1640-check-array-formation-through-concatenation.py

"""
Problem Link: https://leetcode.com/problems/check-array-formation-through-concatenation/

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).
"""
class Solution:
    def canFormArray(self, arr: List[int], pieces: List[List[int]]) -> bool:
        indexes = {arr[i]: i for i in range(len(arr))}
        
        for piece in pieces:
            prev_index = None
            
            for element in piece:
                cur_index = indexes.get(element)
                if (prev_index is not None and cur_index != prev_index + 1) or cur_index is None:
                    return False
                
                prev_index = cur_index
                indexes[element] = None
        
        return True