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This approach employs DFS for exploring each possible path starting from the given start index. You recursively explore both forward and backward jumps, marking indices as visited to avoid cycles. If you reach an index with a value of 0, return true. If all paths are exhausted without finding a zero, return false.
Time Complexity: O(n), where n is the length of the array, because every index is visited at most once.
Space Complexity: O(n) due to the recursion stack and the visited array.
1import java.util.*;
2
3public class Solution {
4 public boolean canReach(int[] arr, int start) {
5 return dfs(arr, start);
6 }
7
8 private boolean dfs(int[] arr, int pos) {
9 if (pos < 0 || pos >= arr.length || arr[pos] < 0)
10 return false;
11 if (arr[pos] == 0)
12 return true;
13 int jump = arr[pos];
14 arr[pos] = -arr[pos]; // Mark visited
15 boolean result = dfs(arr, pos + jump) || dfs(arr, pos - jump);
16 arr[pos] = -arr[pos]; // Reset
17 return result;
18 }
19}The Java implementation mirrors the C++ solution, using a DFS approach with in-place marking of visited positions to prevent revisiting nodes. This method recursively explores all accessible paths by jumping forward and backward.
This approach adopts a BFS strategy to explore all reachable indices from the starting point iteratively. Using a queue, enqueue all possible jump indices from a given position if they haven't been visited yet. Pop indices from the queue, and for each pop, check whether the position contains a zero, marking indices visited as you go.
Time Complexity: O(n), as all reachable indices are visited.
Space Complexity: O(n), given the storage requirements for the queue and the visited array.
1
The Python solution uses a deque to manage the exploration queue, marking indices as visited in an auxiliary boolean list. By iterating over the two possible jumps from each current index, the solution entirely embodies a typical BFS workflow.