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.

 

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.

Algorithm

In the DP Approach, we make use of a hashmap $map$ which contains $key:value$ pairs such that $key$ refers to the position at which a stone is present and $value$ is a set containing the $jumpsize$ which can lead to the current stone position. We start by making a hashmap whose $key$s are all the positions at which a stone is present and the $value$s are all empty except position 0 whose value contains 0. Then, we start traversing the elements(positions) of the given stone array in sequential order. For the $currentPosition$, for every possible $jumpsize$ in the $value$ set, we check if $currentPosition + newjumpsize$ exists in the $map$, where $newjumpsize$ can be either $jumpsize-1$, $jumpsize$, $jumpsize+1$. If so, we append the corresponding $value$ set with $newjumpsize$. We continue in the same manner. If at the end, the $value$ set corresponding to the last position is non-empty, we conclude that reaching the end is possible, otherwise, it isn't.

Solution:

class Solution {
    public boolean canCross(int[] stones) {
        HashMap<Integer, Set<Integer>> map = new HashMap<>();
        for (int i = 0; i < stones.length; i++) {
            map.put(stones[i], new HashSet<Integer>());
        }
        map.get(0).add(0);
        for (int i = 0; i < stones.length; i++) {
            for (int k : map.get(stones[i])) {
                for (int step = k - 1; step <= k + 1; step++) {
                    if (step > 0 && map.containsKey(stones[i] + step)) {
                        map.get(stones[i] + step).add(step);
                    }
                }
            }
        }
        return map.get(stones[stones.length - 1]).size() > 0;
    }
}