<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:
 vector < vector <int>> dp;
   bool recfun(  vector<int>& a , int k ,  int i   ){
                  if( dp[ i ][k] !=-1) return dp[i][k];
                     if( i == a.size()-1) return dp [i ][ k ]   =  true;
                   int curr = a[i];
                   bool flg = false;
                   int n = a.size() ;
                   for( int j = i+1 ; j <n;j++){
                          int num = a[ j ];
                          if( num - curr > k + 1 ){ break;}
                          if( num - curr == k - 1) flg =  flg | recfun( a,k - 1,j  );
                        else  if( num - curr == k ) flg =  flg | recfun( a, k , j);
                          else if( num - curr == k+1 ) flg =  flg | recfun( a, k+1 , j);
                   }

              return  dp [ i ][ k ]   =  flg;
   }
    bool canCross(  vector<int>& st ) {
             dp.resize(st.size() + 1 , vector <int> ( 2300, -1) );
             if( st[1] != 1) return false;
              return  recfun( st , 1 , 1  );
             
              
    } 
};
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