Sort an integer array keeping first
You can store the value of the first item and use it in the state for the first delta sort. Then sort by standard delta.
How it works (sort order 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 sections mean that one of the elements goes to the first and the second to the second section, the value is taken by the delta condition.
Elements of the same section mean that both elements belong to the same section. Delta values are returned for sorting.
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; }
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An easy solution might be to sort the entire array and then split the resulting array by the original first element.
those.
- first sort
[10, 7, 12, 3]
before[3, 7, 10, 12]
- then split it by
>= 10
and< 10
:[10, 12]
and[3, 7]
- and finally combine both with
[10, 12, 3, 7]
An example of implementation without polishing:
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([]));
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Check out below snippet.
It's really simple. Just remove the first item from arrary into a new array.
Detach the rest of the array and check if the max of this array is greater than the first element.
If so, paste it into the result. Otherwise concat rest array with result array
var input1 = [10, 7, 12, 3, 5, 6];
var expected1 = [10, 12, 3, 5, 6, 7];
var input2 = [12, 8, 5, 9, 6, 10];
var expected2 = [12, 5, 6, 8, 9, 10];
function customSort(aInput) {
var aResult = [aInput.shift()];
aInput = aInput.sort(function(a, b) { return a - b;});
if (aInput[aInput.length - 1] > aResult[0]) {
var iMax = aInput.pop();
aResult.push(iMax);
}
aResult = aResult.concat(aInput);
return aResult;
}
console.log("Expected: ", expected1.toString());
console.log("Sorted: ", customSort(input1).toString());
console.log("Expected: ", expected2.toString());
console.log("Sorted: ", customSort(input2).toString());
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Updated **
it doesn't work with array.concat, but with this merge function, no idea why ....
Here's a kind of solution,
Array.prototype.merge = function(/* variable number of arrays */){
for(var i = 0; i < arguments.length; i++){
var array = arguments[i];
for(var j = 0; j < array.length; j++){
if(this.indexOf(array[j]) === -1) {
this.push(array[j]);
}
}
}
return this;
};
var $a = [10, 7, 12, 3, 5, 6];
var $b = [12, 8, 5, 9, 6, 10];
function reorganize($array){
var $reorganize = [];
$reorganize.push($array.shift());
$array.sort(function(a, b) { return a - b; });
$reorganize.merge($array);
return $reorganize;
}
console.log(reorganize($a));
console.log(reorganize($b));
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My suggestion is based on:
- reduce the original array to two, containing the bottom and top numbers, down to the first.
- sort each array and concatenate
Thus, the original task is divided into two sub-processes that are easier to work with.
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([]));
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