Jump Game IV

Description

Given an array of integers arr, you are initially positioned at the first index of the array.

In one step you can jump from index i to index:

• i + 1 where: i + 1 < arr.length.
• i - 1 where: i - 1 >= 0.
• j where: arr[i] == arr[j] and i != j.

Return the minimum number of steps to reach the last index of the array.

Notice that you can not jump outside of the array at any time.

 

Example 1:

Input: arr = [100,-23,-23,404,100,23,23,23,3,404]
Output: 3
Explanation: You need three jumps from index 0 --> 4 --> 3 --> 9. Note that index 9 is the last index of the array.

Example 2:

Input: arr = [7]
Output: 0
Explanation: Start index is the last index. You don't need to jump.

Example 3:

Input: arr = [7,6,9,6,9,6,9,7]
Output: 1
Explanation: You can jump directly from index 0 to index 7 which is last index of the array.

Example 4:

Input: arr = [6,1,9]
Output: 2

Example 5:

Input: arr = [11,22,7,7,7,7,7,7,7,22,13]
Output: 3

 

Constraints:

• 1 <= arr.length <= 5 * 10^4
• -10^8 <= arr[i] <= 10^8

Solution(javascript)

/**
 * @param {number[]} arr
 * @return {number}
 */
const minJumps = function (arr) {
  const map = arr.reduce((acc, num, index) => {
    acc[num] = acc[num] || []
    acc[num].push(index)
    return acc
  }, {})
  let current = [0]
  const visited = {
    0: true,
  }
  let steps = 0
  while (current.length > 0) {
    const next = []
    const push = (index) => {
      if (index > 0 && index < arr.length && !visited[index]) {
        visited[index] = true
        next.push(index)
      }
    }
    for (const index of current) {
      if (index === arr.length - 1) {
        return steps
      }
      push(index + 1)
      push(index - 1)
      const indices = map[arr[index]] || []
      for (const index2 of indices) {
        push(index2)
      }
      delete map[arr[index]] // 注意重复数字的情况
    }
    steps += 1
    current = next
  }
  return steps
}