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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;
    }
}




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