LeetCode in Kotlin
975. Odd Even Jump
Hard
You are given an integer array arr. 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
jsuch 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
jsuch 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 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 * 1040 <= arr[i] < 105
Solution
class Solution {
private lateinit var valToPos: IntArray
fun oddEvenJumps(arr: IntArray): Int {
val size = arr.size
val odd = BooleanArray(size)
val even = BooleanArray(size)
valToPos = IntArray(100001)
valToPos.fill(-1)
valToPos[arr[size - 1]] = size - 1
even[size - 1] = true
odd[size - 1] = even[size - 1]
var count = 1
for (i in size - 2 downTo 0) {
val curVal = arr[i]
val maxS = findMaxS(curVal)
val minL = findMinL(curVal)
if (minL != -1 && even[minL]) {
// System.out.println("find minL is true at: "+minL+" start from "+i);
odd[i] = even[minL]
count++
}
if (maxS != -1) {
even[i] = odd[maxS]
}
valToPos[arr[i]] = i
}
return count
}
private fun findMaxS(`val`: Int): Int {
for (i in `val` downTo 0) {
if (valToPos[i] != -1) {
return valToPos[i]
}
}
return -1
}
private fun findMinL(`val`: Int): Int {
for (i in `val`..100000) {
if (valToPos[i] != -1) {
return valToPos[i]
}
}
return -1
}
}