2624. Snail Traversal

Problem Description

The problem asks for a function that transforms a one-dimensional array into a two-dimensional array that represents a matrix. This matrix should be structured in a "snail traversal order". To achieve this, we are directed to enhance all arrays in a way that they have a snail method. This snail method takes two parameters, rowsCount and colsCount, which stand for the number of rows and number of columns in the intended 2D array, respectively.

The "snail traversal order" is a specific way to fill the 2D array. It starts from the top-left corner of the matrix and fills the first column from the top to the bottom. Then, for the next column, it fills from the bottom to the top, and this pattern alternates for each column until all elements from the 1D array are placed into the 2D matrix. If the number of elements in the 1D array does not match the total number of cells in the 2D array (rowsCount * colsCount), the input is considered invalid, and the method should return an empty array.

Intuition

To solve this problem, we can follow these steps:

1. Validate if the number of elements in the original array matches the rowsCount * colsCount. If not, return an empty array.
2. Initialize a 2D array (ans) of size rowsCount x colsCount to store the snail order traversal.
3. Iterate over each element in the 1D array, placing each element into the appropriate cell of the 2D array.
4. Use a direction variable (k) that will alternate between 1 and -1 to control the traversal direction (top to bottom or bottom to top).
5. Traversal starts at the first cell of ans and moves vertically. When reaching the end or the start of a column, change direction and proceed to the next column.

Implementing this approach in TypeScript involves extending the global Array interface with a snail method and adding the logic described above in this method. The variable h tracks the current index in the 1D array, i and j represent the current row and column indices in the 2D array respectively, and k is the vertical traversal direction (1 for downwards, -1 for upwards). The loop continues until all elements in the input array have been placed in ans.

Solution Approach

To implement this solution, several key components are used in the snail method definition on the Array prototype:

1. Input Validation

The very first step is to check whether the given array can be transformed into a 2D array with the given rowsCount and colsCount. This is done by multiplying these two values and checking if the product equals the length of the original array. If it doesn't, it's an invalid input, and we should return an empty array:

1if (rowsCount * colsCount !== this.length) {
2    return [];
3}

2. 2D Array Initialization

The method starts by initializing a 2D array, ans, with the size of rowsCount x colsCount. This is the structure where the snail ordered elements will be placed:

1const ans: number[][] = Array.from({ length: rowsCount }, () => Array(colsCount));

3. Traversal

As the 1D array is traversed element by element, these elements are placed into ans at the appropriate positions:

1for (let h = 0, i = 0, j = 0, k = 1; h < this.length; ++h) {
2    ans[i][j] = this[h];

h is used as the index for the input array, i and j are the indices for the rows and columns in the 2D array, and k represents the vertical movement direction (1 for moving down, -1 for moving up).

4. Direction Control and Incrementing Column Index

The vertical direction is controlled by incrementing or decrementing i based on k. When i reaches the bounds of the row indices (rowsCount or -1), it's time to switch to the next column. This means reversing the direction by negating k and incrementing j to move to the next column:

1i += k;
2if (i === rowsCount || i === -1) {
3    i -= k;
4    k = -k;
5    ++j;
6}

After filling the last element, the loop ends, and the method returns the filled 2D array ans.

5. Returning the Result

Finally, the completed 2D array is returned:

1return ans;

Algorithm Complexity

The time complexity of the algorithm is O(n), where n is the number of elements in the input array since every element is placed exactly once into the 2D array. The space complexity is also O(n), considering the space taken by the 2D array ans.

With this approach, we are able to enhance all arrays to have a snail method that successfully transforms an array into the snail traversal order in an efficient and precise manner.

Example Walkthrough

Let's take a small example to illustrate the solution approach. Assume we have a one-dimensional array [1, 2, 3, 4, 5, 6] and we want to transform it into a 2D array with rowsCount = 3 and colsCount = 2.

First, let's validate the input:

• rowsCount * colsCount equals 3 * 2, which is 6.
• The original array has 6 elements.
• Since 6 equals 6, the input is valid.

Now let's initialize the 2D array ans of size 3 x 2.

The 2D array before the traversal:

1[ [undefined, undefined],
2  [undefined, undefined],
3  [undefined, undefined] ]

We will fill this array using the snail order described, starting from the top-left corner.

Starting the traversal, the direction k is initially set to 1 (moving down), and we fill the first column:

First iteration (h=0, i=0, j=0, k=1):

1[ [1,          undefined],
2  [undefined, undefined],
3  [undefined, undefined] ]

Second iteration (h=1, i=1, j=0, k=1):

1[ [1,          undefined],
2  [2,          undefined],
3  [undefined, undefined] ]

Third iteration (h=2, i=2, j=0, k=1):

1[ [1,          undefined],
2  [2,          undefined],
3  [3,          undefined] ]

Now we need to change the direction since i would go out of bounds. We switch k to -1 and increment j, moving to the next column.

Fourth iteration (h=3, i=2, j=1, k=-1):

1[ [1,          6],
2  [2, undefined],
3  [3, undefined] ]

Fifth iteration (h=4, i=1, j=1, k=-1):

1[ [1, 6],
2  [2, 5],
3  [3, undefined] ]

Sixth and final iteration (h=5, i=0, j=1, k=-1):

1[ [1, 6],
2  [2, 5],
3  [3, 4] ]

Having placed all elements, the traversal is complete. The final snail ordered 2D array ans is:

1[ [1, 6],
2  [2, 5],
3  [3, 4] ]

We return this array, which reflects the desired snail traversal ordering of the original array.

The time complexity is O(n) as each element of the array is placed once in the 2D array, and the space complexity is also O(n) for storing the 2D array ans. With these steps, any array with a correct number of elements can be transformed into this snail order matrix efficiently.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the snail function is determined by the nested loop structure where the outer loop, guarded by h < this.length, iterates over all elements of the array once. Since the array size is rowsCount * colsCount, which is strictly checked to be equal to this.length, the loop runs rowsCount * colsCount times.

Hence, the time complexity of the snail function is O(n), where n is the number of elements in the input array or equivalently rowsCount * colsCount.

Space Complexity

The space complexity involves the additional space taken up by the ans variable, which is an rowsCount x colsCount matrix. The space used by ans is directly proportional to the number of elements in the matrix, which is rowsCount * colsCount. No additional significant space is used as the indices (i, j, k) and the counter h use a constant amount of space.

Therefore, the space complexity for the snail function is also O(n), which is the space required for the output matrix where n is the same as specified in the time complexity analysis.