<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 canCrossII(int sum, int prev, vector<int>& stones, map<int, int>& m,
                    map<vector<int>, bool>& dp) {
        if (sum == stones[stones.size() - 1]) {
            return true;
        }
        if (m.count(sum) == 0) {
            return false;
        }
        if (prev - 1 != 0) {
            if (dp.count({sum + prev - 1, prev - 1}) == 0) {
                dp[{sum + prev - 1, prev - 1}] =
                    canCrossII(sum + prev - 1, prev - 1, stones, m, dp);
            }
        }
        if (dp.count({sum + prev, prev}) == 0) {
            dp[{sum + prev, prev}] =
                canCrossII(sum + prev, prev, stones, m, dp);
        }
        if (dp.count({sum + prev + 1, prev + 1}) == 0) {
            dp[{sum + prev + 1, prev + 1}] =
                canCrossII(sum + prev + 1, prev + 1, stones, m, dp);
        }
        if (prev - 1 != 0) {
            dp[{sum, prev}] = dp[{sum + prev, prev}] ||
                              dp[{sum + prev - 1, prev - 1}] ||
                              dp[{sum + prev + 1, prev + 1}];
        } else {
            dp[{sum, prev}] =
                dp[{sum + prev, prev}] || dp[{sum + prev + 1, prev + 1}];
        }
        return dp[{sum, prev}];
    }
    bool canCross(vector<int>& stones) {
        map<int, int> m;
        map<vector<int>, bool> dp;
        for (unsigned int i = 0; i < stones.size(); i++) {
            m[stones[i]]++;
        }
        return canCrossII(1, 1, stones, m, dp);
    }
};
242583