How do you solve Jump Game III on LeetCode?

Jump Game III (LeetCode #1306) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: BFS is more efficient than DFS for this problem due to its iterative nature and ability to avoid deep recursion.

What is the brute force solution for Jump Game III?

The brute force approach for Jump Game III is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves exploring all possible jumps from the starting index using recursion or backtracking. We will check if we can reach an index with a value of 0 by trying every possible jump until we either find a 0 or exhaust all options. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Jump Game III?

Jump Game III uses the following concepts: Array, Depth-First Search, Breadth-First Search. The recommended patterns to study are: BFS, Graph Traversal, Array.

What are the interview tips for Jump Game III?

Always clarify the problem constraints and edge cases. Think about using BFS for problems involving exploration of paths. Practice explaining your thought process as you code.

What are common mistakes when solving Jump Game III?

Not using a visited set, leading to infinite loops. Confusing the direction of jumps, leading to incorrect indices.

#1306

Jump Game III

Medium

ArrayDepth-First SearchBreadth-First SearchBFSGraph TraversalArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal approach uses a Breadth-First Search (BFS) to explore all possible jumps iteratively. This ensures that we efficiently check each index without revisiting, leading to a faster solution.

⚙️

Algorithm

5 steps

1Step 1: Initialize a queue and add the starting index to it.
2Step 2: Use a set to keep track of visited indices to avoid cycles.
3Step 3: While the queue is not empty, dequeue an index and check if its value is 0.
4Step 4: Calculate the next possible indices to jump to (i + arr[i] and i - arr[i]).
5Step 5: If the next index is valid and not visited, add it to the queue and mark it as visited.

solution.py19 lines

1# Full working Python code
2
3from collections import deque
4
5def canReach(arr, start):
6    queue = deque([start])
7    visited = set()
8    while queue:
9        index = queue.popleft()
10        if arr[index] == 0:
11            return True
12        visited.add(index)
13        next_index = index + arr[index]
14        if next_index < len(arr) and next_index not in visited:
15            queue.append(next_index)
16        next_index = index - arr[index]
17        if next_index >= 0 and next_index not in visited:
18            queue.append(next_index)
19    return False

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Complexity note: The BFS approach processes each index at most once, leading to O(n) time complexity. The space complexity is O(n) due to the queue and visited set.

1BFS is more efficient than DFS for this problem due to its iterative nature and ability to avoid deep recursion.
2Visited tracking is crucial to prevent infinite loops.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.