<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();
      if (stones[1] != 1) {
        return false;
      }
      std::vector<std::unordered_map<int, bool>> dp(n + 1);
      dp[0][0] = true;
      for (int i = 1; i < n; ++i) {
        int k = i - 1;
        while (k >= 0) {
          int dist = stones[i] - stones[k];
          bool cur=false, prev=false, next= false;
          if (dp[k].find(dist) != dp[k].end()) {
            cur = dp[k][dist];
          }
          if (dp[k].find(dist + 1) != dp[k].end()) {
            next = dp[k][dist + 1];
          }
          if (dp[k].find(dist - 1) != dp[k].end()) {
            prev = dp[k][dist - 1];
          }
          cur |= next || prev;
          if (cur) {
            dp[i][dist] = true;
          }
          if (dist > 2e6) {
            break;
          }
          --k;
        }
      }  
      bool ans = false;
      for (auto it : dp[n - 1]) {
        ans |= it.second;
      }
      return ans;
    }
};

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