Introduction

In this blog post, we will explore how to solve the LeetCode problem "2400" 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 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

Explanation

Here's the breakdown of the solution:

• Dynamic Programming: The core idea is to use dynamic programming to store and reuse the number of ways to reach a particular position from the starting position in a certain number of steps. We build a table dp[steps][position] representing the number of ways to reach position in steps moves.

• Base Case and Transitions: The base case is dp[0][startPos] = 1, as there's one way to be at the starting position with zero steps. The transition involves considering moving left or right from a previous position: dp[i][j] = dp[i-1][j-1] + dp[i-1][j+1].

• Optimization and Constraints: We only need to consider positions within a relevant range to avoid unnecessary computations. The range is defined by [startPos - k, startPos + k]. Also, we apply the modulo operator % (10**9 + 7) to avoid integer overflow. If the absolute difference between startPos and endPos is greater than k, it is impossible to reach endPos in k steps. Similarly, if k is less than the distance and k is of different parity than the distance, it is impossible to get there.

• Runtime Complexity: O(k^2) where k is the number of steps.

• Storage Complexity: O(k^2)

Code

    def numberOfWays(startPos: int, endPos: int, k: int) -> int:
    if abs(startPos - endPos) > k:
        return 0

    if (k - abs(startPos - endPos)) % 2 != 0:
        return 0

    MOD = 10**9 + 7
    dp = [[0] * (2 * k + 1) for _ in range(k + 1)]
    dp[0][startPos + k] = 1

    for steps in range(1, k + 1):
        for pos in range(startPos - k, startPos + k + 1):
            index = pos + k
            if index >= 1:
                dp[steps][index] = (dp[steps][index] + dp[steps - 1][index - 1]) % MOD
            if index < 2 * k:
                dp[steps][index] = (dp[steps][index] + dp[steps - 1][index + 1]) % MOD

    return dp[k][endPos + k]