You are given an integer array arr. From some starting index, you can make a series of jumps. The (1^st, 3^rd, 5^th, ...) jumps in the series are called odd-numbered jumps, and the (2^nd, 4^th, 6^th, ...) 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.

Let’s solve the challenging and elegant "Odd Even Jump" problem using your structured, interview-friendly approach — with clear logic and optimized code 🧠💪

✅ 1. Problem Explanation (In Simple Terms)

You're given an array arr of integers.

You start at any index i and want to reach the end of the array by jumping.

Jump Rules:

You alternate between odd and even jumps.
On the odd jump:
- You must jump to the smallest index j > i such that arr[j] ≥ arr[i]
On the even jump:
- You must jump to the smallest index j > i such that arr[j] ≤ arr[i]

You win if you can reach the last index.

❓ Return the total number of starting indices i where you can reach the end by obeying the odd-even rules.

💡 2. Key Insights

It's a DP problem from the end of the array backwards:
- We track: "can we reach the end from i with an odd jump?" and "with an even jump?"
At each position:
- Precompute the next index for an odd jump
- Precompute the next index for an even jump
Use monotonic stacks and TreeMap-style sorting to efficiently compute jump destinations.

🧠 3. Steps to Solve

Initialize n = arr.length
Create two arrays:
- odd[i] = can we reach end from i starting with odd jump?
- even[i] = can we reach end from i starting with even jump?
For last index n-1 → both odd[n-1] = even[n-1] = true
Precompute:
- nextHigher[i] = index to jump to on an odd jump
- nextLower[i] = index to jump to on an even jump
Process backwards:
- If nextHigher[i] is defined → odd[i] = even[nextHigher[i]]
- If nextLower[i] is defined → even[i] = odd[nextLower[i]]
Count how many odd[i] == true → those are valid starting points

✅ 4. JavaScript Code (Optimized & Clean)

/**
 * @param {number[]} arr
 * @return {number}
 */
function oddEvenJumps(arr) {
  const n = arr.length;

  // Step 1: Helper to build jump maps
  const makeJumpMap = (compareFn) => {
    const result = Array(n).fill(-1);
    const stack = [];
    const indices = Array.from({ length: n }, (_, i) => i)
      .sort((a, b) => arr[a] - arr[b] || a - b); // sort indices by value, then index

    if (compareFn === 'desc') indices.sort((a, b) => arr[b] - arr[a] || a - b);

    for (const i of indices) {
      while (stack.length && i > stack[stack.length - 1]) {
        result[stack.pop()] = i;
      }
      stack.push(i);
    }

    return result;
  };

  // Step 2: Precompute jump destinations
  const nextHigher = makeJumpMap('asc');
  const nextLower = makeJumpMap('desc');

  // Step 3: DP arrays
  const odd = Array(n).fill(false);
  const even = Array(n).fill(false);
  odd[n - 1] = even[n - 1] = true;

  // Step 4: Fill DP bottom-up
  for (let i = n - 2; i >= 0; i--) {
    if (nextHigher[i] !== -1) {
      odd[i] = even[nextHigher[i]];
    }
    if (nextLower[i] !== -1) {
      even[i] = odd[nextLower[i]];
    }
  }

  // Step 5: Count how many odd[i] are true
  return odd.filter(Boolean).length;
}

🔍 5. Dry Run Example

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

→ From 0 (10):
   - Odd jump → 2 (12) → even → 3 (14) → odd → 4 (15) ✅

→ From 1 (13):
   - Odd → 3 (14) → even → 4 ✅

→ From 2 (12):
   - Odd → 3 (14) → even → 4 ✅

→ From 3 (14):
   - Odd → 4 ✅

→ From 4: already at end ✅

✅ Total = 5

⏱️ 6. Time & Space Complexity

Sorting each jump list: O(n log n)
Stack processing: O(n)
DP fill: O(n)

✅ Total Time: O(n log n)
✅ Space: O(n)

✅ Final Verdict

A beautiful mix of:
- Preprocessing with monotonic stacks
- DP to track jump reachability
Commonly asked in top-tier interviews (Google, Amazon, etc.)
Excellent to show off sorting, DP, and stack skills

Odd Even Jump (Hard)