<p>You are given an integer array <code>arr</code>. From some starting index, you can make a series of jumps. The (1<sup>st</sup>, 3<sup>rd</sup>, 5<sup>th</sup>, ...) jumps in the series are called <strong>odd-numbered jumps</strong>, and the (2<sup>nd</sup>, 4<sup>th</sup>, 6<sup>th</sup>, ...) jumps in the series are called <strong>even-numbered jumps</strong>. Note that the <strong>jumps</strong> are numbered, not the indices.</p>

<p>You may jump forward from index <code>i</code> to index <code>j</code> (with <code>i &lt; j</code>) in the following way:</p>

<ul>
	<li>During <strong>odd-numbered jumps</strong> (i.e., jumps 1, 3, 5, ...), you jump to the index <code>j</code> such that <code>arr[i] &lt;= arr[j]</code> and <code>arr[j]</code> is the smallest possible value. If there are multiple such indices <code>j</code>, you can only jump to the <strong>smallest</strong> such index <code>j</code>.</li>
	<li>During <strong>even-numbered jumps</strong> (i.e., jumps 2, 4, 6, ...), you jump to the index <code>j</code> such that <code>arr[i] &gt;= arr[j]</code> and <code>arr[j]</code> is the largest possible value. If there are multiple such indices <code>j</code>, you can only jump to the <strong>smallest</strong> such index <code>j</code>.</li>
	<li>It may be the case that for some index <code>i</code>, there are no legal jumps.</li>
</ul>

<p>A starting index is <strong>good</strong> if, starting from that index, you can reach the end of the array (index <code>arr.length - 1</code>) by jumping some number of times (possibly 0 or more than once).</p>

<p>Return <em>the number of <strong>good</strong> starting indices</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> arr = [10,13,12,14,15]
<strong>Output:</strong> 2
<strong>Explanation:</strong> 
From starting index i = 0, we can make our 1st jump to i = 2 (since arr[2] is the smallest among arr[1], arr[2], arr[3], arr[4] that is greater or equal to arr[0]), then we cannot jump any more.
From starting index i = 1 and i = 2, we can make our 1st jump to i = 3, then we cannot jump any more.
From starting index i = 3, we can make our 1st jump to i = 4, so we have reached the end.
From starting index i = 4, we have reached the end already.
In total, there are 2 different starting indices i = 3 and i = 4, where we can reach the end with some number of
jumps.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> arr = [2,3,1,1,4]
<strong>Output:</strong> 3
<strong>Explanation:</strong> 
From starting index i = 0, we make jumps to i = 1, i = 2, i = 3:
During our 1st jump (odd-numbered), we first jump to i = 1 because arr[1] is the smallest value in [arr[1], arr[2], arr[3], arr[4]] that is greater than or equal to arr[0].
During our 2nd jump (even-numbered), we jump from i = 1 to i = 2 because arr[2] is the largest value in [arr[2], arr[3], arr[4]] that is less than or equal to arr[1]. arr[3] is also the largest value, but 2 is a smaller index, so we can only jump to i = 2 and not i = 3
During our 3rd jump (odd-numbered), we jump from i = 2 to i = 3 because arr[3] is the smallest value in [arr[3], arr[4]] that is greater than or equal to arr[2].
We can&#39;t jump from i = 3 to i = 4, so the starting index i = 0 is not good.
In a similar manner, we can deduce that:
From starting index i = 1, we jump to i = 4, so we reach the end.
From starting index i = 2, we jump to i = 3, and then we can&#39;t jump anymore.
From starting index i = 3, we jump to i = 4, so we reach the end.
From starting index i = 4, we are already at the end.
In total, there are 3 different starting indices i = 1, i = 3, and i = 4, where we can reach the end with some
number of jumps.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> arr = [5,1,3,4,2]
<strong>Output:</strong> 3
<strong>Explanation:</strong> We can reach the end from starting indices 1, 2, and 4.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= arr.length &lt;= 2 * 10<sup>4</sup></code></li>
	<li><code>0 &lt;= arr[i] &lt; 10<sup>5</sup></code></li>
</ul>


class Solution {
public:
    int oddEvenJumps(vector<int>& arr) {
        if (arr.empty()) return 0;
        
        int n = arr.size();
        vector<int> odd_next(n, -1), even_next(n, -1);

        auto next_greater_element = [&](vector<int>& sorted_indices) {
            vector<int> result(n, -1);
            stack<int> stk;
            for (int i : sorted_indices) {
                while (!stk.empty() && stk.top() < i) {
                    result[stk.top()] = i;
                    stk.pop();
                }
                stk.push(i);
            }
            return result;
        };

        // Calculate odd_next (next greater element) based on sorted indices
        vector<int> sorted_indices(n);
        for (int i = 0; i < n; ++i) sorted_indices[i] = i;
        sort(sorted_indices.begin(), sorted_indices.end(), [&](int a, int b) {
            return arr[a] < arr[b] || (arr[a] == arr[b] && a < b);
        });
        odd_next = next_greater_element(sorted_indices);

        // Calculate even_next (next smaller element) based on sorted indices (in descending order)
        sort(sorted_indices.begin(), sorted_indices.end(), [&](int a, int b) {
            return arr[a] > arr[b] || (arr[a] == arr[b] && a < b);
        });
        even_next = next_greater_element(sorted_indices);

        // dp[i][0] indicates if starting from index i with an odd jump can reach the end
        // dp[i][1] indicates if starting from index i with an even jump can reach the end
        vector<vector<int>> dp(n, vector<int>(2, 0));
        dp[n - 1][0] = dp[n - 1][1] = 1; // The last index is always reachable

        // Traverse from the second last index to the beginning
        for (int i = n - 2; i >= 0; --i) {
            // If odd jump is possible
            if (odd_next[i] != -1) {
                dp[i][0] = dp[odd_next[i]][1];
            }
            // If even jump is possible
            if (even_next[i] != -1) {
                dp[i][1] = dp[even_next[i]][0];
            }
        }

        // The result is the count of starting indices where an odd jump leads to the end
        int result = 0;
        for (int i = 0; i < n; ++i) {
            result += dp[i][0];
        }
        return result;
    }
};

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