Loop through looped sequence of numbers

Question

There is an array of numbers [1,2,3,4,5,6,7,8,9,10]

I need to get all numbers from this sequence that are different from current for more than 2 items, but looped.

For example if current number is one, so new list should have everything except 9,10,1,2,3, or if current number is four so new list should be everything except 2,3,4,5,6.

Is there any technique how to make this, without creating multiple loops for items at start and at the end?

Thank you.

Answer 1

var a = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];          

var exclude = function (start, distance, array) { 
    var result = [];                              

    for (var i = 0; i < array.length; i++) {      

        var d = Math.min(                         
            Math.abs(start - i - 1),              
            Math.abs(array.length + start - i - 1)
        )                                         

        if (d > distance) {                      
            result.push(array[i]);                
        }                                         
    }                                             

    return result;                                
}                                                 

Answer 2

I think this performs what you asked:

// Sorry about the name
function strangePick(value, array) {
  var n = array.length
    , i = array.indexOf(value);

  if (i >= 0) {
    // Picked number
    var result = [value];

    // Previous 2 numbers
    result.unshift(array[(i + n - 1) % n]);
    result.unshift(array[(i + n - 2) % n]);

    // Next 2 numbers
    result.push(array[(i + 1) % n]);
    result.push(array[(i + 2) % n]);

    return result;
  } else {
    return [];
  }
}

Some tests:

var array = [1,2,3,4,5,6,7,8,9,10];

console.log(strangePick(1, array)); // [9,10,1,2,3]
console.log(strangePick(4, array)); // [2,3,4,5,6]