Frog Jump

Hard

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ArraysDynamic Programming

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.

Try coding it in LeetCode

Frog Jump

Hard

Asked by:

27 views

Topics:

ArraysDynamic Programming

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.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of values within the stones array? Can the stone values be negative, zero, or non-integers?
If the frog cannot reach the last stone, what should the function return? Should it return false, an error, or some other indicator?
Are the stone positions guaranteed to be in strictly increasing order?
Can the input array be empty or contain only one stone?
Is the starting position of the frog always the first stone, and does the jump always start with a jump size of 1?

Brute Force Solution

Approach

The frog wants to cross a river by hopping on stones. The brute force method is like trying every single possible sequence of jumps the frog could make to see if it works.

Here's how the algorithm would work step-by-step:

Start with the first stone.
Consider all possible jump lengths from the current stone to the next stone. This means try jumping one unit, two units, and so on.
For each possible jump, check if there is a stone at the landing spot. If there isn't, that jump is invalid, so try a different jump length.
If the jump is valid, move the frog to the new stone and repeat the process: again consider all possible jump lengths from this new stone.
Keep doing this until the frog either reaches the last stone (success!) or gets stuck (no stones to jump to, so this path doesn't work).
If the frog gets stuck, go back to a previous stone where there were other jump lengths to try and explore those different paths.
Continue exploring all possible jumping paths until you find one that successfully reaches the last stone. If you explore all paths and none work, then it's impossible for the frog to cross.

Code Implementation

def can_cross_river_brute_force(stones): 
    stones_set = set(stones)
    last_stone = stones[-1]

    def can_reach_destination(current_stone, current_jump_size): 
        if current_stone == last_stone:
            return True

        # Try all possible jump sizes: k-1, k, k+1
        for next_jump_size in range(current_jump_size - 1, current_jump_size + 2): 
            if next_jump_size > 0:
                next_stone = current_stone + next_jump_size

                # Ensure the next stone exists
                if next_stone in stones_set: 

                    # Recursively check if frog can reach the destination
                    if can_reach_destination(next_stone, next_jump_size): 
                        return True
        return False

    # The frog always starts at the first stone with a jump size of 1
    if stones[0] == 0 and 1 in stones_set:
        # Initial jump of 1 from the first stone
        return can_reach_destination(1, 1)
    else:
        return False

Big(O) Analysis

Time Complexity

O(3^n) – The frog jump problem, when solved with a brute force approach, explores every possible jump sequence. At each stone (except the first and last), the frog has approximately three jump options (k-1, k, k+1 where k is the previous jump's length). Thus, in the worst case, for each stone, it branches out into three possibilities. If n is the number of stones, the time complexity is proportional to the number of possible paths, which is roughly 3 multiplied by itself n times. This leads to an exponential time complexity of O(3^n).

Space Complexity

O(N^2) – The provided brute force approach implicitly uses recursion. In the worst-case scenario, the frog explores a large portion of possible jump sequences. Each jump corresponds to a recursive call and the depth of the recursion can grow proportionally to the number of stones N, where N is the number of stones. For each stone, the frog can potentially make multiple jump attempts to other stones, leading to branching. This creates a call stack where the maximum depth is related to N, and where each call potentially stores jump length options from that position. The maximum number of active calls can be proportional to N and at each level there could potentially be up to N possible jumps from that stone, and hence the space complexity would be O(N^2).

Optimal Solution

Approach

The frog can jump to a stone only if it has hopped that exact distance before. The optimal solution tracks possible jump lengths the frog can make from each stone. This way, we can determine if the frog can reach the final stone without trying all combinations.

Here's how the algorithm would work step-by-step:

First, create a way to remember all possible jump lengths the frog can make when standing on each stone.
Start at the first stone, and mark that the frog can start with a jump of length 0.
Now, move through the stones one by one.
For each stone, look at the jump lengths the frog can make from that stone.
For each of those jump lengths, check if the frog can jump to the next stone by jumping one unit shorter, the same length, or one unit longer.
If the frog *can* reach the next stone with any of those jump lengths, record that the frog *can* make that jump from the new stone.
Repeat this until you get to the last stone.
If the frog can make *any* jump to the last stone, it means the frog can reach it, and we can say 'yes'. Otherwise, it means 'no'.

Code Implementation

def can_cross(stones):    stone_positions = set(stones)
    if stones[1] != 1:
        return False

    # Keep track of possible jump lengths
    jump_lengths = {stone: set() for stone in stones}
    jump_lengths[stones[0]].add(0)

    for stone in stones:
        for jump_length in jump_lengths[stone]:
            for next_jump in [jump_length - 1, jump_length, jump_length + 1]:
                if next_jump > 0 and (stone + next_jump) in stone_positions:

                    # The frog can jump to the next stone
                    jump_lengths[stone + next_jump].add(next_jump)

    # Check if the frog can reach the last stone
    return bool(jump_lengths[stones[-1]])

Big(O) Analysis

Time Complexity

O(n²) – The dominant operation is iterating through each stone (n stones) and, for each stone, considering all possible jump lengths originating from it. For each of these jump lengths, the algorithm checks at most three potential next stones. In the worst case, each stone can have up to n possible jump lengths (where n is related to the number of stones), resulting in a nested loop structure. Therefore the overall time complexity is O(n * n), which simplifies to O(n²).

Space Complexity

O(N*M) – The primary auxiliary space comes from storing possible jump lengths for each stone. We maintain a data structure (likely a set or list) for each of the N stones, recording the jump lengths possible from that stone. In the worst case, the maximum possible jump length is related to the distance to the furthest stone, so we denote the maximum jump length for a given stone as M. Thus, each stone might store up to M jump lengths, leading to O(N*M) space overall, where N is the number of stones and M is the maximum possible jump length from any stone.

Edge Cases

Case	How to Handle
Empty stones array	Return true if the array is empty, as the frog starts at position 0, which is considered a success.
Stones array with only the first stone (index 0)	Return true, as the frog starts at the first stone and has already succeeded.
Stones array where the second stone is not at position 1	Return false, because the frog can only jump 1 unit from the initial position.
Stones are not strictly increasing	Handle this edge case by sorting the stones array and checking for valid jumps in the sorted array.
Large difference between consecutive stones	Limit the jump size to a reasonable range (e.g., k-1, k, k+1), and if the gap is too large, backtrack or return false.
Integer overflow when calculating jumps or stone positions	Use long integers for calculations or add overflow checks when calculating jumps to prevent unexpected behavior.
Maximum number of stones causing stack overflow in recursive implementations	Use dynamic programming (memoization) instead of recursion to avoid stack overflow with large inputs.
No path exists to the last stone	Return false if the dynamic programming table indicates that the frog cannot reach the last stone from any reachable position.