> ## Documentation Index
> Fetch the complete documentation index at: https://leetcode-py.wisl.dev/llms.txt
> Use this file to discover all available pages before exploring further.

> ## Agent Instructions
> leetcode-py is a Python LeetCode practice environment generator with one CLI: lcpy. It is not a service or platform.
> Each problem is a directory under leetcode/ with README.md, solution.py, test_solution.py, helpers.py, and playground.ipynb. lcpy gen creates them from JSON templates bundled with the package.
> Examples are backed by tests; copy them verbatim.

# Frog Jump Python Solution with Tests

> Tested Python solution for LeetCode 403 with 17 pytest cases. Generate a practice environment with lcpy.

LeetCode 403, [Hard](/catalog/hard). Topics: [Array](/catalog/topics/array), [Dynamic Programming](/catalog/topics/dynamic-programming). [View on LeetCode](https://leetcode.com/problems/frog-jump/description/).

Generate this problem as a practice environment: tested reference solution, 17 [parametrized pytest cases](/practice/testing), and a playground notebook:

```bash theme={"theme":{"light":"github-light","dark":"github-dark"}}
lcpy gen -n 403   # by problem number
lcpy gen -s frog_jump   # by problem name
```

## Problem

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.

### Examples

```
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.

## Solution

Reference implementation from [solution.py on GitHub](https://github.com/wislertt/leetcode-py/blob/main/leetcode/frog_jump/solution.py), full suite in [test\_solution.py](https://github.com/wislertt/leetcode-py/blob/main/leetcode/frog_jump/test_solution.py):

```python theme={"theme":{"light":"github-light","dark":"github-dark"}}
class Solution:
    # Time: O(n^2)
    # Space: O(n^2)
    def can_cross(self, stones: list[int]) -> bool:
        if stones[1] != 1:
            return False
        positions = set(stones)
        last = stones[-1]
        if last == 1:
            return True
        jumps: dict[int, set[int]] = {pos: set() for pos in stones}
        jumps[1].add(1)
        for pos in stones[1:]:
            for k in jumps[pos]:
                for step in (k - 1, k, k + 1):
                    nxt = pos + step
                    if step > 0 and nxt in positions:
                        if nxt == last:
                            return True
                        jumps[nxt].add(step)
        return False
```

## Complexity

| Time | Space |
| - | - |
| O(n^2) | O(n^2) |

## Tags


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