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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:
Breadth-first Search, Graph
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;
}
};
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class Solution { public boolean canReach(int[] arr, int start) { int length = arr.length; boolean[] canReach = new boolean[length]; canReach[start] = true; Queue<Integer> queue = new LinkedList<Integer>(); 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; } }
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// 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]); } };
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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
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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 }
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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; }