Check Array Formation Through Concatenation

Author: anantkaushik

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

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+"""
+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