Sort an integer array, keeping first in place

Question

How would I sort arrays as follows:

[10, 7, 12, 3, 5, 6] --> [10, 12, 3, 5, 6, 7]

[12, 8, 5, 9, 6, 10] --> [12, 5, 6, 8, 9, 10] 

• keeping array[0] in place
• with the next highest integer(s) following (if there are any)
• then ascending from the lowest integer

Answer 1

You could save the value of the first element and use it in a condition for the first sorting delta. Then sort by the standard delta.

How it works (the sort order is from Edge)

              condition  numerical     sortFn
   a      b       delta      delta     result  comment
-----  -----  ---------  ---------  ---------  -----------------
   7     10*          1                     1  different section
  12*     7          -1                    -1  different section
  12*    10*          0          2          2  same section
  12*     7          -1                    -1  same section
   3      7           0         -4         -4  same section
   3     12*          1                     1  different section
   3      7           0         -4         -4  same section
   5      7           0         -2         -2  same section
   5     12*          1                     1  different section
   5      3           0          2          2  same section
   5      7           0         -2         -2  same section
   6      7           0         -1         -1  same section
   6      3           0          3          3  same section
   6      5           0          1          1  same section
   6      7           0         -1         -1  same section

* denotes elements who should be in the first section

Elements of different section means one of the elements goes into the first and the other into the second section, the value is taken by the delta of the condition.

Elements of the same section means, both elements belongs to the same section. For sorting the delta of the values is returned.

function sort(array) {
    var first = array[0];
    array.sort(function (a, b) {
       return (a < first) - (b < first) || a - b;
    });
    return array;
}

console.log(sort([10, 7, 12, 3, 5, 6]));
console.log(sort([12, 8, 5, 9, 6, 10]));

.as-console-wrapper { max-height: 100% !important; top: 0; }

Answer 2

An easy solution could be to sort the entire array and then partition the resulting array on basis of your initial first element.

I.e.

• first sort [10, 7, 12, 3] to [3, 7, 10, 12]
• then split it into >= 10 and < 10: [10, 12] and [3, 7]
• and finally combine both to [10, 12, 3, 7]

Sample implementation without polish:

function customSort(input) {
  var firstElem = input[0];
  var sortedInput = input.sort(function(a, b) { return a-b; });
  var firstElemPos = sortedInput.indexOf(firstElem);
  var greaterEqualsFirstElem = sortedInput.splice(firstElemPos);
  var lessThanFirstElem = sortedInput.splice(0, firstElemPos);
  return greaterEqualsFirstElem.concat(lessThanFirstElem);
}

console.log(customSort([10, 7, 12, 3, 5, 6]));
console.log(customSort([12, 8, 5, 9, 6, 10]));
console.log(customSort([12, 8, 5, 9, 12, 6, 10]));
console.log(customSort([12]));
console.log(customSort([]));