Introduction

In this blog post, we will explore how to solve the LeetCode problem "2498" 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 a 0-indexed integer array stones sorted in strictly increasing order representing the positions of stones in a river. A frog, initially on the first stone, wants to travel to the last stone and then return to the first stone. However, it can jump to any stone at most once. The length of a jump is the absolute difference between the position of the stone the frog is currently on and the position of the stone to which the frog jumps. More formally, if the frog is at stones[i] and is jumping to stones[j], the length of the jump is |stones[i] - stones[j]|. The cost of a path is the maximum length of a jump among all jumps in the path. Return the minimum cost of a path for the frog. Example 1: Input: stones = [0,2,5,6,7] Output: 5 Explanation: The above figure represents one of the optimal paths the frog can take. The cost of this path is 5, which is the maximum length of a jump. Since it is not possible to achieve a cost of less than 5, we return it. Example 2: Input: stones = [0,3,9] Output: 9 Explanation: The frog can jump directly to the last stone and come back to the first stone. In this case, the length of each jump will be 9. The cost for the path will be max(9, 9) = 9. It can be shown that this is the minimum achievable cost. Constraints: 2 <= stones.length <= 105 0 <= stones[i] <= 109 stones[0] == 0 stones is sorted in a strictly increasing order.

Explanation

Here's the solution to the frog jumping problem:

• Binary Search: The problem asks for the minimum cost. This strongly suggests a binary search approach. We binary search over the possible costs (maximum jump lengths).
• Feasibility Check (DFS/Recursion): For a given cost mid, we need to determine if it's possible for the frog to reach the last stone and return to the first stone without any jump exceeding mid. We can check this using a Depth-First Search (DFS) approach. The DFS explores possible paths, pruning branches if a jump exceeds the allowed cost or if a stone is visited twice.

• Memoization (Optimization): To optimize the DFS, we use memoization (dynamic programming). We store whether a state (combination of the frog's forward and backward positions) has been visited and whether it can lead to a successful path. This avoids redundant calculations.

• Time Complexity: O(n² log M), where n is the number of stones and M is the maximum possible jump length (stones[-1] - stones[0]). The binary search takes O(log M) time, and the DFS (with memoization) takes O(n²) time in the worst case because there are n² possible states (forward and backward positions). Space Complexity: O(n²) for the memoization table.

Code

    def solve():
    def min_cost(stones):
        n = len(stones)

        def is_possible(max_jump):
            memo = {}  # (forward_idx, backward_idx): bool

            def dfs(forward_idx, backward_idx, forward_direction):
                if (forward_idx, backward_idx) in memo:
                    return memo[(forward_idx, backward_idx)]

                # Base Case: Reached destination and returning
                if forward_idx == n - 1 and backward_idx == 0 and not forward_direction:
                    return True

                # Moving forward towards the end
                if forward_direction:
                    for next_idx in range(forward_idx + 1, n):
                        jump_length = abs(stones[forward_idx] - stones[next_idx])
                        if jump_length <= max_jump:
                            if dfs(next_idx, backward_idx, True):
                                memo[(forward_idx, backward_idx)] = True
                                return True

                    # If can reach the end, now turn around
                    if forward_idx == n - 1:
                        if dfs(forward_idx, backward_idx, False):
                            memo[(forward_idx, backward_idx)] = True
                            return True

                    memo[(forward_idx, backward_idx)] = False
                    return False
                # Returning to the start
                else:
                    for next_idx in range(backward_idx - 1, -1, -1):
                        jump_length = abs(stones[backward_idx] - stones[next_idx])

                        if jump_length <= max_jump:
                            if dfs(forward_idx, next_idx, False):
                                memo[(forward_idx, backward_idx)] = True
                                return True

                    # If can reach the start
                    if backward_idx == 0:
                        memo[(forward_idx, backward_idx)] = True
                        return True

                    memo[(forward_idx, backward_idx)] = False
                    return False

            return dfs(0, n - 1, True)

        low = 0
        high = stones[-1] - stones[0]
        ans = high  # Initialize with maximum possible cost

        while low <= high:
            mid = (low + high) // 2
            if is_possible(mid):
                ans = mid
                high = mid - 1
            else:
                low = mid + 1

        return ans

    stones = [0, 2, 5, 6, 7]
    print(min_cost(stones))  # Output: 5

    stones = [0, 3, 9]
    print(min_cost(stones))  # Output: 9

solve()