<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:
    bool canCross(vector<int>& stones) {
        int n = stones.size();
        // vector<unordered_set<long long>> dp(n);
        // dp[0].insert(0);
        // for(int i = 1; i < n; ++i){
        //     for(int j = 0; j < i; ++j){
        //         long long diff = stones[i]-stones[j];
        //         if(dp[j].count(diff) || dp[j].count(diff-1) || dp[j].count(diff+1)){
        //             dp[i].insert(diff);
        //         }
        //     }
        // }
        // // for(int steps = 1; steps <= 2000; ++steps){
        // //     if(helper(stones, n-1, steps)){
        // //         return true;
        // //     }
        // // }
        // // for(int j = 0; j <= 2000; ++j){
        // //     if(dp[n-1][j]){
        // //         return true;
        // //     }
        // // }
        // return !dp[n-1].empty();
        unordered_map<int,unordered_map<long long,bool>> cache;
        for(int i = 0; i < n - 1; i++){
            int steps = stones[n-1]-stones[i];
            if(helper(stones, n-1, steps,cache)){
                return true;
            }
        }
        // for(int steps = 1; steps <= 2000; ++steps){
        //     if(helper(stones, n-1, steps,cache)){
        //         return true;
        //     }
        // }
        return false;
    }
    bool helper(vector<int>& stones, int cur, int steps, unordered_map<int,unordered_map<long long,bool>>& cache){
        if(cur == 0 ){
            return steps == 0;
        }
        if(steps < 1){
            return false;
        }
        if(cache.count(cur) && cache[cur].count(steps)){
            return cache[cur][steps];
        }
        for(int i = cur-1; i >= 0; --i){
            long long dif = stones[cur]-stones[i];
            if(dif != steps ){
                continue;
            }else{
                if(helper(stones,i, dif,cache) || helper(stones,i, dif-1,cache) || helper(stones,i, dif+1,cache)){
                    cache[cur][steps] = true;
                    return true;
                }else{
                    break;
                }
            }
            
            
            
        }
        cache[cur][steps] = false;
        return false;
    }
};
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