LeetCode 2624 - Snail Traversal

The problem asks us to take a 1D array of integers and convert it into a 2D matrix of specified dimensions, rowsCount and colsCount, following a snail traversal order by columns.

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LeetCode Problem 2624

Difficulty: 🟡 Medium
Topics: —

Solution

Problem Understanding

The problem asks us to take a 1D array of integers and convert it into a 2D matrix of specified dimensions, rowsCount and colsCount, following a snail traversal order by columns. Unlike a typical row-wise filling, the snail traversal order here moves column by column: the first column is filled top to bottom, the second column bottom to top, the third column top to bottom again, and so on, alternating direction for each successive column.

The input consists of a 1D array nums and two integers rowsCount and colsCount. The output is a 2D array (list of lists in Python, slice of slices in Go) where the elements of nums are placed according to this alternating column traversal. If the number of elements in nums does not equal rowsCount * colsCount, the input is invalid, and an empty array must be returned.

The constraints indicate the array can be up to length 250, and each number is within [1, 1000]. rowsCount and colsCount can be up to 250. This means the algorithm does not need extreme optimization but must handle empty arrays, single-row, single-column, and uneven shapes carefully.

Important edge cases include empty input, mismatched dimensions, a single row, a single column, and the smallest non-empty arrays.

Approaches

A brute-force approach would attempt to simulate the traversal explicitly: repeatedly slicing the input array for each column and inserting it into the matrix in the correct order. While correct, this approach may involve unnecessary slicing, temporary lists, or reversal operations for each column, leading to slightly inefficient code.

The optimal approach is to pre-allocate the matrix and use simple index arithmetic to fill each cell directly from the 1D array. By keeping a running index through nums and using the parity of the column index to determine the direction (top-down vs bottom-up), we can fill the matrix in a single pass without extra memory for temporary slices.

Approach Time Complexity Space Complexity Notes
Brute Force O(n) O(n) Slice and reverse columns separately; slightly redundant
Optimal O(n) O(n) Single-pass filling of pre-allocated matrix with direction check

Algorithm Walkthrough

  1. Validate input: Check if rowsCount * colsCount equals len(nums). If not, return an empty matrix.
  2. Initialize the matrix: Create a 2D matrix of dimensions rowsCount x colsCount filled with placeholders (e.g., zeros or None).
  3. Initialize index: Start an index i = 0 to track the position in nums.
  4. Iterate over columns: For each column c from 0 to colsCount - 1, determine the traversal direction.
  5. Determine direction: If the column index is even, traverse top-down (row 0 to rowsCount - 1). If odd, traverse bottom-up (row rowsCount - 1 to 0).
  6. Fill the column: For each row in the chosen order, assign matrix[row][c] = nums[i] and increment i.
  7. Return result: After all columns are filled, return the matrix.

Why it works: The algorithm ensures each element of nums is placed exactly once, and the traversal alternates as required by the snail order. The index arithmetic guarantees the sequential reading of the original array while respecting the directional pattern.

Python Solution

from typing import List

class Solution:
    def snail(self, nums: List[int], rowsCount: int, colsCount: int) -> List[List[int]]:
        if len(nums) != rowsCount * colsCount:
            return []
        
        matrix = [[0 for _ in range(colsCount)] for _ in range(rowsCount)]
        idx = 0
        
        for c in range(colsCount):
            if c % 2 == 0:
                for r in range(rowsCount):
                    matrix[r][c] = nums[idx]
                    idx += 1
            else:
                for r in range(rowsCount - 1, -1, -1):
                    matrix[r][c] = nums[idx]
                    idx += 1
        
        return matrix

This implementation first checks for invalid input. It then pre-allocates the matrix to avoid dynamic resizing. The for c in range(colsCount) loop fills each column in either top-down or bottom-up order based on parity. Each element of nums is inserted exactly once in order.

Go Solution

package main

func Snail(nums []int, rowsCount int, colsCount int) [][]int {
	if len(nums) != rowsCount*colsCount {
		return [][]int{}
	}
	
	matrix := make([][]int, rowsCount)
	for i := range matrix {
		matrix[i] = make([]int, colsCount)
	}
	
	idx := 0
	for c := 0; c < colsCount; c++ {
		if c%2 == 0 {
			for r := 0; r < rowsCount; r++ {
				matrix[r][c] = nums[idx]
				idx++
			}
		} else {
			for r := rowsCount - 1; r >= 0; r-- {
				matrix[r][c] = nums[idx]
				idx++
			}
		}
	}
	return matrix
}

In Go, slices are pre-allocated similarly. Unlike Python, Go requires explicit initialization of each row slice. Otherwise, the logic mirrors the Python version with direction determined by the parity of the column index.

Worked Examples

Example 1

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

Iteration table:

Column Direction Filled values
0 top-down 19,10,3,7,9
1 bottom-up 8,5,2,1,17
2 top-down 16,14,12,18,6
3 bottom-up 13,11,20,4,15

Result:

[
 [19,17,16,15],
 [10,1,14,4],
 [3,2,12,20],
 [7,5,18,11],
 [9,8,6,13]
]

Example 2

Single row nums = [1,2,3,4], rowsCount = 1, colsCount = 4

All columns are traversed top-down (only one row), so output is [[1,2,3,4]].

Example 3

Invalid dimensions nums = [1,3], rowsCount = 2, colsCount = 2

Since 2*2 != 2, return [].

Complexity Analysis

Measure Complexity Explanation
Time O(n) Each element of nums is visited exactly once.
Space O(n) The output matrix stores all nums elements in a new 2D array.

The algorithm runs in linear time relative to the number of elements and uses additional space proportional to the size of the output matrix, which is unavoidable.

Test Cases

# Provided examples
assert Solution().snail([19,10,3,7,9,8,5,2,1,17,16,14,12,18,6,13,11,20,4,15], 5, 4) == [
    [19,17,16,15],
    [10,1,14,4],
    [3,2,12,20],
    [7,5,18,11],
    [9,8,6,13]
]  # Complex 5x4 matrix
assert Solution().snail([1,2,3,4], 1, 4) == [[1,2,3,4]]  # Single row
assert Solution().snail([1,3], 2, 2) == []  # Invalid dimensions

# Additional edge cases
assert Solution().snail([], 0, 0) == []  # Empty array
assert Solution().snail([1], 1, 1) == [[1]]  # Single element
assert Solution().snail([1,2,3,4], 2, 2) == [[1,4],[2,3]]  # Small square
assert Solution().snail([1,2,3], 3, 1) == [[1],[2],[3]]  # Single column
Test Why
5x4 complex Validates snail order across multiple columns
Single row Tests minimal row traversal
Invalid dimensions Ensures function returns empty list on mismatch
Empty array Edge case for no elements
Single element Minimal non-empty array
2x2 small square Alternating direction in a small matrix
Single column Verifies top-down only traversal

Edge Cases

  1. Empty input: When nums is empty,