1447 Jump Game Iv
1447 Jump Game Iv
Jump Game IV 
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:
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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:
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**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:
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**Input:** arr = [7]
**Output:** 0
**Explanation:** Start index is the last index. You do not need to jump.
Example 3:
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**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:
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1 <= arr.length <= 5 * 104
-108 <= arr[i] <= 108
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class Solution:
def minJumps(self, arr) -> int:
n = len(arr)
if n <= 1:
return 0
graph = {}
for i in range(n):
if arr[i] in graph:
graph[arr[i]].append(i)
else:
graph[arr[i]] = [i]
curs = set([0]) # store layers from start
visited = {0, n-1}
step = 0
other = set([n-1]) # store layers from end
# when current layer exists
while curs:
# search from the side with fewer nodes
if len(curs) > len(other):
curs, other = other, curs
nex = set()
# iterate the layer
for node in curs:
# check same value
for child in graph[arr[node]]:
if child in other:
return step + 1
if child not in visited:
visited.add(child)
nex.add(child)
# clear the list to prevent redundant search
graph[arr[node]].clear()
# check neighbors
for child in [node-1, node+1]:
if child in other:
return step + 1
if 0 <= child < len(arr) and child not in visited:
visited.add(child)
nex.add(child)
curs = nex
step += 1
return -1
This post is licensed under CC BY 4.0 by the author.