<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 {
public:
    
    bool solve(vector<int>&nums,int start,vector<int>&dp){
        int n = nums.size();
        if(start == n-1){
            return true;
        }
        if(start>=n){
            return false;
        }

        if(dp[start]!=-1){
            return dp[start];
        }

        int maxi = min(nums[start] + start, n-1);

        for(int i=start+1;i<=maxi;i++){
            if(solve(nums,i,dp)){
                return dp[start] = true;
            }
        }

        return dp[start] = false;
    }

    bool canJump(vector<int>& nums) {
        vector<int>dp(nums.size()+1,-1);
        return solve(nums,0,dp);
    }
};
54000