Teaching Kids Programming: Videos on Data Structures and Algorithms
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 1×1 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 1×2 2D array, so return an empty 2D array.Hints:
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.
Reshape Algorithm to Convert 1D Array to 2D Matrix
Need to check if the sizes match, and then fill/reshape the 2D matrix row by row.
1 2 3 4 5 6 7 8 9 10 11 | class Solution: def construct2DArray(self, original: List[int], m: int, n: int) -> List[List[int]]: if len(original) != m * n: return [] t = 0 ans = [[0 for _ in range(n)] for _ in range(m)] for r in range(m): for c in range(n): ans[r][c] = original[t] t += 1 return ans |
class Solution: def construct2DArray(self, original: List[int], m: int, n: int) -> List[List[int]]: if len(original) != m * n: return [] t = 0 ans = [[0 for _ in range(n)] for _ in range(m)] for r in range(m): for c in range(n): ans[r][c] = original[t] t += 1 return ans
Time/Space complexity is O(RC) or O(N) where N is the length of the original 1D array.
–EOF (The Ultimate Computing & Technology Blog) —
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