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Formatted question description: https://leetcode.ca/all/1306.html

# 1306. Jump Game III (Medium)

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

Related Topics:

Similar Questions:

## Solution 1. DFS

// OJ: https://leetcode.com/problems/jump-game-iii/
// Time: O(N)
// Space: O(N)
class Solution {
public:
bool canReach(vector<int>& A, int start) {
if (start < 0 || start >= A.size() || A[start] < 0) return false;
if (A[start] == 0) return true;
A[start] *= -1;
return canReach(A, start + A[start]) || canReach(A, start - A[start]);
}
};


## Solution 2. BFS

// OJ: https://leetcode.com/problems/jump-game-iii/
// Time: O(N)
// Space: O(N)
class Solution {
public:
bool canReach(vector<int>& A, int start) {
queue<int> q;
q.push(start);
while (q.size()) {
int i = q.front();
q.pop();
if (A[i] == 0) return true;
if (A[i] < 0) continue;
if (i + A[i] < A.size()) q.push(i + A[i]);
if (i - A[i] >= 0) q.push(i - A[i]);
A[i] *= -1;
}
return false;
}
};

• class Solution {
public boolean canReach(int[] arr, int start) {
int length = arr.length;
boolean[] canReach = new boolean[length];
canReach[start] = true;
queue.offer(start);
while (!queue.isEmpty()) {
int index = queue.poll();
int step = arr[index];
int left = index - step, right = index + step;
if (left >= 0) {
if (arr[left] == 0)
return true;
else if (!canReach[left]) {
canReach[left] = true;
queue.offer(left);
}
}
if (right < length) {
if (arr[right] == 0)
return true;
else if (!canReach[right]) {
canReach[right] = true;
queue.offer(right);
}
}
}
return false;
}
}

############

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;
}
}

• // OJ: https://leetcode.com/problems/jump-game-iii/
// Time: O(N)
// Space: O(N)
class Solution {
public:
bool canReach(vector<int>& A, int start) {
if (start < 0 || start >= A.size() || A[start] < 0) return false;
if (A[start] == 0) return true;
A[start] *= -1;
return canReach(A, start + A[start]) || canReach(A, start - A[start]);
}
};

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


• 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
}

• function canReach(arr: number[], start: number): boolean {
const q: number[] = [start];
while (q.length) {
const i: number = q.shift()!;
if (arr[i] === 0) {
return true;
}
const x: number = arr[i];
arr[i] = -1;
for (const j of [i + x, i - x]) {
if (j >= 0 && j < arr.length && arr[j] !== -1) {
q.push(j);
}
}
}
return false;
}