<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:
    static bool cmp(const pair<int, int>& a, const pair<int, int>& b) {
        if (a.first != b.first) {
            return a.first < b.first;
        } else {
            return a.second > b.second;
        }
   }

    int dp[20010][2];
    int n ;
    int rec(int level, int jump, vector<int>&nextsmaller, vector<int>&nextgreater){
        if(level == n-1){
            return 1;
        }
         
        if(dp[level][jump] != -1)return dp[level][jump];

        int ans =0 ;
        if(jump == 1){
            //odd
            
            if(nextgreater[level] != -1){
                ans = ans|rec(nextgreater[level],0,nextsmaller,nextgreater);
            }
        }
        else{
            //even
            if(nextsmaller[level] != -1){
                ans = ans|rec(nextsmaller[level],1,nextsmaller,nextgreater);
            }
        }

        return dp[level][jump] = ans ;
    }

    int oddEvenJumps(vector<int>& arr) {
        n = arr.size();
        vector<pair<int,int>>vec;
        for(int i =0 ; i<n ; i++){
            vec.push_back({arr[i],i});
        }
        sort(vec.begin(),vec.end());
        vector<int>nums(n,0);
        for(int i =0 ; i<n  ; i++){
            nums[i] = vec[i].second ;
        }
        

        vector<int>nextsmaller(n,0);
        vector<int>nextgreater(n,0);
        vector<int>nxt(n,0);
        stack<int>st;
        for(int i = n-1 ; i>=0 ; i--){
            while(!st.empty() && nums[st.top()] < nums[i]){
                st.pop();
            }

            if(!st.empty()){
                nextgreater[i] = nums[st.top()];
            }
            else{
                nextgreater[i] = -1 ;
            }

            st.push(i);
        }
        
         for(int i =0 ;i<n ;i++){
            nxt[nums[i]] = nextgreater[i];
         }
    
        


        while(!st.empty()){
            st.pop();
        }


        for(int i =0 ; i<n  ; i++){
            nums[i] = vec[i].second ;
        }

        vector<int>smll(n,0);

        sort(vec.begin(),vec.end(),cmp);

        for(int i =0 ; i<n  ; i++){
            nums[i] = vec[i].second ;
        }

        for(int i = 0 ; i<n ; i++){
            while(!st.empty() && nums[st.top()] < nums[i]){
                st.pop();
            }

            if(!st.empty()){
                nextsmaller[i] = nums[st.top()];
            }
            else{
                nextsmaller[i] = -1 ;
            }

            st.push(i);
        }

         for(int i =0 ;i<n ;i++){
            smll[nums[i]] = nextsmaller[i];
         }

        //  for(int i =0 ;i<n ;i++){
        //     cout<<nxt[i]<<" ";
        //  }
        //  cout<<endl;
        //  for(int i =0 ;i<n ;i++){
        //     cout<<smll[i]<<" ";
        //  }
        //  cout<<endl;
        

         memset(dp,-1,sizeof(dp));
         int ans =0 ;
         for(int i =0 ; i<n ;i++){
            if(rec(i,1,smll,nxt) == 1){
                // cout<<i<<" ";
                ans++;
            }
         }
         return ans ;
    }
};
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