<p>You are given an integer array <code>nums</code>. You are initially positioned at the array&#39;s <strong>first index</strong>, and each element in the array represents your maximum jump length at that position.</p>

<p>Return <code>true</code><em> if you can reach the last index, or </em><code>false</code><em> otherwise</em>.</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> true
<strong>Explanation:</strong> 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 = [3,2,1,0,4]
<strong>Output:</strong> false
<strong>Explanation:</strong> You will always arrive at index 3 no matter what. Its maximum jump length is 0, which makes it impossible to reach the last index.
</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;= 10<sup>5</sup></code></li>
</ul>

class Solution {

int dp[10001];

private:
    bool dfs(int i, vector<int>& nums) {
        if (i >= nums.size()) { return false; }
        if (i == nums.size() - 1) {
            return true;
        }
        if (nums[i] == 0) { return false; }
        if (dp[i] != -1) { return dp[i]; }

        for (int j = 1; j <= nums[i]; j++) {
            if (dfs(i + j, nums)) {
                dp[i] = true;
                return true; 
            }
        }
        
        dp[i] = false;
        return false;
    }

public:
    bool canJump(vector<int>& nums) {
        memset(dp, -1, sizeof(dp));
        return dfs(0, nums);
    }
};


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