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Java-based LeetCode algorithm problem solutions, regularly updated
package g0901_1000.s0975_odd_even_jump;
// #Hard #Array #Dynamic_Programming #Stack #Ordered_Set #Monotonic_Stack
// #2022_03_31_Time_49_ms_(98.38%)_Space_69.1_MB_(5.18%)
import java.util.Arrays;
/**
* 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 `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
**/
public class Solution {
private int[] valToPos;
public int oddEvenJumps(int[] arr) {
int size = arr.length;
boolean[] odd = new boolean[size];
boolean[] even = new boolean[size];
valToPos = new int[100001];
Arrays.fill(valToPos, -1);
valToPos[arr[size - 1]] = size - 1;
odd[size - 1] = even[size - 1] = true;
int count = 1;
for (int i = size - 2; i >= 0; i--) {
int curVal = arr[i];
int maxS = findMaxS(curVal);
int 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 int findMaxS(int val) {
for (int i = val; i >= 0; i--) {
if (valToPos[i] != -1) {
return valToPos[i];
}
}
return -1;
}
private int findMinL(int val) {
for (int i = val; i < 100001; i++) {
if (valToPos[i] != -1) {
return valToPos[i];
}
}
return -1;
}
}