2624. Snail Traversal LeetCode Solution

In this guide, you will get 2624. Snail Traversal LeetCode Solution with the best time and space complexity. The solution to Snail Traversal problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Snail Traversal solution in C++
Snail Traversal solution in Java
Snail Traversal solution in Python
Additional Resources

2624. Snail Traversal LeetCode Solution image

Problem Statement of Snail Traversal

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.

Complexity Analysis

Time Complexity: Google AdSense
Space Complexity: Google Analytics

2624. Snail Traversal LeetCode Solution in C++

declare global {
  interface Array<T> {
    snail(rowsCount: number, colsCount: number): number[][];
  }
}

Array.prototype.snail = function (
  rowsCount: number,
  colsCount: number
): number[][] {
  if (rowsCount * colsCount != this.length) {
    return [];
  }
  const ans: number[][] = [];
  for (let i = 0; i < rowsCount; ++i) {
    ans.push([]);
  }
  for (let i = 0; i < rowsCount; ++i)
    for (let j = 0; j < colsCount; ++j) {
      const k = j * rowsCount + (j % 2 == 0 ? i : rowsCount - 1 - i);
      ans[i][j] = this[k];
    }
  return ans;
};
/* code provided by PROGIEZ */