Jump Game IV - LeetCode solutions

Problem

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 * 10^4
-10^8 <= arr[i] <= 10^8

Solution

class Solution {
    public int minJumps(int[] arr) {
        if (arr.length == 1) {
            return 0;
        }

        int len = arr.length;
        HashMap<Integer, List<Integer>> myHash = new HashMap<>();

        int i = 0;
        while (i < arr.length) {
            List<Integer> curList = myHash.getOrDefault(arr[i], new ArrayList<>());
            curList.add(i);
            int tempNum = arr[i];
            int tempIndex = i;
            while (i < arr.length && arr[i] == tempNum) {
                i++;
            }
            if (i != tempIndex + 1) {
                curList.add(i - 1);
            }
            myHash.put(tempNum, curList);
        }

        Deque<Integer> myQueue = new LinkedList<>();
        int step = 0;
        myQueue.offerLast(0);

        boolean[] visited = new boolean[arr.length];
        visited[0] = true;

        while (!myQueue.isEmpty()) {
            int curCount = myQueue.size();
            int j = 0;
            while (j < curCount) {
                int 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;
                }
                List<Integer> tempList = myHash.getOrDefault(arr[curIndex], new ArrayList<>());
                for (Integer integer : tempList) {
                    if (!visited[integer]) {
                        myQueue.offerLast(integer);
                        visited[integer] = true;
                    }
                }
                myHash.remove(arr[curIndex]);
                j++;
            }
            step++;
        }

        return step;
    }
}

Explain:

nope.

Complexity:

Time complexity : O(n).
Space complexity : O(n).