Write code that enhances all arrays such that you can call the snail(rowsCount, colsCount) method that transforms the 1D array into a 2D array organised in the pattern known as snail traversal order. Invalid input values should output an empty array. If rowsCount * colsCount !== nums.length, the input is considered invalid.
Snail traversal order starts at the top left cell with the first value of the current array. It then moves through the entire first column from top to bottom, followed by moving to the next column on the right and traversing it from bottom to top. This pattern continues, alternating the direction of traversal with each column, until the entire current array is covered. For example, when given the input array [19, 10, 3, 7, 9, 8, 5, 2, 1, 17, 16, 14, 12, 18, 6, 13, 11, 20, 4, 15] with rowsCount = 5 and colsCount = 4, the desired output matrix is shown below. Note that iterating the matrix following the arrows corresponds to the order of numbers in the original array.
Example 1:
Input: nums = [19, 10, 3, 7, 9, 8, 5, 2, 1, 17, 16, 14, 12, 18, 6, 13, 11, 20, 4, 15] rowsCount = 5 colsCount = 4 Output: [ [19,17,16,15], [10,1,14,4], [3,2,12,20], [7,5,18,11], [9,8,6,13] ]
Example 2:
Input: nums = [1,2,3,4] rowsCount = 1 colsCount = 4 Output: [[1, 2, 3, 4]]
Example 3:
Input: nums = [1,3] rowsCount = 2 colsCount = 2 Output: [] Explanation: 2 multiplied by 2 is 4, and the original array [1,3] has a length of 2; therefore, the input is invalid.
Constraints:
0 <= nums.length <= 2501 <= nums[i] <= 10001 <= rowsCount <= 2501 <= colsCount <= 250In #2624 Snail Traversal, the goal is to convert a 1D array into a 2D matrix with rowsCount rows and colsCount columns using a vertical zigzag (snail-like) pattern. Before constructing the matrix, verify that rowsCount * colsCount == arr.length. If this condition fails, it is impossible to build the matrix.
The key idea is to fill the matrix column by column. For even-indexed columns, insert elements from top to bottom. For odd-indexed columns, insert elements from bottom to top. This alternating direction creates the characteristic "snail" traversal pattern. Maintain a pointer in the input array and update it as elements are placed into the matrix.
This approach processes each element exactly once, making it efficient. The algorithm runs in O(n) time where n is the length of the array, and uses O(rowsCount × colsCount) space for the resulting matrix.
| Approach | Time Complexity | Space Complexity |
|---|---|---|
| Column-wise Zigzag Matrix Construction | O(n) | O(rowsCount × colsCount) |
Ashish Pratap Singh
Use these hints if you're stuck. Try solving on your own first.
Different ways to approach this problem. Perhaps store a boolean if you are moving up or down and a current column. Reverse the direction and increment the column every time you hits a wall.
Is there a way way to do this without storing state - by just using math?
This approach involves directly filling a 2D array using the input array nums by iterating over the columns in tandem with rows, alternating between top-down and bottom-up directions for each new column.
The key idea is to iterate over the given nums array, and fill the current column based on whether its index is even or odd. For even indices we fill from the top row to the bottom row; for odd indices, from the bottom row to the top row.
The algorithm involves:
Time Complexity: O(ROWS * COLS) as each element must be traversed once.
Space Complexity: O(ROWS * COLS) for storing the result matrix.
1public class SnailTraversal {
2 public int[][] Snail(int[] nums, int rowsCount, int colsCount) {
3 if (rowsCount * colsCount != nums.Length) {
4 return new int[0][];
5 }
6 int[][] result = new int[rowsCount][];
7 for (int r = 0; r < rowsCount; ++r) {
8 result[r] = new int[colsCount];
9 }
10 int index = 0;
11 for (int c = 0; c < colsCount; ++c) {
12 if (c % 2 == 0) {
13 for (int r = 0; r < rowsCount; ++r) {
result[r][c] = nums[index++];
}
} else {
for (int r = rowsCount - 1; r >= 0; --r) {
result[r][c] = nums[index++];
}
}
}
return result;
}
}
This C# implementation follows the same logical structure as the Java solution. It constructs a matrix based on a check of nums length against desired matrix size and fills columns in alternating directions.
This second approach draws on precomputing indices for traversing the nums array according to snail traversal rules using a two-pointer technique. Here, two pointers always mark top and bottom traversals of a given column.
In this strategy:
Time Complexity: O(ROWS * COLS)
Space Complexity: O(ROWS * COLS) due to resulting vector matrix.
1#include <vector>
2using namespace std;
3class Solution {
4public:
vector<vector<int>> snail(vector<int>& nums, int rowsCount, int colsCount) {
if (rowsCount * colsCount != nums.size()) return {};
vector<vector<int>> result(rowsCount, vector<int>(colsCount));
int idx = 0;
for (int c = 0; c < colsCount; ++c) {
if (c % 2 == 0) {
int top = 0;
while (top < rowsCount) {
result[top++][c] = nums[idx++];
}
} else {
int bot = rowsCount - 1;
while (bot >= 0) {
result[bot--][c] = nums[idx++];
}
}
}
return result;
}
};
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Jot down your thoughts, approach, and key learnings
Yes, variations of zigzag or patterned matrix construction appear in coding interviews. They test understanding of matrix traversal, indexing logic, and careful control of iteration direction.
A 2D array (matrix) is the most suitable structure because the problem requires arranging elements into rows and columns. The input array is iterated sequentially while elements are placed into the matrix following the zigzag column pattern.
The optimal approach is to construct the matrix column by column while alternating the direction of filling. Even columns are filled from top to bottom, while odd columns are filled from bottom to top. This ensures the required snail-like traversal pattern in linear time.
The matrix must contain exactly rowsCount × colsCount elements. If the array length does not match this value, it is impossible to fill the matrix completely without missing or extra elements, so the correct response is to return an empty matrix.
The C++ solution uses vectors to dynamically allocate and manage the rows of matrices. With two pointers, iterating up or down depending on even or odd column indices results in efficient walking through each column without extra list operations.