# 975. Odd Even Jump

## Description

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 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

## Solutions

• class Solution {
private int n;
private Integer[][] f;
private int[][] g;

public int oddEvenJumps(int[] arr) {
TreeMap<Integer, Integer> tm = new TreeMap<>();
n = arr.length;
f = new Integer[n][2];
g = new int[n][2];
for (int i = n - 1; i >= 0; --i) {
var hi = tm.ceilingEntry(arr[i]);
g[i][1] = hi == null ? -1 : hi.getValue();
var lo = tm.floorEntry(arr[i]);
g[i][0] = lo == null ? -1 : lo.getValue();
tm.put(arr[i], i);
}
int ans = 0;
for (int i = 0; i < n; ++i) {
ans += dfs(i, 1);
}
return ans;
}

private int dfs(int i, int k) {
if (i == n - 1) {
return 1;
}
if (g[i][k] == -1) {
return 0;
}
if (f[i][k] != null) {
return f[i][k];
}
return f[i][k] = dfs(g[i][k], k ^ 1);
}
}

• class Solution {
public:
int oddEvenJumps(vector<int>& arr) {
int n = arr.size();
map<int, int> d;
int f[n][2];
int g[n][2];
memset(f, 0, sizeof(f));
for (int i = n - 1; ~i; --i) {
auto it = d.lower_bound(arr[i]);
g[i][1] = it == d.end() ? -1 : it->second;
it = d.upper_bound(arr[i]);
g[i][0] = it == d.begin() ? -1 : prev(it)->second;
d[arr[i]] = i;
}
function<int(int, int)> dfs = [&](int i, int k) -> int {
if (i == n - 1) {
return 1;
}
if (g[i][k] == -1) {
return 0;
}
if (f[i][k] != 0) {
return f[i][k];
}
return f[i][k] = dfs(g[i][k], k ^ 1);
};
int ans = 0;
for (int i = 0; i < n; ++i) {
ans += dfs(i, 1);
}
return ans;
}
};

• from sortedcontainers import SortedDict

class Solution:
def oddEvenJumps(self, arr: List[int]) -> int:
@cache
def dfs(i: int, k: int) -> bool:
if i == n - 1:
return True
if g[i][k] == -1:
return False
return dfs(g[i][k], k ^ 1)

n = len(arr)
g = [[0] * 2 for _ in range(n)]
sd = SortedDict()
for i in range(n - 1, -1, -1):
j = sd.bisect_left(arr[i])
g[i][1] = sd.values()[j] if j < len(sd) else -1
j = sd.bisect_right(arr[i]) - 1
g[i][0] = sd.values()[j] if j >= 0 else -1
sd[arr[i]] = i
return sum(dfs(i, 1) for i in range(n))

• func oddEvenJumps(arr []int) (ans int) {
n := len(arr)
rbt := redblacktree.NewWithIntComparator()
f := make([][2]int, n)
g := make([][2]int, n)
for i := n - 1; i >= 0; i-- {
if v, ok := rbt.Ceiling(arr[i]); ok {
g[i][1] = v.Value.(int)
} else {
g[i][1] = -1
}
if v, ok := rbt.Floor(arr[i]); ok {
g[i][0] = v.Value.(int)
} else {
g[i][0] = -1
}
rbt.Put(arr[i], i)
}
var dfs func(int, int) int
dfs = func(i, k int) int {
if i == n-1 {
return 1
}
if g[i][k] == -1 {
return 0
}
if f[i][k] != 0 {
return f[i][k]
}
f[i][k] = dfs(g[i][k], k^1)
return f[i][k]
}
for i := 0; i < n; i++ {
if dfs(i, 1) == 1 {
ans++
}
}
return
}