Frog Jump Problem

Problem Description

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.

Naive Approach (Brute Force with Recursion)

The most straightforward approach is to use recursion to explore all possible jump combinations. Starting from the first stone, we explore jumps of size k-1, k, and k+1 where k is the size of the previous jump. If any of these jumps leads us to the last stone, we return true. If we reach a point where no valid jump is possible, we return false.

Code (Python):

def can_cross_recursive(stones):
    def solve(current_position, last_jump):
        if current_position == stones[-1]:
            return True

        for next_jump in [last_jump - 1, last_jump, last_jump + 1]:
            if next_jump > 0:
                next_position = current_position + next_jump
                if next_position in stones:
                    if next_position > current_position and solve(next_position, next_jump):
                        return True
        return False

    if stones[1] != 1:
        return False

    return solve(1, 1)

Time Complexity: O(3^N) in the worst case, where N is the number of stones. This is because at each stone, we explore up to 3 possible jump lengths.

Space Complexity: O(N) due to the maximum depth of the recursion call stack.

This naive approach will likely result in a "Time Limit Exceeded" error for larger input sizes.

Optimal Approach (Dynamic Programming or Memoization)

We can optimize the brute force approach by using dynamic programming (memoization) to avoid redundant calculations. The key idea is to store the results of subproblems so that we can reuse them later. We can use a dictionary or a 2D array to store the results. The state of the subproblem can be defined by the current stone index and the last jump size.

Code (Python):

def can_cross_dp(stones):
    n = len(stones)
    stone_map = {stone: i for i, stone in enumerate(stones)}
    dp = {}  # (stone_index, last_jump) -> boolean

    def solve(stone_index, last_jump):
        if stone_index == n - 1:
            return True

        if (stone_index, last_jump) in dp:
            return dp[(stone_index, last_jump)]

        for next_jump in [last_jump - 1, last_jump, last_jump + 1]:
            if next_jump > 0:
                next_stone = stones[stone_index] + next_jump
                if next_stone in stone_map:
                    next_index = stone_map[next_stone]
                    if next_index > stone_index:
                        if solve(next_index, next_jump):
                            dp[(stone_index, last_jump)] = True
                            return True

        dp[(stone_index, last_jump)] = False
        return False

    if stones[1] != 1:
        return False

    return solve(1, 1)

Time Complexity: O(N²), where N is the number of stones. We iterate through each stone and possible jump sizes.

Space Complexity: O(N²) because of the dp dictionary which at worse can store a result for each combination of stone index and possible jump size.

Edge Cases

• If the second stone is not at position 1, the frog cannot cross the river.
• An empty input list is invalid based on the problem statement. The list must contain at least two stones.

Summary

The dynamic programming approach provides a much more efficient solution than the brute-force recursive approach. By storing the results of subproblems, we significantly reduce the number of calculations required.