Description
https://leetcode.com/problems/check-if-array-is-sorted-and-rotated/
You are given an array of distinct integers arr
and an array of integer arrays pieces
, where the integers in pieces
are distinct. Your goal is to form arr
by concatenating the arrays in pieces
in any order. However, you are not allowed to reorder the integers in each array pieces[i]
.
Return true
if it is possible to form the array arr
from pieces
. Otherwise, return false
.
Example 1:
Input: arr = [85], pieces = [[85]] Output: true
Example 2:
Input: arr = [15,88], pieces = [[88],[15]] Output: true Explanation: Concatenate[15]
then[88]
Example 3:
Input: arr = [49,18,16], pieces = [[16,18,49]] Output: false Explanation: Even though the numbers match, we cannot reorder pieces[0].
Example 4:
Input: arr = [91,4,64,78], pieces = [[78],[4,64],[91]] Output: true Explanation: Concatenate[91]
then[4,64]
then[78]
Example 5:
Input: arr = [1,3,5,7], pieces = [[2,4,6,8]] Output: false
Constraints:
1 <= pieces.length <= arr.length <= 100
sum(pieces[i].length) == arr.length
1 <= pieces[i].length <= arr.length
1 <= arr[i], pieces[i][j] <= 100
- The integers in
arr
are distinct. - The integers in
pieces
are distinct (i.e., If we flatten pieces in a 1D array, all the integers in this array are distinct).
Explanation
Check if piece character is in the original arr set. If piece length greater than 1, additionally check if the order in original arr matches.
Python Solution
class Solution:
def canFormArray(self, arr: List[int], pieces: List[List[int]]) -> bool:
arr_set = set(arr)
for piece in pieces:
for p in piece:
if p not in arr_set:
return False
if len(piece) > 1:
arr_index = arr.index(piece[0])
for i in range(len(piece)):
if arr_index + i > len(arr) - 1 or arr[arr_index + i] != piece[i]:
return False
return True
- Time Complexity: O(N).
- Space Complexity: O(N).