<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 solve(vector<int> &stones, int index, int lastJump, int n, vector<vector<int>> &dp) {
        if (index == n - 1) // If we reach the last stone, return true
            return true;

        // If this state has already been computed, return the stored result
        if(dp[index][lastJump] != -1)
            return dp[index][lastJump];

        // For k-1, k, and k+1
        bool case1 = false, case2 = false, case3 = false;
        
        for (int i = index + 1; i < n; i++) {
            int gap = stones[i] - stones[index]; // Calculate the distance between stones

            // Check for k-1
            if (gap == lastJump - 1 && lastJump - 1 > 0) { 
                case1 = solve(stones, i, gap, n, dp);
            }
            // Check for k
            else if (gap == lastJump) { 
                case2 = solve(stones, i, gap, n, dp);
            }
            // Check for k+1
            else if (gap == lastJump + 1) { 
                case3 = solve(stones, i, gap, n, dp);
            }

            // If any of the cases succeed, no need to continue further
            if (case1 || case2 || case3) 
                return dp[index][lastJump] = true;
        }

        // If none of the cases succeed, return false
        return dp[index][lastJump] = false;
    }

    bool canCross(vector<int>& stones) {
        int lastJump = 1; // The first jump must be 1 unit
        int n = stones.size();

        // If the second stone is not 1 unit away, it's impossible to cross
        if (stones[1] != 1) 
            return false;

        // Initialize dp table with -1 (indicating the state has not been computed yet)
        vector<vector<int>> dp(n, vector<int>(n + 1, -1));

        return solve(stones, 1, lastJump, n, dp); // Start from the first stone with the first jump of 1
    }
};

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