ash0x0's Obsidian Quartz

Odd Even Jump

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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.

Examples

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:

Solution

using namespace std;
 
class Solution {
private:
	struct Candidate {
		int val;
		int index;
		Candidate(x, y): index(x), val(y) {};
	};
	
	int getJumpCandidates(vector<int>& arr, int starting, int jump) {
		unordered_set<Candidate> candidates;
		for (int j = 0; j < arr.size(); j++) {
			if (jump % 2 == 0 && starting >= arr[j]) {
				candidates.push_back(Candidate(j, arr[j]));
			} else if (jump % 2 == 1 && arr[i] <= arr[j]) {
				candidates.push_back(Candidate(j, arr[j]));
			}
		}
		Candidate bestCandidate = Candidate(-1, -1);
		for (auto candidate : candidates) {
			if (jumps % 2 == 0 && candidate.val < bestCandidate.val) {
				bestCandidate = candidate;
			} else if (jumps % 2 == 0 && candidate.val < bestCandidate.val) {
				
			}
		}
		int minIt;
		if (jump % 2 == 0) {
			minIt = min_element(
		        myMap.begin(), myMap.end(),
		        [](const auto& a, const auto& b) {
		            return a.first < b.first;
		        }
		    );
		} else {
			minIt = max_element(
		        myMap.begin(), myMap.end(),
		        [](const auto& a, const auto& b) {
		            return a.first > b.first;
		        }
		    );
		}
		return minIt;
	}
public:
	int goodJumps(vector<int>& arr) {
		int goodJumps = 0;
		for (int i = 0; i < arr.size(); i++) {
			unordered_set<int> candidates;
			int jump = 0;
			int bestIndex = -1;
			int bestValue = -1;
			for (int j = 1; i < arr.size); j++) {
				
				if (j == arr.size - 1) goodJumps++;
			}
		}
		return goodJumps;
	}
}

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