g0901_1000.s0975_odd_even_jump

Class Solution

java.lang.Object

g0901_1000.s0975_odd_even_jump.Solution

public class Solution
extends Object

975 - Odd Even Jump.

Hard

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 * 104
• 0 <= arr[i] < 105

Constructor Summary

Constructors

Constructor and Description
Solution()

Method Summary

Modifier and Type	Method and Description
int	oddEvenJumps(int[] arr)

Methods inherited from class java.lang.Object

clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait

Constructor Detail

Solution

public Solution()

Method Detail

oddEvenJumps

public int oddEvenJumps(int[] arr)