2022. Convert 1D Array Into 2D Array #

Problem #

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

Problem Summary #

Given a 0-indexed one-dimensional integer array original and two integers m and n. You need to use all elements in original to create a 2D array with m rows and n columns.

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

Return an m x n 2D array according to the above procedure. If such a 2D array cannot be constructed, return an empty 2D array.

Solution Approach #

• Easy problem. Take n elements at a time from the 1D array original and place them in order into m rows. In this problem, if m*n is greater than or less than the length of original, output an empty array.

Code #

package leetcode

func construct2DArray(original []int, m int, n int) [][]int {
	if m*n != len(original) {
		return [][]int{}
	}
	res := make([][]int, m)
	for i := 0; i < m; i++ {
		res[i] = original[n*i : n*(i+1)]
	}
	return res
}