# 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 array arr from pieces. 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].


Example 3:

Input: arr = [91,4,64,78], pieces = [[78],[4,64],[91]]
Output: true
Explanation: Concatenate [91] then [4,64] then [78]


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).

## Solutions

• class Solution {
public boolean canFormArray(int[] arr, int[][] pieces) {
for (int i = 0; i < arr.length;) {
int k = 0;
while (k < pieces.length && pieces[k][0] != arr[i]) {
++k;
}
if (k == pieces.length) {
return false;
}
int j = 0;
while (j < pieces[k].length && arr[i] == pieces[k][j]) {
++i;
++j;
}
}
return true;
}
}

• class Solution {
public:
bool canFormArray(vector<int>& arr, vector<vector<int>>& pieces) {
for (int i = 0; i < arr.size();) {
int k = 0;
while (k < pieces.size() && pieces[k][0] != arr[i]) {
++k;
}
if (k == pieces.size()) {
return false;
}
int j = 0;
while (j < pieces[k].size() && arr[i] == pieces[k][j]) {
++i;
++j;
}
}
return true;
}
};

• class Solution:
def canFormArray(self, arr: List[int], pieces: List[List[int]]) -> bool:
i = 0
while i < len(arr):
k = 0
while k < len(pieces) and pieces[k][0] != arr[i]:
k += 1
if k == len(pieces):
return False
j = 0
while j < len(pieces[k]) and arr[i] == pieces[k][j]:
i, j = i + 1, j + 1
return True


• func canFormArray(arr []int, pieces [][]int) bool {
for i := 0; i < len(arr); {
k := 0
for k < len(pieces) && pieces[k][0] != arr[i] {
k++
}
if k == len(pieces) {
return false
}
j := 0
for j < len(pieces[k]) && arr[i] == pieces[k][j] {
i, j = i+1, j+1
}
}
return true
}

• function canFormArray(arr: number[], pieces: number[][]): boolean {
const n = arr.length;
let i = 0;
while (i < n) {
const target = arr[i];
const items = pieces.find(v => v[0] === target);
if (items == null) {
return false;
}
for (const item of items) {
if (item !== arr[i]) {
return false;
}
i++;
}
}
return true;
}


• /**
* @param {number[]} arr
* @param {number[][]} pieces
* @return {boolean}
*/
var canFormArray = function (arr, pieces) {
const d = new Map();
for (const p of pieces) {
d.set(p[0], p);
}
for (let i = 0; i < arr.length; ) {
if (!d.has(arr[i])) {
return false;
}
const p = d.get(arr[i]);
for (const v of p) {
if (arr[i++] != v) {
return false;
}
}
}
return true;
};


• use std::collections::HashMap;
impl Solution {
pub fn can_form_array(arr: Vec<i32>, pieces: Vec<Vec<i32>>) -> bool {
let n = arr.len();
let mut map = HashMap::new();
for (i, v) in pieces.iter().enumerate() {
map.insert(v[0], i);
}
let mut i = 0;
while i < n {
match map.get(&arr[i]) {
None => {
return false;
}
Some(&j) => {
for &item in pieces[j].iter() {
if item != arr[i] {
return false;
}
i += 1;
}
}
}
}
true
}
}