<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 rec(set<int>& stone_hai, int i, int last_step, unordered_map<int, unordered_map<int, int>>& memo, int target) {
        // Pruning
        if (i > target || i < 0 || last_step <= 0) return false;
        if (stone_hai.find(i) == stone_hai.end()) return false;
        
        // Base case
        if (i == target) return true;
        
        // Check memoization
        if (memo[i].find(last_step) != memo[i].end()) return memo[i][last_step];
        
        // Save and call
        bool first_step = rec(stone_hai, i + last_step, last_step, memo, target);
        bool second_step = rec(stone_hai, i + last_step - 1, last_step - 1, memo, target);
        bool third_step = rec(stone_hai, i + last_step + 1, last_step + 1, memo, target);
        
        // Save and return
        return memo[i][last_step] = first_step || second_step || third_step;
    }
    
    bool canCross(vector<int>& stones) {
        int n = stones.size();
        set<int> stone_hai;
        for (auto it : stones) stone_hai.insert(it);
        int target = stones[n - 1];
        
        // Memoization table
        unordered_map<int, unordered_map<int, int>> memo;
        
        return rec(stone_hai, 1, 1, memo, target);
    }
};
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