975. Odd Even Jump

Description

You are given an integer array A. 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 A[i] <= A[j] and A[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 A[i] >= A[j] and A[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 A.length - 1) by jumping some number of times (possibly 0 or more than once).

Return the number of good starting indices.

Constraints

Approach

Links

• GeeksforGeeks

• Leetcode

• ProgramCreek

• YouTube

Examples

Example 1

Input: A = [10, 13, 12, 14, 15]

Output: 2

Explanation:

From starting index i = 0, we can make our 1st jump to i = 2 (since A[2] is the smallest among A[1], A[2], A[3], A[4] that is greater or equal to A[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

Example 3

Solutions

Solution 1

/**
 * Time complexity : O(NlogN), where NN is the length of A.
 * Space complexity : O(N)
 */

class Solution {
    public int oddEvenJumps(int[] A) {
        if(A == null || A.length == 0) return 0;
        
        int n = A.length;
        boolean[] lower = new boolean[n];
        boolean[] higher = new boolean[n];
        
        higher[n-1] = true;
        lower[n-1] = true;
        
        int goodStartingIndices = 1;
        TreeMap<Integer, Integer> map = new TreeMap();
        map.put(A[n-1], n-1);
        
        for(int i = n-2; i >= 0; i--) {
            Map.Entry<Integer, Integer> higherPair = map.ceilingEntry(A[i]);
            if(higherPair != null) {
                higher[i] = lower[(int) higherPair.getValue()];
            }
            
            Map.Entry<Integer, Integer> lowerPair = map.floorEntry(A[i]);
            if(lowerPair != null) {
                lower[i] = higher[(int) lowerPair.getValue()];
            }
            
            if(higher[i]) goodStartingIndices++;
            map.put(A[i], i);
        }
        return goodStartingIndices;
    }
}

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