<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:
//greeedy algorithm is very different from recursion and it also can be done using recursion and memoization. In this question. I cannot know whether or not i can use the recursion or greedy over here. 
//FIRST Let me code the normal recursion here. using memoization.
    int dp[10001];
    bool recursion(vector<int> &nums, int idx, int n){
        if(idx >=n-1) return true; 
        if(dp[idx]!=-1){
            return dp[idx];
        }
        for(int i = 1; i<= nums[idx]; i++){
            if(recursion(nums, idx + i, n) == true){
                return dp[idx] = true;
            }
        }
        return dp[idx] = false;
        
    }
    bool canJump(vector<int>& nums) {

        int n = nums.size();
        if(n == 1) return true;
        memset(dp, -1, sizeof(dp));
        return recursion(nums, 0, n);
    }
};
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