Snail §

Given an n x n array, return the array elements arranged from outermost elements to the middle element, traveling clockwise.

array = [[1,2,3],
         [4,5,6],
         [7,8,9]]
snail(array) #=> [1,2,3,6,9,8,7,4,5]

For better understanding, please follow the numbers of the next array consecutively:

array = [[1,2,3],
         [8,9,4],
         [7,6,5]]
snail(array) #=> [1,2,3,4,5,6,7,8,9]

This image will illustrate things more clearly:

NOTE: The idea is not sort the elements from the lowest value to the highest; the idea is to traverse the 2-d array in a clockwise snailshell pattern.

NOTE 2: The 0x0 (empty matrix) is represented as en empty array inside an array [[]].

My solution §

snail = function(array) {
    // enjoy
    let arr = array
    let snail = []
    while (arr[0] != null && arr.length !== 0){ // Continue till the array is empty
    let cr = []; let cl = []; // Reset the coloumn right and left (cr, cl)
 
      for (i=1; i<arr.length-1; i++){
              cr.push(arr[i].pop())     // Takes the right most value of column
              cl.unshift(arr[i].shift())// Takes the left most value of column
            }
  
    snail.push(arr.shift()) // Inputs Top
 
    snail.push(cr) // Inputs Right
 
    if(arr[0] != null){         //
          let last_r = arr.pop()// Inputs Bottom Value
          last_r.reverse()      //
          snail.push(last_r)}   //
    
      snail.push(cl)// Inputs Left
 
      arr.length == 1 ? snail.push(arr.shift()) : null // Inputs Center Value
      cr = []
      cl= [] 
    }
  return snail.flat()
}
 
console.log(snail([[1, 2, 3, 4, 5], [6, 7, 8, 9, 10], [11, 12, 13, 14, 15], [16, 17, 18, 19, 20], [21, 22, 23, 24, 25]]))

%%🖋 Edit in Excalidraw, and the dark exported image%%

Algorithm §

Snail = Top + Right + Bottom.reversed() + Left - Repeat till Array is empty

In order to do this question, I made the array it self a condition. The algorithm will stop when the array is empty So the while loop:

while (arr[0] != null && arr.length !== 0)

Makes sure that happens After words you We will make a new loop That goes through the array, excluding the first and last subarray (They are the top and bottom pieces) Hence the lines

 let cr = []; let cl = []; // Reset the coloumn right and left (cr, cl)
 
      for (i=1; i<arr.length-1; i++){
              cr.push(arr[i].pop())     // Takes the right most value of column
              cl.unshift(arr[i].shift())// Takes the left most value of column
            }

arr[i] represents a subarray, and so we can use this logic

Lets look at this step by step

• cr (Column Right)
  • the array.pop() method “pops” The last value of array and returns it
  • We take this value and “push” it to cr which will push its last value
• cl(Column Left)
  • The array.shift() method “shifts” the first value and returns it
  • We are taking this value and “unshifting” it the first value of Column left This can be illustrated in a diagram like so %%🖋 Edit in Excalidraw, and the dark exported image%% Now we have left and right columns in an array cl and cr Now it’s easy!

snail.push(arr.shift()) // Inputs Top
 
snail.push(cr) // Inputs Right
 
if(arr[0] != null){         //
    let last_r = arr.pop()// Inputs Bottom Value
	last_r.reverse()      //
	snail.push(last_r)}   //
    
snail.push(cl)// Inputs Left
 
arr.length == 1 ? snail.push(arr.shift()) : null // Inputs Center Value
cr = []
cl= [] 
    }
return snail.flat()
}

1. Inputs the top/first sub array in the array
2. Inputs the Right sub array in the array
3. Inputs the Bottom Value IF array is not yet empty
4. Inputs the Left Value
5. Checks if array only has 1 subarray remaining, This would mean a centter value. So just add that
6. array.flat() Flattens the subarray in to their elements so

let array = [1, [2], [3,4]]
console.log(array.flat())

Just better ! §

snail = function(array) {
  var result;
  while (array.length) {
    // Steal the first row.
    result = (result ? result.concat(array.shift()) : array.shift());
    // Steal the right items.
    for (var i = 0; i < array.length; i++)
      result.push(array[i].pop());
    // Steal the bottom row.
    result = result.concat((array.pop() || []).reverse());
    // Steal the left items.
    for (var i = array.length - 1; i >= 0; i--)
      result.push(array[i].shift());
  }
  return result;
}