LeetCode 1306. Jump Game III Solution in Java, C++, Python & More | Explanation + Code

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.

PythonJavaC++GoTypeScript

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(code-box)

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; } }(code-box)

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; } };(code-box)

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 }(code-box)

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; }(code-box)