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] 

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Answer 1

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([]));