<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:
    int bs(int l, int h, vector<int>&nums, int key){
        
        while(l<=h){
            
            int mid=(h-l)/2+l;
            if(nums[mid]==key){
                // cout<<"yay";
                // cout<<mid;
                return mid;
            }
            else if(nums[mid]>key){
                h=mid-1;
            }
            else{
                l=mid+1;
            }
        }
        return -1;
    }
    int solve(int key, int k, vector<int>&stones, vector<vector<int>>&vj, int n){
         int a=bs(0, n-1, stones, key);
        //  return 1;
        //  cout<<a<<endl;
        //  return 1;
        if(a==-1)return 0;
        if(a==stones.size()-1){cout<<"YAY";return 1;}
       
        if(vj[a][k]!=-1)return vj[a][k];
        int ans=0;
        if(k>0)ans=ans||solve(key+k, k, stones, vj, n);
        if(k-1>0)ans=ans||solve(key+k-1, k-1, stones, vj, n);
        if(k+1>0)ans=ans||solve(key+k+1, k+1, stones, vj, n);
        return vj[a][k]=ans;
    }
    bool canCross(vector<int>& stones) {
        vector<vector<int>>vj(stones.size(), vector<int>(2002, -1));
        int a= solve(stones[0], 0, stones, vj, stones.size());
        // for(int i=0; i<stones.size(); i++)cout<<vj[i]<<" ";

        return (bool)a;
    }
};
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