975. Odd Even Jump

Problem

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 * 10^4

• 0 <= arr[i] < 10^5

Solution

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;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).