<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 solve(vector<int>& s, unordered_map<int,int>&m, int i, int j, vector<vector<int>>&dp){
    //     if(i>=s.size()-1) return 1;
    //     if(dp[i][j]!=-1) return dp[i][j];
    //     int ans = 0;
    //     if(m.count(s[i]+j-1) && j!=1) ans = ans | solve(s,m,m[s[i]+j-1],j-1,dp);
    //     if(m.count(s[i]+j)) ans = ans | solve(s,m,m[s[i]+j],j,dp);
    //     if(m.count(s[i]+j+1)) ans = ans | solve(s,m,m[s[i]+j+1],j+1,dp);
    //     return dp[i][j]=ans;
    // }

    bool canCross(vector<int>& s) {
        int n = s.size();
        if(n==1) return 1;
        if(s[1]-s[0]>1) return 0;
        vector<vector<int>>dp(s.size()+1, vector<int>(s.size()+1,0));
        for(int i=1; i<=n; i++) dp[n-1][i]=1;
        unordered_map<int,int>m;
        for(int i=0; i<s.size(); i++){
            m[s[i]]=i;
        }
        for(int i=n-2; i>=1; i--){
            for(int j=n-1; j>=1; j--){
                int ans = 0;
                if(m.count(s[i]+j-1) && j!=1) ans = ans | dp[m[s[i]+j-1]][j-1];
                if(m.count(s[i]+j)) ans = ans | dp[m[s[i]+j]][j];
                if(m.count(s[i]+j+1)) ans = ans | dp[m[s[i]+j+1]][j+1];
                dp[i][j]=ans;
            }
        }
        return dp[1][1];
    }
};
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