1640. Check Array Formation Through Concatenation

LeetCode 1640. Check Array Formation Through Concatenation

Description

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 arrayarr frompieces. Otherwise, return false.

Example 1:

Input: arr = [15,88], pieces = [[88],[15]]
Output: true
Explanation: Concatenate [15] then [88]

Example 2:

Input: arr = [49,18,16], pieces = [[16,18,49]]
Output: false
Explanation: Even though the numbers match, we cannot reorder pieces[0].

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).

Tags

Array, Hash Table, Sort

Solution

Employ a hash table to index pieces using the first character of each piece as the key to speedup lookup. Then, we use each character in arr to query a piece and join them together to obtain a new array. Return the result that if arr and the new array are the same.

Complexity

• Time complexity: $O(n)$, n for the length of arr;

• Space complexity: $O(m)$, m for the length of pieces.

Code

func canFormArray(arr []int, pieces [][]int) bool {
	connected, dict := make([]int, 0, len(arr)), map[int][]int{} // pieces[i][0]:pieces[i]
	for _, piece := range pieces {
		dict[piece[0]] = piece
	}
	for _, a := range arr {
		connected = append(connected, dict[a]...)
	}
	if len(arr) != len(connected) {
		return false
	}
	for i := 0; i < len(arr); i++ {
		if arr[i] != connected[i] {
			return false
		}
	}
	return true
}

Reference