Question

975. Odd Even Jump

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Description

You are given an integer array arr. From some starting index, you can make a series of jumps. The (1^st, 3^rd, 5^th, ...) jumps in the series are called odd-numbered jumps, and the (2^nd, 4^th, 6^th, ...) 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

Solution 1

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))

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;
  }
};

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
}