Frog Jump Leetcode Problem | Solution | 2022

 Frog Jump Leetcode 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.

 
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.

Solution1:

public boolean canCross(int[] a) {

int n=a.length;

if(n<=1) return true;
// linkedlist to store the previous jumps to reach at ith jump
LinkedList<Integer>[] al=new LinkedList[n];

for(int i=0;i<n;i++){
al[i]=new LinkedList<Integer>();
}
// 0 jump before first stone
al[0].add(0);
for(int i=1;i<n;i++){
for(int j=0;j<i;j++){
LinkedList<Integer> l=al[j];
// number of jumps required to reach ith stone from jth stone
int jump=a[i]-a[j];
// we can reach at ith stone if jth stone contains jump, jump+1 or jump-1
if(l.contains(jump)||l.contains(jump+1)||l.contains(jump-1)) al[i].add(jump);
}
}
// return true if we can reach at last stone
return al[n-1].size()!=0;
}

Following is my backtracking solution using dict for memorization.
The memo dict is used to save those dead ends. So when we get to the same stone with the same speed we don't need to search further.

class Solution(object):
def canCross(self, stones):
self.memo = set()
target = stones[-1]
stones = set(stones)

res = self.bt(stones, 1, 1, target)
return res

def bt(self, stones, cur, speed, target):
# check memo
if (cur, speed) in self.memo:
return False

if cur==target:
return True

if cur>target or cur<0 or speed<=0 or cur not in stones:
return False
# dfs
candidate = [speed-1, speed, speed+1]
for c in candidate:
if (cur + c) in stones:
if self.bt(stones, cur+c, c, target):
return True

self.memo.add((cur,speed))
return False

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