LeetCode 1640. Check Array Formation Through Concatenation

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).

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