g0901_1000.s0975_odd_even_jump.Solution

public class Solution extends Object

975 - Odd Even Jump.Hard You are given an integer array <code>arr</code>. 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 <code>i</code> to index <code>j</code> (with <code>i < j</code>) in the following way: <ul> <li>During odd-numbered jumps (i.e., jumps 1, 3, 5, …), you jump to the index <code>j</code> such that <code>arr[i] <= arr[j]</code> and <code>arr[j]</code> is the smallest possible value. If there are multiple such indices <code>j</code>, you can only jump to the smallest such index <code>j</code>.</li> <li>During even-numbered jumps (i.e., jumps 2, 4, 6, …), you jump to the index <code>j</code> such that <code>arr[i] >= arr[j]</code> and <code>arr[j]</code> is the largest possible value. If there are multiple such indices <code>j</code>, you can only jump to the smallest such index <code>j</code>.</li> <li>It may be the case that for some index <code>i</code>, there are no legal jumps.</li> </ul> A starting index is good if, starting from that index, you can reach the end of the array (index <code>arr.length - 1</code>) 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: <ul> <li><code>1 <= arr.length <= 2 * 104</code></li> <li><code>0 <= arr[i] < 105</code></li> </ul>

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Solution()
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int

oddEvenJumps(int[] arr)

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clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait

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- Solution
  
  public Solution()
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- oddEvenJumps
  
  public int oddEvenJumps(int[] arr)

Class Solution

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Solution

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oddEvenJumps