Frog Jump
<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>
Array; Dynamic Programming
class Solution {
public:
    bool canCross(vector<int>& stones) {
        int n = stones.size();
        int f[n][n];
        memset(f, -1, sizeof(f));
        unordered_map<int, int> pos;
        for (int i = 0; i < n; ++i) {
            pos[stones[i]] = i;
        }
        function<bool(int, int)> dfs = [&](int i, int k) -> bool {
            if (i == n - 1) {
                return true;
            }
            if (f[i][k] != -1) {
                return f[i][k];
            }
            for (int j = k - 1; j <= k + 1; ++j) {
                if (j > 0 && pos.count(stones[i] + j) && dfs(pos[stones[i] + j], j)) {
                    return f[i][k] = true;
                }
            }
            return f[i][k] = false;
        };
        return dfs(0, 0);
    }
};