Jump Game III

#leetcode #dsa #theabbie

Given an array of non-negative integers arr, you are initially positioned at start index of the array. When you are at index i, you can jump to i + arr[i] or i - arr[i], check if you can reach to any index with value 0.

Notice that you can not jump outside of the array at any time.

Example 1:

Input: arr = [4,2,3,0,3,1,2], start = 5
Output: true
Explanation:
All possible ways to reach at index 3 with value 0 are:
index 5 -> index 4 -> index 1 -> index 3
index 5 -> index 6 -> index 4 -> index 1 -> index 3

Example 2:

Input: arr = [4,2,3,0,3,1,2], start = 0
Output: true
Explanation:
One possible way to reach at index 3 with value 0 is:
index 0 -> index 4 -> index 1 -> index 3

Example 3:

Input: arr = [3,0,2,1,2], start = 2
Output: false
Explanation: There is no way to reach at index 1 with value 0.

Constraints:

1 <= arr.length <= 5 * 104
0 <= arr[i] < arr.length
0 <= start < arr.length

SOLUTION:

class Solution:
    def canReach(self, arr: List[int], start: int) -> bool:
        n = len(arr)
        paths = [[start]]
        visited = set()
        while len(paths) > 0:
            curr = paths.pop(0)
            currIndex = curr[-1]
            jump = arr[currIndex]
            if jump == 0:
                return True
            visited.add(currIndex)
            if currIndex - jump >= 0 and (currIndex - jump) not in visited:
                paths.append(path + [currIndex - jump])
            if currIndex + jump <= n - 1 and (currIndex + jump) not in visited:
                paths.append(path + [currIndex + jump])
        return False