Introduction

In this blog post, we will explore how to solve the LeetCode problem "1269" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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

Explanation

Here's the breakdown of the approach, complexity, and the Python code:

• Approach: We can use dynamic programming to solve this problem. dp[i][j] will represent the number of ways to reach index j after i steps. We can build up this table iteratively. The base case is dp[0][0] = 1. For each step i, we can transition from dp[i-1][j] to dp[i][j-1], dp[i][j], and dp[i][j+1] if the indices j-1 and j+1 are within the array bounds.

• Complexity: O(steps min(steps, arrLen)) time, O(steps min(steps, arrLen)) space

Code

    def numWays(steps: int, arrLen: int) -> int:
    """
    Calculates the number of ways to return to index 0 after a given number of steps.

    Args:
        steps: The number of steps to take.
        arrLen: The length of the array.

    Returns:
        The number of ways to return to index 0 modulo 10^9 + 7.
    """

    arrLen = min(steps, arrLen)  # Optimization: We can never reach beyond steps index.
    dp = [[0] * arrLen for _ in range(steps + 1)]
    dp[0][0] = 1
    mod = 10**9 + 7

    for i in range(1, steps + 1):
        for j in range(arrLen):
            dp[i][j] = dp[i - 1][j]  # Stay

            if j > 0:
                dp[i][j] = (dp[i][j] + dp[i - 1][j - 1]) % mod  # Left

            if j < arrLen - 1:
                dp[i][j] = (dp[i][j] + dp[i - 1][j + 1]) % mod  # Right

    return dp[steps][0]