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

My proposal is based on:

• reduce original array into two containing the lower and upper numbers to the first one
• sort each array and concatenate

In this way the initial problem is divided in two sub problems that are more simple to work on.

function cSort(arr) {
    var retval = arr.reduce(function (acc, val) {
        acc[(val < arr[0]) ? 1 : 0].push(val);
        return acc;
    }, [[], []]);
    return retval[0].sort((a, b) => a-b).concat(retval[1].sort((a, b) => a-b));
}

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