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.

Brute Force Solution

A brute-force approach would involve exploring all possible jump combinations from each stone. This can be implemented using recursion.

1. Start at the first stone (index 0) with a jump size of 1.
2. For each stone, try jumps of size k-1, k, and k+1, where k is the size of the last jump.
3. If a jump leads to the last stone, return true.
4. If no jump combination leads to the last stone, return false.

This approach will likely result in a Time Limit Exceeded (TLE) error for larger inputs due to its exponential time complexity.

Code (Python):

def can_cross_brute_force(stones):
    n = len(stones)
    if n <= 1:
        return True

    def can_jump(index, jump_size):
        if index == n - 1:
            return True

        for next_jump in [jump_size - 1, jump_size, jump_size + 1]:
            if next_jump > 0:
                next_stone = stones[index] + next_jump
                if next_stone in stones:
                    next_index = stones.index(next_stone)
                    if can_jump(next_index, next_jump):
                        return True

        return False

    return can_jump(0, 1)

Time Complexity: O(3^N), where N is the number of stones. For each stone, we explore up to 3 possible jumps.

Space Complexity: O(N) due to the recursion depth.

Optimal Solution: Dynamic Programming

A more efficient solution involves using dynamic programming to avoid redundant calculations. We can use a hash map (dictionary in Python) to store the possible jump sizes for each stone.

1. Create a hash map where the keys are the stone positions and the values are sets of possible jump sizes that can reach that stone.
2. Initialize the hash map with the first stone (position 0) and a jump size of 0 (since the initial jump is 1, we can think of it as starting with a jump of zero to the first stone).
3. Iterate through the stones. For each stone, iterate through the possible jump sizes that can reach it.
4. For each jump size k, calculate the positions reachable with jumps k-1, k, and k+1.
5. If a reachable position is a stone, add the corresponding jump size to the set of possible jump sizes for that stone.
6. If the last stone can be reached (i.e., its set of possible jump sizes is not empty), return true. Otherwise, return false.

Code (Python):

def can_cross(stones):
    n = len(stones)
    if n <= 1:
        return True

    stone_jumps = {stone: set() for stone in stones}
    stone_jumps[0].add(0)

    for i in range(n):
        stone = stones[i]
        for jump_size in stone_jumps[stone]:
            for next_jump in [jump_size - 1, jump_size, jump_size + 1]:
                if next_jump > 0:
                    next_stone = stone + next_jump
                    if next_stone in stone_jumps:
                        stone_jumps[next_stone].add(next_jump)

    return len(stone_jumps[stones[-1]]) > 0

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

Space Complexity: O(N), where N is the number of stones, to store the stone_jumps hash map.

Edge Cases

1. Empty or single-stone river: Should return true.
2. First jump not possible: If the second stone is not at position 1, it's impossible to cross.
3. Large gaps between stones: If the gaps between stones are too large, the frog won't be able to cross.
4. All stones are in a straight line: The frog might still not be able to cross if the jump sizes don't allow it.

Summary

The dynamic programming approach provides a more efficient solution compared to the brute-force method. It avoids redundant calculations by storing the possible jump sizes for each stone, resulting in a time complexity of O(N²) and a space complexity of O(N). Handling edge cases is crucial to ensure the correctness of the solution.