<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 {
    vector<int>a;
    int n;
   int dp[2010][5010];
   unordered_map<int,int>mp;

    int rec(int x,int j){
        if(x==n-1){
            return 1;
            
        }
        
        //if(x==n)return 1;

        if(dp[x][j]!=-1)return dp[x][j];

        int ans=0;
       if(mp.find(a[x]+j)!=mp.end())ans=rec(mp[a[x]+j],j);
       if(mp.find(a[x]+j+1)!=mp.end()) ans=max(ans,rec(mp[a[x]+j+1],j+1));
       if(j>1&&(mp.find(a[x]+j-1)!=mp.end())) ans=max(ans,rec(mp[a[x]+j-1],j-1));
       //use of mp[a[x]+j+1] in rec(mp[a[x]+j+1],j+1)) to find the index at which a[x]+j+1 value is there

        return dp[x][j]=ans;

    }
public:
    bool canCross(vector<int>& stones) {
        n=stones.size();
        for(int i=0;i<n;i++){
            a.push_back(stones[i]);
        }

        //a=stones;
        memset(dp,-1,sizeof(dp));
        if(a[1]!=1)return false;
        //return true;
        for(int i=0;i<n;i++){
            mp[stones[i]]=i;
        }

        int k=rec(1,1);
        if(k==1)return true;
        else return false;
        
    }
};
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