Jump game IV

Time: O(N); Space: O(N); 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

Notes:

  • 1 <= len(arr) <= 5 * 10^4

  • -10^8 <= arr[i] <= 10^8

Hints:

  1. Build a graph of n nodes where nodes are the indices of the array and edges for node i are nodes i+1, i-1, j where arr[i] == arr[j].

  2. Start bfs from node 0 and keep distance, answer is the distance when you reach onode n-1.

[1]:

import collections

class Solution1(object):
    """
    Time:  O(n)
    Space: O(n)
    """
    def minJumps(self, arr):
        """
        :type arr: List[int]
        :rtype: int
        """
        groups = collections.defaultdict(list)

        for i, x in enumerate(arr):
            groups[x].append(i)
        q = collections.deque([(0, 0)])
        lookup = set([0])

        while q:
            pos, step = q.popleft()
            if pos == len(arr)-1:
                break
            neighbors = set(groups[arr[pos]] + [pos-1, pos+1])
            groups[arr[pos]] = []

            for p in neighbors:
                if p in lookup or not 0 <= p < len(arr):
                    continue
                lookup.add(p)
                q.append((p, step+1))
        return step

[2]:

s = Solution1()

arr = [100,-23,-23,404,100,23,23,23,3,404]
assert s.minJumps(arr) == 3

arr = [7]
assert s.minJumps(arr) == 0

arr = [7,6,9,6,9,6,9,7]
assert s.minJumps(arr) == 1

arr = [6,1,9]
assert s.minJumps(arr) == 2

arr = [11,22,7,7,7,7,7,7,7,22,13]
assert s.minJumps(arr) == 3