<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 util(vector<int> &nums, int ind, vector<int>& dp)
    {
        if (ind >= nums.size()-1)
        {
            return true;
        }

        if (dp[ind] != -1) return dp[ind];
        bool ans = false;
        for (int i=1; i<= nums[ind]; i++)
        {
            ans |= util(nums, ind + i, dp);
            if (ans) break;
        }

        return dp[ind]=ans;
    }
    bool canJump(vector<int>& nums) {
        int n = nums.size();
        if (n == 1) return true;
        vector<int> dp(n+5, -1);

        // return util(nums, 0, dp);
        vector<bool> t(n+5, false); // t[i] -> I can reach till ith index from starting

        t[0] = true; // can reach 0th index always

        for (int i=1; i<n;i++)
        {
            for (int j=i-1; j>=0;j--)
            {
                if (j+nums[j] >= i && t[j])
                {
                    t[i] = true;
                    break;
                }
            }
        }

        return t[n-1];
    }
};
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