Class Solution
-
- All Implemented Interfaces:
public final class Solution
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<sup>st</sup>, 3<sup>rd</sup>, 5<sup>th</sup>, ...) jumps in the series are called odd-numbered jumps , and the (2<sup>nd</sup>, 4<sup>th</sup>, 6<sup>th</sup>, ...) 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 indexj
(withi < j
) in the following way:During odd-numbered jumps (i.e., jumps 1, 3, 5, ...), you jump to the index
j
such thatarr[i] <= arr[j]
andarr[j]
is the smallest possible value. If there are multiple such indicesj
, you can only jump to the smallest such indexj
.During even-numbered jumps (i.e., jumps 2, 4, 6, ...), you jump to the index
j
such thatarr[i] >= arr[j]
andarr[j]
is the largest possible value. If there are multiple such indicesj
, you can only jump to the smallest such indexj
.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 arr2 is the smallest among arr1,
arr2, arr3, arr4 that is greater or equal to arr0), 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 arr1 is the smallest value in [arr1,
arr2, arr3, arr4] that is greater than or equal to arr0.
During our 2nd jump (even-numbered), we jump from i = 1 to i = 2 because arr2 is the largest value in
[arr2, arr3, arr4] that is less than or equal to arr1. arr3 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 arr3 is the smallest value in
[arr3, arr4] that is greater than or equal to arr2.
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:
<code>1 <= arr.length <= 2 * 10<sup>4</sup></code>
<code>0 <= arri< 10<sup>5</sup></code>
-
-
Constructor Summary
Constructors Constructor Description Solution()
-
Method Summary
Modifier and Type Method Description final Integer
oddEvenJumps(IntArray arr)
-
-
Method Detail
-
oddEvenJumps
final Integer oddEvenJumps(IntArray arr)
-
-
-
-