LeetCode in Kotlin

1345. Jump Game IV

Hard

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 do not 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.

Constraints:

• 1 <= arr.length <= 5 * 104
• -108 <= arr[i] <= 108

Solution

import java.util.Deque
import java.util.LinkedList

class Solution {
    fun minJumps(arr: IntArray): Int {
        if (arr.size == 1) {
            return 0
        }
        val len = arr.size
        val myHash = HashMap<Int, MutableList<Int>>()
        var i = 0
        while (i < arr.size) {
            val curList = myHash.getOrDefault(arr[i], ArrayList())
            curList.add(i)
            val tempNum = arr[i]
            val tempIndex = i
            while (i < arr.size && arr[i] == tempNum) {
                i++
            }
            if (i != tempIndex + 1) {
                curList.add(i - 1)
            }
            myHash[tempNum] = curList
        }
        val myQueue: Deque<Int> = LinkedList()
        var step = 0
        myQueue.offerLast(0)
        val visited = BooleanArray(arr.size)
        visited[0] = true
        while (myQueue.isNotEmpty()) {
            val curCount = myQueue.size
            var j = 0
            while (j < curCount) {
                val curIndex = myQueue.pollFirst()
                if (curIndex == len - 1) {
                    return step
                }
                if (curIndex + 1 < len && !visited[curIndex + 1]) {
                    myQueue.offerLast(curIndex + 1)
                    visited[curIndex + 1] = true
                }
                if (curIndex - 1 >= 0 && !visited[curIndex - 1]) {
                    myQueue.offerLast(curIndex - 1)
                    visited[curIndex - 1] = true
                }
                val tempList: List<Int> = myHash.getOrDefault(arr[curIndex], ArrayList())
                for (integer in tempList) {
                    if (!visited[integer]) {
                        myQueue.offerLast(integer)
                        visited[integer] = true
                    }
                }
                myHash.remove(arr[curIndex])
                j++
            }
            step++
        }
        return step
    }
}

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