
        else:
            dp[i][j] = p * dp[i+1][j-1] + (1-p) * dp[i-1][j-1]

# Return the answer
return dp[0][n]

Example usage

print(numWays(3, 0.5)) # Output: 4

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