<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) {
        
        int length  = arr.size();

        set<pair<int,int>> st1;
        set<pair<int,int>> st2;

        vector<bool> odd(length,false);
        vector<bool> even(length,false);
        odd[length-1] = true;
        even[length-1] = true;
        st1.insert({arr[length-1],length-1});
        st2.insert({-arr[length-1],length-1});
        int i = length - 2;
        while(i>=0){

            auto successor = st1.lower_bound({arr[i],INT_MIN});
            auto predecessor = st2.lower_bound({-arr[i],INT_MIN});

            if(successor!=st1.end()){
                odd[i] = even[successor->second];
            }

            if(predecessor!=st2.end()){
                even[i] = odd[predecessor->second];
            }

            st1.insert({arr[i],i});
            st2.insert({-arr[i],i});

            i--;
        }
        int count =0;
        for(int i=0;i<length;i++){

            if(odd[i]){
                count++;
            }
        }

        return count;
    }
};
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