LeetCode #1640 — EASY

Check Array Formation Through Concatenation

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

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).

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map

Example 1

[15,88]
[[88],[15]]

Example 2

[49,18,16]
[[16,18,49]]

Example 3

[91,4,64,78]
[[78],[4,64],[91]]

Step 02

Core Insight

What unlocks the optimal approach

Note that the distinct part means that every position in the array belongs to only one piece
Note that you can get the piece every position belongs to naively

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1640: Check Array Formation Through Concatenation
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;
    }
}

// Accepted solution for LeetCode #1640: Check Array Formation Through Concatenation
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
}

# Accepted solution for LeetCode #1640: Check Array Formation Through Concatenation
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

// Accepted solution for LeetCode #1640: Check Array Formation Through Concatenation
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
    }
}

// Accepted solution for LeetCode #1640: Check Array Formation Through Concatenation
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;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.