main.go

// Source: https://leetcode.com/problems/odd-even-jump
// Title: Odd Even Jump
// Difficulty: Hard
// Author: Mu Yang <http://muyang.pro>

////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
// You are given an integer array `arr`. From some starting index, you can make a series of jumps. The (1^st, 3^rd, 5^th, ...) jumps in the series are called **odd-numbered jumps**, and the (2^nd, 4^th, 6^th, ...) 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 `arr[i] <= arr[j]` and `arr[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 `arr[i] >= arr[j]` and `arr[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 `arr.length - 1`) by jumping some number of times (possibly 0 or more than once).
//
// Return the number of **good** starting indices.
//
// **Example 1:**
//
// ```
// Input: arr = [10,13,12,14,15]
// Output: 2
// Explanation:
// 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.
// ```
//
// **Example 2:**
//
// ```
// Input: arr = [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 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'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.
// ```
//
// **Example 3:**
//
// ```
// Input: arr = [5,1,3,4,2]
// Output: 3
// Explanation: We can reach the end from starting indices 1, 2, and 4.
// ```
//
// **Constraints:**
//
// - `1 <= arr.length <= 2 * 10^4`
// - `0 <= arr[i] < 10^5`
//
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

package main

import (
	"github.com/emirpasic/gods/v2/trees/redblacktree"
)

// Use Tree Map
//
// Use an array to store if a node can reach the end.
// Loop from right to left.
// For each node, try both odd and event cases, and find the min-larger / max-smaller node.
// If that node can reach the end, so do current node.
func oddEvenJumps(arr []int) int {
	n := len(arr)
	set := redblacktree.New[int, int]() // val -> idx

	odds := make([]bool, n)
	evens := make([]bool, n)
	odds[n-1] = true
	evens[n-1] = true
	set.Put(arr[n-1], n-1)

	for i := n - 2; i >= 0; i-- {
		if node, found := set.Ceiling(arr[i]); found {
			odds[i] = evens[node.Value]
		}
		if node, found := set.Floor(arr[i]); found {
			evens[i] = odds[node.Value]
		}
		set.Put(arr[i], i)
	}

	ans := 0
	for i := range n {
		if odds[i] {
			ans++
		}
	}

	return ans
}