D
DevExCode

LeetCode 2624: Snail Traversal Solution

Master LeetCode problem 2624 (Snail Traversal), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2624. Snail Traversal

Medium

Problem Explanation

Explanation

The problem requires us to transform a given 1D array into a 2D array following the snail traversal pattern. We need to iterate through the array in a specific order to form the 2D array. The algorithm involves determining the boundaries of the current traversal and moving in a clockwise direction to populate the 2D matrix in the required order.

  1. Initialize variables to keep track of the boundaries for row and column traversal.
  2. Perform a clockwise traversal of the 2D array, updating the boundaries after each row or column traversal.
  3. Populate the 2D array with values from the input 1D array in the snail traversal pattern.

Time Complexity: O(rowsCount * colsCount) - We need to visit each cell of the 2D array once. Space Complexity: O(rowsCount * colsCount) - Space is required to store the 2D array.

Solution Code

class Solution {
    public int[][] snail(int rowsCount, int colsCount, int[] nums) {
        if (rowsCount * colsCount != nums.length) {
            return new int[0][0];
        }

        int[][] result = new int[rowsCount][colsCount];
        int top = 0, bottom = rowsCount - 1, left = 0, right = colsCount - 1;
        int index = 0;

        while (top <= bottom && left <= right) {
            for (int i = left; i <= right; i++) {
                result[top][i] = nums[index++];
            }
            top++;

            for (int i = top; i <= bottom; i++) {
                result[i][right] = nums[index++];
            }
            right--;

            if (top <= bottom) {
                for (int i = right; i >= left; i--) {
                    result[bottom][i] = nums[index++];
                }
                bottom--;
            }

            if (left <= right) {
                for (int i = bottom; i >= top; i--) {
                    result[i][left] = nums[index++];
                }
                left++;
            }
        }

        return result;
    }
}

Try It Yourself

Loading code editor...

Frequently Asked Questions

How to solve LeetCode 2624 (Snail Traversal)?

This page provides optimized solutions for LeetCode problem 2624 (Snail Traversal) in Java, C++, and Python, along with a detailed explanation and an interactive code editor to test your code.

What is the time complexity of LeetCode 2624 (Snail Traversal)?

The time complexity for LeetCode 2624 (Snail Traversal) varies by solution. Check the detailed explanation section for specific complexities in Java, C++, and Python implementations.

Can I run code for LeetCode 2624 on DevExCode?

Yes, DevExCode provides an interactive code editor where you can write, test, and run your code for LeetCode 2624 in Java, C++, or Python.

Back to LeetCode Solutions