1345. Jump Game IV(BFS) 20.7.8 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 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: O(n) T and S
class Solution(object):
    def minJumps(self, arr):
        """
        :type arr: List[int]
        :rtype: int
        """
        if not arr:
            return -1
        
        number_to_index = collections.defaultdict(list)
        for i, num in enumerate(arr):
            number_to_index[num].append(i)
        visited = set()
        visited.add(0)
        queue = collections.deque([0])
        step = 0
        
        while queue:
            for _ in range(len(queue)):
                curr = queue.popleft()
                if curr == len(arr) - 1:
                    return step
                if curr - 1 >= 0 and not (curr - 1) in visited:
                    visited.add(curr - 1)
                    queue.append(curr - 1)
                if curr + 1 < len(arr) and not (curr + 1) in visited:
                    visited.add(curr + 1)
                    queue.append(curr + 1)
                for other_index in number_to_index[arr[curr]]:
                    if other_index == curr or other_index in visited:
                        continue
                    visited.add(other_index)
                    queue.append(other_index)
                del number_to_index[arr[curr]]                                  # Remove un-needed resourses
            step += 1
        
        return -1