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Formatted question description: https://leetcode.ca/all/2022.html

2022. Convert 1D Array Into 2D Array (Easy)

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.

Example 4:

Input: original = [3], m = 1, n = 2
Output: []
Explanation:
There is 1 element in original.
It is impossible to make 1 element fill all the spots in a 1x2 2D array, so return an empty 2D array.

 

Constraints:

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

Similar Questions:

Solution 1. Brute Force

  • // OJ: https://leetcode.com/problems/convert-1d-array-into-2d-array/
// Time: O(K) where `K` is the length of the original array
// Space: O(1)
class Solution {
public:
    vector<vector<int>> construct2DArray(vector<int>& A, int m, int n) {
        if (A.size() != m * n) return {};
        vector<vector<int>> ans(m, vector<int>(n));
        int k = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                ans[i][j] = A[k++];
            }
        }
        return ans;
    }
};

  • 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)]

############

# 2022. Convert 1D Array Into 2D Array
# https://leetcode.com/problems/convert-1d-array-into-2d-array/

class Solution:
    def construct2DArray(self, original: List[int], m: int, n: int) -> List[List[int]]:
        if len(original) != m * n: return []
        
        index = 0
        res = [[0] * n for _ in range(m)]
        
        for i in range(m):
            for j in range(n):
                res[i][j] = original[index]
                index += 1
                
        return res


  • 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;
    }
}

  • 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
}

  • 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;
}


  • /**
 * @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;
};

Discuss

https://leetcode.com/problems/convert-1d-array-into-2d-array/discuss/1499226/C%2B%2B-Brute-Force

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