<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>

class Solution:
    def jump(self, nums: List[int]) -> int:
        n = len(nums)
        dp = [10001] * n
        dp[0] = 0
        def do_dp(i):
            # print("do_dp",i)
            if dp[i] != 10001:
                return dp[i]
            for j in range(i):
                if j + nums[j] >= i:
                    dp[i] = min(dp[i], do_dp(j)+1)
            return dp[i]
        # do_dp(n-1)
        # print(dp)
        return do_dp(n-1)
19900