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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:

• Jump Game II (Hard)
• Jump Game (Medium)

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;
    }
};

• Java
• C++
• Python
• Go
• TypeScript

• 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;
      }
  }
  

• // 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;
  }
  
  

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