1345. 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 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

jump-game-iv.py

class Solution:
    def minJumps(self, arr: List[int]) -> int:
        n = len(arr)
        graph = collections.defaultdict(list)

        for i, x in enumerate(arr):
            graph[x].append(i)

        queue = collections.deque([(0, 0)])
        num_seen, pos_seen = set([0]), set()

        while queue:
            index, steps = queue.popleft()
            num = arr[index]

            if index == n - 1: 
                return steps

            for p in [index - 1, index + 1] + (graph[num] * (num not in num_seen)):
                if 0 <= p < n and p not in pos_seen:
                    queue.append((p, steps + 1))
                    pos_seen.add(p)

            num_seen.add(num)