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# 975. Odd Even Jump

### Description

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

You may jump forward from index `i` to index `j` (with `i < j`) in the following way:

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

A starting index is **good** if, starting from that index, you can reach the end of the array (index `A.length - 1`) by jumping some number of times (possibly 0 or more than once).

Return *the number of **good** starting indices*.

### Constraints

### Approach

### Links

* GeeksforGeeks
* [Leetcode](https://leetcode.com/problems/odd-even-jump/)
* ProgramCreek
* [YouTube](https://youtu.be/r2I7KIqHTPU)

### **Examples**

{% tabs %}
{% tab title="Example 1" %}
**Input:** A = \[10, 13, 12, 14, 15]

**Output:** 2

**Explanation:**

From starting index i = 0, we can make our 1st jump to i = 2 (since A\[2] is the smallest among A\[1], A\[2], A\[3], A\[4] that is greater or equal to A\[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.
{% endtab %}

{% tab title="Example 2" %}
**Input:** A = \[2,3,1,1,4]

**Output:** 3

**Explanation:**

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 A\[1] is the smallest value in \[A\[1], A\[2], A\[3], A\[4]] that is greater than or equal to A\[0].

During our 2nd jump (even-numbered), we jump from i = 1 to i = 2 because A\[2] is the largest value in \[A\[2], A\[3], A\[4]] that is less than or equal to A\[1]. A\[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 A\[3] is the smallest value in \[A\[3], A\[4]] that is greater than or equal to A\[2].

We can'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'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.
{% endtab %}

{% tab title="Example 3" %}
**Input:** A = \[5, 1, 3, 4, 2]

**Output:** 3

**Explanation:**

We can reach the end from starting indices 1, 2, and 4.
{% endtab %}
{% endtabs %}

### **Solutions**

{% tabs %}
{% tab title="Solution 1" %}

```java
/**
 * Time complexity : O(NlogN), where NN is the length of A.
 * Space complexity : O(N)
 */

class Solution {
    public int oddEvenJumps(int[] A) {
        if(A == null || A.length == 0) return 0;
        
        int n = A.length;
        boolean[] lower = new boolean[n];
        boolean[] higher = new boolean[n];
        
        higher[n-1] = true;
        lower[n-1] = true;
        
        int goodStartingIndices = 1;
        TreeMap<Integer, Integer> map = new TreeMap();
        map.put(A[n-1], n-1);
        
        for(int i = n-2; i >= 0; i--) {
            Map.Entry<Integer, Integer> higherPair = map.ceilingEntry(A[i]);
            if(higherPair != null) {
                higher[i] = lower[(int) higherPair.getValue()];
            }
            
            Map.Entry<Integer, Integer> lowerPair = map.floorEntry(A[i]);
            if(lowerPair != null) {
                lower[i] = higher[(int) lowerPair.getValue()];
            }
            
            if(higher[i]) goodStartingIndices++;
            map.put(A[i], i);
        }
        return goodStartingIndices;
    }
}
```

{% endtab %}
{% endtabs %}

### **Follow up**

*