README_EN.md

2022. Convert 1D Array Into 2D Array

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Description

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

Solutions

Python3

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)]

Java

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;
    }
}

C++

class Solution {
public:
    vector<vector<int>> construct2DArray(vector<int>& original, int m, int n) {
        if (m * n != original.size()) {
            return {};
        }
        vector<vector<int>> ans(m, vector<int>(n));
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                ans[i][j] = original[i * n + j];
            }
        }
        return ans;
    }
};

Go

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
}

JavaScript

/**
 * @param {number[]} original
 * @param {number} m
 * @param {number} n
 * @return {number[][]}
 */
var construct2DArray = function (original, m, n) {
    if (m * n != original.length) {
        return [];
    }
    const ans = [];
    for (let i = 0; i < m * n; i += n) {
        ans.push(original.slice(i, i + n));
    }
    return ans;
};

TypeScript

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;
}

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