# 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. 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 `A[i] <= A[j]` and `A[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 `A[i] >= A[j]` and `A[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 `A.length - 1`) by jumping some number of times (possibly 0 or more than once).

Return  the number of good starting indices.

Example 1:

``````Input: A = [10,13,12,14,15]
Output: 2
Explanation:
From starting index i = 0, we can make our 1st jump to i = 2 (since A is the smallest among A, A, A,
A that is greater or equal to A), 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: A = [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 A is the smallest value in [A, A,
A, A] that is greater than or equal to A.

During our 2nd jump (even-numbered), we jump from i = 1 to i = 2 because A is the largest value in [A, A,
A] that is less than or equal to A. A 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 A is the smallest value in [A, A]
that is greater than or equal to A.

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: A = [5,1,3,4,2]
Output: 3
Explanation:
We can reach the end from starting indices 1, 2, and 4.
``````

Constraints:

• `1 <= A.length <= 2 * 104`
• `0 <= A[i] < 105`

``````class Solution {
public:
int oddEvenJumps(vector<int>& A) {
int res = 1, n = A.size();
vector<bool> higher(n), lower(n);
higher[n - 1] = lower[n - 1] = true;
map<int, int> num2idx;
num2idx[A[n - 1]] = n - 1;
for (int i = n - 2; i >= 0; --i) {
auto hi = num2idx.lower_bound(A[i]), lo = num2idx.upper_bound(A[i]);
if (hi != num2idx.end()) higher[i] = lower[hi->second];
if (lo != num2idx.begin()) lower[i] = higher[(--lo)->second];
if (higher[i]) ++res;
num2idx[A[i]] = i;
}
return res;
}
};
``````

Github 同步地址:

https://github.com/grandyang/leetcode/issues/975

https://leetcode.com/problems/odd-even-jump/

https://leetcode.com/problems/odd-even-jump/discuss/217974/Java-solution-DP-%2B-TreeMap

https://leetcode.com/problems/odd-even-jump/discuss/217981/JavaC%2B%2BPython-DP-using-Map-or-Stack

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