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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
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
Solutions
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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 }