std way to remove multiple indices from an array at once

[Nicholas Wilson]

On Thursday, 21 December 2017 at 00:23:08 UTC, Steven 
Schveighoffer wrote:
> On 12/20/17 6:01 PM, aliak wrote:
>> Hi, is there a way to remove a number of elements from an 
>> array by a range of indices in the standard library somewhere?
>> 
>> I wrote one (code below), but I'm wondering if there's a 
>> better way?
>> 
>> Also, can the below be made more efficient?
>> 
>> auto without(T, R)(T[] array, R indices) if (isForwardRange!R 
>> && isIntegral!(ElementType!R) && !isInfinite!R) {
>>      T[] newArray;
>>      ElementType!R start = 0;
>>      foreach (i; indices) {
>>          newArray ~= array[start .. i];
>>          start = i + 1;
>>      }
>>      newArray ~= array[start .. $];
>>      return newArray;
>> }
>> 
>> // Usage
>> long[] aa = [1, 2, 3, 4]
>> aa = aa.without([1, 3])
>> 
>> Thanks!
>
> I'm assuming here indices is sorted? Because it appears you 
> expect that in your code. However, I'm going to assume it isn't 
> sorted at first.
>
> Now:
>
> import std.range;
> import std.algorithm;
>
> auto indices = [1,2,3,4];
> aa = aa.enumerate.filter!(a => !indicies.canFind(a[0])).map(a 
> => a[1]).array;
>
> Now, this is going to be O(n^2), but if indices is sorted, you 
> could use binary search:
>
> auto sortedIdxs = indices.assumeSorted; // also could be = 
> indices.sort()
>
> arr = arr.enumerate.filter!(a => 
> !sortedIdxs.contains(a[0])).map(a => a[1]).array;
>
> Complexity would be O(n lg(n))
>
> It's not going to be as good as hand-written code, complexity 
> wise, but it's definitely shorter to write :)
>
> -Steve

isn't that n log(m), where n is length of array and m is length 
of indices?

If indices is sorted with no duplicates and random access then 
you can do it in linear time.

int i;
int ii;
int[] indicies = ...;
arr = arr.filter!((T t)
{
     scope(exit) i++;
     if (i == indicies[ii])
     {
         ii++;
         return false;
     }
     return true;
}).array;