<p>A frog is crossing a river. The river is divided into some number of units, and at each unit, there may or may not exist a stone. The frog can jump on a stone, but it must not jump into the water.</p>

<p>Given a list of <code>stones</code>&nbsp;positions (in units) in sorted <strong>ascending order</strong>, determine if the frog can cross the river by landing on the last stone. Initially, the frog is on the first stone and assumes the first jump must be <code>1</code> unit.</p>

<p>If the frog&#39;s last jump was <code>k</code> units, its next jump must be either <code>k - 1</code>, <code>k</code>, or <code>k + 1</code> units. The frog can only jump in the forward direction.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> stones = [0,1,3,5,6,8,12,17]
<strong>Output:</strong> true
<strong>Explanation:</strong> The frog can jump to the last stone by jumping 1 unit to the 2nd stone, then 2 units to the 3rd stone, then 2 units to the 4th stone, then 3 units to the 6th stone, 4 units to the 7th stone, and 5 units to the 8th stone.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> stones = [0,1,2,3,4,8,9,11]
<strong>Output:</strong> false
<strong>Explanation:</strong> There is no way to jump to the last stone as the gap between the 5th and 6th stone is too large.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= stones.length &lt;= 2000</code></li>
	<li><code>0 &lt;= stones[i] &lt;= 2<sup>31</sup> - 1</code></li>
	<li><code>stones[0] == 0</code></li>
	<li><code>stones</code>&nbsp;is sorted in a strictly increasing order.</li>
</ul>

class Solution {
public:
    bool canCross(vector<int>& a) {
        long n = a.size();
        
        unordered_map<long,long>idx;


        for(int i=0;i<n;i++)
        {
            idx[a[i]] = i+1;
        }

        long dp[2000][2000];
        memset(dp,0,sizeof(dp));
        long N = INT_MAX;

        for(int i=0;i<n;i++)dp[i][n-1] = 1;

        for(int i=n-3;i>=0;i--)
        {
            for(int j=i+1;j<n;j++)
            {   
                int dis = a[j]-a[i];

                if(dis-1>2000)continue;
                
                int id1 = idx[a[j]+dis-1];
                int id2 = idx[a[j]+dis];
                int id3 = idx[a[j]+dis+1];

                

                if(id1&&id1!=j)
                {
                    dp[i][j] = dp[i][j]||dp[j][id1-1];

                    
                }

                if(id2&&id2!=j)
                {
                    dp[i][j] = dp[i][j]||dp[j][id2-1];
                    
                }


                  if(id3&&id3!=j)
                {
                    dp[i][j] = dp[i][j]||dp[j][id3-1];
                }

                
            }
        }

        

        return dp[0][1]&&a[1]==1;
        


    
    }
};
67938