1345. Jump Game IV

Problem Link

Description

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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

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Python3

class Solution:
    def minJumps(self, arr: List[int]) -> int:
        N = len(arr)
        mp = defaultdict(list)
 
        for i, x in enumerate(arr):
            mp[x].append(i)
        
        num_seen = set()
        pos_seen = set([0])
        queue = deque([0])
        steps = 0
 
        while queue:
            M = len(queue)
 
            for _ in range(M):
                index = queue.popleft()
                num = arr[index]
 
                if index == N - 1: return steps
 
                for nxt in [index + 1, index - 1] + (mp[num] * (num not in num_seen)):
                    if 0 <= nxt < N and nxt not in pos_seen:
                        pos_seen.add(nxt)
                        queue.append(nxt)
                
                num_seen.add(num)
 
            
            steps += 1
 
        return -1