Press "Enter" to skip to content

花花酱 LeetCode 975. Odd Even Jump

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.

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

  • During odd numbered jumps (ie. 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 indexes j, you can only jump to the smallest such index j.
  • During even numbered jumps (ie. 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 indexes 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 indexes.

Example 1:

Input: [10,13,12,14,15] 
Output: 2

Example 2:

Input: [2,3,1,1,4] 
Output: 3

Example 3:

Input: [5,1,3,4,2] 
Output: 3

Note:

  1. 1 <= A.length <= 20000
  2. 0 <= A[i] < 100000

Solution: Binary Search + DP

Time complexity: O(nlogn)
Space complexity: O(n)

C++

请尊重作者的劳动成果,转载请注明出处!花花保留对文章/视频的所有权利。
如果您喜欢这篇文章/视频,欢迎您捐赠花花。
If you like my articles / videos, donations are welcome.

Buy anything from Amazon to support our website
您可以通过在亚马逊上购物(任意商品)来支持我们

Paypal
Venmo
huahualeetcode
微信打赏

Be First to Comment

Leave a Reply