Introduction

In this blog post, we will explore how to solve the LeetCode problem "403" 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

A frog is crossing a river. The river is divided into some number of units, and at each unit, there may or may not exist a stone. The frog can jump on a stone, but it must not jump into the water. Given a list of stones positions (in units) in sorted ascending order, determine if the frog can cross the river by landing on the last stone. Initially, the frog is on the first stone and assumes the first jump must be 1 unit. If the frog's last jump was k units, its next jump must be either k - 1, k, or k + 1 units. The frog can only jump in the forward direction. Example 1: Input: stones = [0,1,3,5,6,8,12,17] Output: true Explanation: The frog can jump to the last stone by jumping 1 unit to the 2nd stone, then 2 units to the 3rd stone, then 2 units to the 4th stone, then 3 units to the 6th stone, 4 units to the 7th stone, and 5 units to the 8th stone. Example 2: Input: stones = [0,1,2,3,4,8,9,11] Output: false Explanation: There is no way to jump to the last stone as the gap between the 5th and 6th stone is too large. Constraints: 2 <= stones.length <= 2000 0 <= stones[i] <= 231 - 1 stones[0] == 0 stones is sorted in a strictly increasing order.

Explanation

Here's a breakdown of the solution:

• Dynamic Programming (Memoization): We'll use a recursive function with memoization (caching) to explore all possible jump sequences. The memoization will avoid recomputing the same subproblems, which is crucial for efficiency.
• State Definition: The state of our recursion will be defined by the current stone index and the length of the previous jump. This defines the current configuration the frog is in.

• Base Case & Transition: The base case is when the frog reaches the last stone. The transitions involve trying jumps of length k-1, k, and k+1 from the current stone, where k is the length of the previous jump. We check if the next stone exists at the calculated position before making the recursive call.

• Time Complexity: O(n*m), where n is the number of stones and m is the maximum possible jump size. In the worst case m can go upto n. This can be expressed as O(n^2).

• Space Complexity: O(n*m) due to the memoization table. Similarly, this can be expressed as O(n^2).

Code

    def can_cross(stones):
    """
    Determines if a frog can cross a river by landing on the last stone.

    Args:
        stones: A list of stone positions in sorted ascending order.

    Returns:
        True if the frog can cross the river, False otherwise.
    """

    n = len(stones)
    if n == 0:
        return True  # Handle empty case

    if stones[1] != 1:
        return False

    memo = {}  # Memoization table to store results of subproblems

    def solve(index, k):
        """
        Recursive helper function to check if the frog can reach the end from a given index with a given jump size.

        Args:
            index: The current stone index.
            k: The length of the previous jump.

        Returns:
            True if the frog can reach the end, False otherwise.
        """

        if (index, k) in memo:
            return memo[(index, k)]

        if index == n - 1:
            return True

        possible_jumps = [k - 1, k, k + 1]
        for jump in possible_jumps:
            if jump > 0:  # Jump must be positive
                next_stone = stones[index] + jump
                next_index = -1
                for i in range(index + 1, n):
                    if stones[i] == next_stone:
                        next_index = i
                        break

                if next_index != -1:
                    if solve(next_index, jump):
                        memo[(index, k)] = True
                        return True

        memo[(index, k)] = False
        return False

    return solve(1, 1)