975. Odd Even Jump

Description

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.

 

Constraints:

• 1 <= arr.length <= 2 * 104
• 0 <= arr[i] < 105

 

Solutions

Solution: Monotonic Stack + Dynamic Programming

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

 

JavaScript

/**
 * @param {number[]} arr
 * @return {number}
 */
const oddEvenJumps = function (arr) {
  const n = arr.length;
  const ascIndices = [...arr.keys()].sort((a, b) => arr[a] - arr[b]);
  const descIndices = [...arr.keys()].sort((a, b) => arr[b] - arr[a]);

  const generateNextJumpIndices = indices => {
    const result = new Array(n).fill(-1);
    const stack = [];

    for (const index of indices) {
      while (index > stack.at(-1)) {
        const previousIndex = stack.pop();

        result[previousIndex] = index;
      }
      stack.push(index);
    }
    return result;
  };

  const oddNextJump = generateNextJumpIndices(ascIndices);
  const evenNextJump = generateNextJumpIndices(descIndices);
  const oddJump = new Array(n).fill(false);
  const evenJump = new Array(n).fill(false);

  oddJump[n - 1] = true;
  evenJump[n - 1] = true;

  for (let index = n - 2; index >= 0; index--) {
    const oddNextIndex = oddNextJump[index];
    const evenNextIndex = evenNextJump[index];

    if (oddNextIndex > -1) {
      oddJump[index] = evenJump[oddNextIndex];
    }
    if (evenNextIndex > -1) {
      evenJump[index] = oddJump[evenNextIndex];
    }
  }
  return oddJump.reduce((total, isReach) => total + isReach, 0);
};