<p>You are given a <strong>0-indexed</strong> array of integers <code>nums</code> of length <code>n</code>. You are initially positioned at <code>nums[0]</code>.</p>

<p>Each element <code>nums[i]</code> represents the maximum length of a forward jump from index <code>i</code>. In other words, if you are at <code>nums[i]</code>, you can jump to any <code>nums[i + j]</code> where:</p>

<ul>
	<li><code>0 &lt;= j &lt;= nums[i]</code> and</li>
	<li><code>i + j &lt; n</code></li>
</ul>

<p>Return <em>the minimum number of jumps to reach </em><code>nums[n - 1]</code>. The test cases are generated such that you can reach <code>nums[n - 1]</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [2,3,1,1,4]
<strong>Output:</strong> 2
<strong>Explanation:</strong> The minimum number of jumps to reach the last index is 2. Jump 1 step from index 0 to 1, then 3 steps to the last index.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [2,3,0,1,4]
<strong>Output:</strong> 2
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>4</sup></code></li>
	<li><code>0 &lt;= nums[i] &lt;= 1000</code></li>
	<li>It&#39;s guaranteed that you can reach <code>nums[n - 1]</code>.</li>
</ul>

from collections import deque

class Solution:
    def jump(self, nums: List[int]) -> int:
        # BFS
        visited = set()
        q = deque([0])
        # key: index / value: distance
        d = {0: 0}

        while len(q) > 0:
            i = q.popleft()
            max_jump = nums[i]
            # print(f"i = {i} ({nums[i]})")

            # for nodes (indices) reachable from i
            for j in range(max_jump):
                target = i + j + 1
                # can't go beyond end of list
                if target >= len(nums) or target in d:
                    continue
                
                this_d = d[i] + 1
                if target == len(nums) - 1:
                    return this_d
                
                # print(f"  target = {target} (this_d = {this_d})")
                
                # update distance to target
                d[target] = this_d
                
                # add to queue if not visited
                if target not in visited:
                    q.append(target)
            
            visited.add(i)

            # print(f"i = {i} ({nums[i]}) // d = {d}\n")
        
        # print(d)
        return d[len(nums) - 1]

18600