[LeetCode in Python] 1306 (M) jump game iii Jump game III

topic

https://leetcode-cn.com/problems/jump-game-iii/

There is an array of non-negative integers, arr, and you are at the beginning of the array. When you are at the subscript i, you can jump to i + arr [i] or i-arr [i].
Please judge whether you can jump to any subscript where the corresponding element value is 0.
Note that no matter what the situation, you cannot jump outside the array.

Example 1:

Input: arr = [4,2,3,0,3,1,2], start = 5
Output: true

Explanation:
Subscript 3 with a value of 0 has the following possible solutions:
subscript 5-> subscript 4-> subscript 1-> subscript 3
subscript 5-> subscript 6-> subscript 4-> subscript 1-> subscript 3

Example 2:

Input: arr = [4,2,3,0,3,1,2], start = 0
Output: true

Explanation:
Subscript 3 with a value of 0 has the following possible solutions:
subscript 0-> subscript 4-> subscript 1-> subscript 3

Example 3:

Input: arr = [3,0,2,1,2], start = 2
Output: false

Explanation: Subscript 1 with a value of 0 cannot be reached.

prompt:

1 <= arr.length <= 5 * 10^4
0 <= arr[i] < arr.length
0 <= start < arr.length

Problem-solving ideas

• Standard BFS process
• Use seen_setset storage to access subscripts
• When traversing the breadth, in addition to judging whether it has been visited, it is also necessary to check whether it has crossed the left and right borders

Code

class Solution:
    def canReach(self, arr: List[int], start: int) -> bool:
        seen_set = set([start])
        q = [start]
        while q:
            p = q.pop(0)
            if arr[p] == 0:
                return True

            for k in [p-arr[p], p+arr[p]]:
                if k in seen_set: continue
                
                if 0 <= k <= len(arr)-1:
                    q.append(k)
                    seen_set.add(k)
        
        return False