README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/1300-1399/1306.Jump%20Game%20III/README_EN.md	1396	Weekly Contest 169 Q3	Depth-First Search Breadth-First Search Array

1306. Jump Game III

中文文档

Description

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 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

Solutions

Solution 1: BFS

We can use BFS to determine whether we can reach the index with a value of $0$.

Define a queue $q$ to store the currently reachable indices. Initially, enqueue the $start$ index.

When the queue is not empty, take out the front index $i$ of the queue. If $arr[i] = 0$, return true. Otherwise, mark the index $i$ as visited. If $i + arr[i]$ and $i - arr[i]$ are within the array range and have not been visited, enqueue them and continue searching.

Finally, if the queue is empty, it means that we cannot reach the index with a value of $0$, so return false.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the length of the array.

Python3

class Solution:
    def canReach(self, arr: List[int], start: int) -> bool:
        q = deque([start])
        while q:
            i = q.popleft()
            if arr[i] == 0:
                return True
            x = arr[i]
            arr[i] = -1
            for j in (i + x, i - x):
                if 0 <= j < len(arr) and arr[j] >= 0:
                    q.append(j)
        return False

Java

class Solution {
    public boolean canReach(int[] arr, int start) {
        Deque<Integer> q = new ArrayDeque<>();
        q.offer(start);
        while (!q.isEmpty()) {
            int i = q.poll();
            if (arr[i] == 0) {
                return true;
            }
            int x = arr[i];
            arr[i] = -1;
            for (int j : List.of(i + x, i - x)) {
                if (j >= 0 && j < arr.length && arr[j] >= 0) {
                    q.offer(j);
                }
            }
        }
        return false;
    }
}

C++

class Solution {
public:
    bool canReach(vector<int>& arr, int start) {
        queue<int> q{{start}};
        while (!q.empty()) {
            int i = q.front();
            q.pop();
            if (arr[i] == 0) {
                return true;
            }
            int x = arr[i];
            arr[i] = -1;
            for (int j : {i + x, i - x}) {
                if (j >= 0 && j < arr.size() && ~arr[j]) {
                    q.push(j);
                }
            }
        }
        return false;
    }
};

Go

func canReach(arr []int, start int) bool {
	q := []int{start}
	for len(q) > 0 {
		i := q[0]
		q = q[1:]
		if arr[i] == 0 {
			return true
		}
		x := arr[i]
		arr[i] = -1
		for _, j := range []int{i + x, i - x} {
			if j >= 0 && j < len(arr) && arr[j] >= 0 {
				q = append(q, j)
			}
		}
	}
	return false
}

TypeScript

function canReach(arr: number[], start: number): boolean {
    const q = [start];
    for (const i of q) {
        if (arr[i] === 0) {
            return true;
        }
        if (arr[i] === -1 || arr[i] === undefined) {
            continue;
        }
        q.push(i + arr[i], i - arr[i]);
        arr[i] = -1;
    }
    return false;
}