1640. Check Array Formation Through Concatenation

Difficulty: Easy
82 / 82 test cases passed.
Runtime: 40 ms
Memory Usage: 14 MB

Author: YuriSpiridonov

Easy/1640.CheckArraFormationThroughConcatenation.py

@@ -0,0 +1,61 @@
+"""
+    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:
+    Input: arr = [85], pieces = [[85]]
+    Output: true
+
+    Example:
+    Input: arr = [15,88], pieces = [[88],[15]]
+    Output: true
+    Explanation: Concatenate [15] then [88]
+
+    Example:
+    Input: arr = [49,18,16], pieces = [[16,18,49]]
+    Output: false
+    Explanation: Even though the numbers match, we cannot reorder pieces[0].
+
+    Example:
+    Input: arr = [91,4,64,78], pieces = [[78],[4,64],[91]]
+    Output: true
+    Explanation: Concatenate [91] then [4,64] then [78]
+
+    Example:
+    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).
+"""
+#Difficulty: Easy
+#82 / 82 test cases passed.
+#Runtime: 40 ms
+#Memory Usage: 14 MB
+
+#Runtime: 40 ms, faster than 86.23% of Python3 online submissions for Check Array Formation Through Concatenation.
+#Memory Usage: 14 MB, less than 100.00% of Python3 online submissions for Check Array Formation Through Concatenation.
+
+class Solution:
+    def canFormArray(self, arr: List[int], pieces: List[List[int]]) -> bool:
+        array = [None] * len(arr)
+        for piece in pieces:
+            lenght = len(piece)
+            if piece[0] in arr:
+                i = arr.index(piece[0])
+                if arr[i:i+lenght] == piece:
+                    for p in piece:
+                        array[i] = p
+                        i += 1
+        return None not in array