Frog Jump

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.

For example: 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.

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] <= 2^31 - 1 stones[0] == 0 stones is sorted in a strictly increasing order.

Here's the breakdown of how to solve this frog jumping problem:

Problem Description

We're given a list of stone positions along a river, sorted in ascending order. A frog starts at the first stone (position 0) and wants to reach the last stone. The frog's first jump is 1 unit. After that, if the frog's last jump was k units, the next jump must be k - 1, k, or k + 1 units. The frog can only jump forward.

The task is to determine if the frog can reach the last stone.

Naive Approach (Brute Force)

A brute-force approach would involve exploring all possible jump sequences from the starting stone. This can be done recursively. At each stone, we try all possible jump lengths (k - 1, k, k + 1) and see if they lead to the next stone. If we reach the last stone, we return true. If we exhaust all possibilities without reaching the last stone, we return false.

However, this approach is highly inefficient due to the overlapping subproblems. We'd end up recalculating the same paths multiple times, leading to exponential time complexity.

Optimal Solution (Dynamic Programming)

We can use dynamic programming to optimize this solution. We can use a hash map to store all the possible jump sizes that can reach a particular stone. The keys of the map are the stone positions, and the values are sets of jump sizes that can reach that stone.

Here's the algorithm:

1. Initialize a hash map dp where keys are stone positions and values are sets of possible jump sizes.
2. Seed the first stone (position 0) with a jump size of 0 (since the first jump is always 1, but we'll use 0 for the initial state).
3. Iterate through the stones:
   • For each stone, iterate through the possible jump sizes that can reach it.
   • For each jump size k, calculate the next possible jump lengths: k - 1, k, and k + 1.
   • For each possible next jump length, check if there's a stone at that position.
   • If there's a stone, add the jump length to the set of possible jump sizes for that stone.
4. Finally, check if the last stone has any possible jump sizes in its set. If it does, the frog can reach the last stone.

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

    for stone in stones:
        for k in dp[stone]:
            for step in range(k - 1, k + 2):
                if step > 0 and stone + step in dp:
                    dp[stone + step].add(step)

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

Here's an example of how the dp map evolves with the input stones = [0,1,3,5,6,8,12,17]:

• Initially: dp = {0: {0}, 1: set(), 3: set(), 5: set(), 6: set(), 8: set(), 12: set(), 17: set()}
• After processing stone 0: dp = {0: {0}, 1: {1}, 3: set(), 5: set(), 6: set(), 8: set(), 12: set(), 17: set()}
• After processing stone 1: dp = {0: {0}, 1: {1}, 3: {1, 2}, 5: set(), 6: set(), 8: set(), 12: set(), 17: set()}
• ...
• Finally, if dp[17] is not empty, it means the frog can reach the last stone.

Big(O) Time Complexity Analysis

The time complexity of this solution is O(N^2), where N is the number of stones. This is because, in the worst case, for each stone, we iterate through all possible jump sizes, which can be at most N.

Specifically:

• The outer loop iterates through each of the N stones.
• The inner loop iterates through the possible jump sizes for each stone. In the worst case, the number of jump sizes for each stone can be proportional to N.
• Inside the inner loop, we have constant-time operations (checking if the next stone exists and adding the jump size to the set).

Thus, the overall time complexity is O(N * N) = O(N^2).

Big(O) Space Complexity Analysis

The space complexity of this solution is O(N^2), where N is the number of stones. This is because the dp hash map stores the stone positions as keys and sets of jump sizes as values. In the worst case, each stone can have a set of jump sizes, and the maximum number of jump sizes for a stone can be proportional to N. Therefore, the space required to store the dp map is O(N * N) = O(N^2).

Edge Cases

1. Empty or Null Input:
   • If the input stones is null or has fewer than 2 stones, return false because the frog needs at least two stones to make a jump.
2. First Stone Not Zero:
   • If the first stone's position is not 0, return false because the frog always starts at position 0.
3. No Possible Jumps:
   • If at any point, a stone has no possible jump sizes, and it's not the last stone, then the frog cannot reach the end. This is handled by the algorithm because the frog will stop exploring paths from that stone.
4. Large Gaps Between Stones:
   • If there are very large gaps between the stones, such that the required jump size exceeds the maximum possible jump size at a given stone, the frog won't be able to cross. This is implicitly handled because if the next stone is too far, it won't be added to the dp map.
5. Single Stone:
   • If there's only one stone [0], return true as the frog is already on the last stone (though this violates the problem's constraint that there must be at least 2 stones).

By using dynamic programming, we can efficiently solve the frog jumping problem, handling various edge cases and optimizing the time and space complexity.