LeetCode #2022 — EASY

Convert 1D Array Into 2D Array

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

The Problem

Problem Statement

You are given a 0-indexed 1-dimensional (1D) integer array original, and two integers, m and n. You are tasked with creating a 2-dimensional (2D) array with m rows and n columns using all the elements from original.

The elements from indices 0 to n - 1 (inclusive) of original should form the first row of the constructed 2D array, the elements from indices n to 2 * n - 1 (inclusive) should form the second row of the constructed 2D array, and so on.

Return an m x n 2D array constructed according to the above procedure, or an empty 2D array if it is impossible.

Example 1:

Input: original = [1,2,3,4], m = 2, n = 2
Output: [[1,2],[3,4]]
Explanation: The constructed 2D array should contain 2 rows and 2 columns.
The first group of n=2 elements in original, [1,2], becomes the first row in the constructed 2D array.
The second group of n=2 elements in original, [3,4], becomes the second row in the constructed 2D array.

Example 2:

Input: original = [1,2,3], m = 1, n = 3
Output: [[1,2,3]]
Explanation: The constructed 2D array should contain 1 row and 3 columns.
Put all three elements in original into the first row of the constructed 2D array.

Example 3:

Input: original = [1,2], m = 1, n = 1
Output: []
Explanation: There are 2 elements in original.
It is impossible to fit 2 elements in a 1x1 2D array, so return an empty 2D array.

Constraints:

1 <= original.length <= 5 * 10⁴
1 <= original[i] <= 10⁵
1 <= m, n <= 4 * 10⁴

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed 1-dimensional (1D) integer array original, and two integers, m and n. You are tasked with creating a 2-dimensional (2D) array with m rows and n columns using all the elements from original. The elements from indices 0 to n - 1 (inclusive) of original should form the first row of the constructed 2D array, the elements from indices n to 2 * n - 1 (inclusive) should form the second row of the constructed 2D array, and so on. Return an m x n 2D array constructed according to the above procedure, or an empty 2D array if it is impossible.

Baseline thinking

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

Pattern signal: Array

Example 1

[1,2,3,4]
2
2

Example 2

[1,2,3]
1
3

Example 3

[1,2]
1
1

Core Insight

What unlocks the optimal approach

When is it possible to convert original into a 2D array and when is it impossible?
It is possible if and only if m * n == original.length
If it is possible to convert original to a 2D array, keep an index i such that original[i] is the next element to add to the 2D array.

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 #2022: Convert 1D Array Into 2D Array
class Solution {
    public int[][] construct2DArray(int[] original, int m, int n) {
        if (m * n != original.length) {
            return new int[0][0];
        }
        int[][] ans = new int[m][n];
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                ans[i][j] = original[i * n + j];
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2022: Convert 1D Array Into 2D Array
func construct2DArray(original []int, m int, n int) (ans [][]int) {
	if m*n != len(original) {
		return [][]int{}
	}
	for i := 0; i < m*n; i += n {
		ans = append(ans, original[i:i+n])
	}
	return
}

# Accepted solution for LeetCode #2022: Convert 1D Array Into 2D Array
class Solution:
    def construct2DArray(self, original: List[int], m: int, n: int) -> List[List[int]]:
        if m * n != len(original):
            return []
        return [original[i : i + n] for i in range(0, m * n, n)]

// Accepted solution for LeetCode #2022: Convert 1D Array Into 2D Array
struct Solution;

impl Solution {
    fn construct2_d_array(original: Vec<i32>, m: i32, n: i32) -> Vec<Vec<i32>> {
        let k = original.len();
        let m = m as usize;
        let n = n as usize;
        if m * n == k {
            let mut res = vec![];
            let mut index = 0;
            for _ in 0..m {
                let mut row = vec![];
                for _ in 0..n {
                    row.push(original[index]);
                    index += 1;
                }
                res.push(row);
            }
            res
        } else {
            vec![]
        }
    }
}

#[test]
fn test() {
    let original = vec![1, 2, 3, 4];
    let m = 2;
    let n = 2;
    let res = vec_vec_i32![[1, 2], [3, 4]];
    assert_eq!(Solution::construct2_d_array(original, m, n), res);
    let original = vec![1, 2, 3];
    let m = 1;
    let n = 3;
    let res = vec_vec_i32![[1, 2, 3]];
    assert_eq!(Solution::construct2_d_array(original, m, n), res);
    let original = vec![1, 2];
    let m = 1;
    let n = 1;
    let res = vec_vec_i32![];
    assert_eq!(Solution::construct2_d_array(original, m, n), res);
    let original = vec![3];
    let m = 1;
    let n = 2;
    let res = vec_vec_i32![];
    assert_eq!(Solution::construct2_d_array(original, m, n), res);
}

// Accepted solution for LeetCode #2022: Convert 1D Array Into 2D Array
function construct2DArray(original: number[], m: number, n: number): number[][] {
    if (m * n != original.length) {
        return [];
    }
    const ans: number[][] = [];
    for (let i = 0; i < m * n; i += n) {
        ans.push(original.slice(i, i + n));
    }
    return ans;
}

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.