<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:
#define ll long long
    bool solve(vector<int>& stones, int index, int k, unordered_map<int, int>& mp, vector<vector<int>>& dp){
        if(index == stones.size()-1) return true;
        if(index >= stones.size()) return false;

        if(dp[index][k] != -1) return dp[index][k];

        bool a = false, b = false, c = false;
        if(k!=1 && mp.find(stones[index]+k-1) != mp.end()){
            a = solve(stones, mp[stones[index]+k-1], k-1, mp, dp);
            if(a) return dp[index][k] = true;
        }
        if(mp.find(stones[index]+k) != mp.end()){
            b = solve(stones, mp[stones[index]+k], k, mp, dp);
            if(b) return dp[index][k] = true;
        }
        if(mp.find(stones[index]+k+1) != mp.end()){
            c = solve(stones, mp[stones[index]+k+1], k+1, mp, dp);
            if(c) return dp[index][k] = true;
        }

        return dp[index][k] = false;
    }

    bool canCross(vector<int>& stones) {
        int n = stones.size();
        if(stones[1] != 1) return false;
        if(stones.back() > 2001000) return false;

        vector<vector<int>> dp(n, vector<int>(2001, -1));

        unordered_map<int, int> mp;
        for(int i = 0; i < n; i++){
            mp[stones[i]] = i;
        }
        return solve(stones, 1, 1, mp, dp);
    }
};
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