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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


  • 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 {
    public boolean canFormArray(int[] arr, int[][] pieces) {
        HashMap<Integer,ArrayList<Integer>> hm = new HashMap();
        for(int[] list:pieces){
            hm.put(list[0],new ArrayList());
            for(int i=1;i<list.length;i++)
        int index = 0;
                return false;
            ArrayList<Integer> list = hm.get(arr[index++]);
            for(int val:list){
                if(index>=arr.length || val!=arr[index])
                    return false;
        return true;


Store the pieces in such a way that each array piece is linked and first element of each piece should be accessible in O(1) time. Hence using a HashMap. Then, traverse the arr and check if each value exits as a key (rest of the values in the arraylist of each key will be indexed over & checked in for loop)

arr = [91,4,64,8,78,2], pieces = [[78,2],[4,64,8],[91]]
78-> [2]
4-> [64,8]
91-> new ArrayList()

[91,4,64,8,78,2] (91 exists as a key)
[91,4,64,8,78,2] (4 exists as a key and [64,8] an array of values)
    ^  * *
[91,4,64,8,78,2] (78 exists as a key and [2] an array of values )
		    ^ * 
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