I would like to know how else I can optimize bubble sort so that it overlooks elements that have already been sorted, even after the first pass.

Eg. [4, 2, 3, 1, 5, 6] --> [2, 3, 1, **4, 5, 6**]

We observe that [4,5,6] are already in sorted order, how can modify my code so that it overlooks this 3 elements in the next pass? (which means the sort would be more efficient?) Do you suggest a recursive method?

public static void bubblesort(int[] a) { for(int i=1; i<a.length; i++) { boolean is_sorted = true; for(int j=0; j<a.length; j++) { if(a[j] > a[j+1]) { int temp = a[j]; a[j] = a[j+1]; a[j+1] = temp; is_sorted = false; } } if(is_sorted) return; } }

Thanks for your time!

## Answer

First of all, you have an out-of-bounds access:

for(int j=0; j<a.length; j++) { if(a[j] > a[j+1]) {

for `j == a.length-1`

, so the loop condition should rather be `j < a.length-1`

.

But, in Bubble sort, you know that after `k`

passes, the largest `k`

elements are sorted at the `k`

last entries of the array, so the conventional Bubble sort uses

public static void bubblesort(int[] a) { for(int i=1; i<a.length; i++) { boolean is_sorted = true; for(int j=0; j < a.length - i; j++) { // skip the already sorted largest elements if(a[j] > a[j+1]) { int temp = a[j]; a[j] = a[j+1]; a[j+1] = temp; is_sorted = false; } } if(is_sorted) return; } }

Now, that would still do a lot of unnecessary iterations when the array has a long sorted tail of largest elements, say you have `k,k-1,...,1`

as the first `k`

elements and `k+1`

to `100000000`

in order after that. The standard Bubble sort will pass `k`

times through (almost) the entire array.

But if you remember where you made your last swap, you know that after that index, there are the largest elements in order, so

public static void bubblesort(int[] a) { int lastSwap = a.length-1; for(int i=1; i<a.length; i++) { boolean is_sorted = true; int currentSwap = -1; for(int j=0; j < lastSwap; j++) { if(a[j] > a[j+1]) { int temp = a[j]; a[j] = a[j+1]; a[j+1] = temp; is_sorted = false; currentSwap = j; } } if(is_sorted) return; lastSwap = currentSwap; } }

would sort the above example with only one pass through the entire array, and the remaining passes only through a (short) prefix.

Of course, in general, that won’t buy you much, but then optimising a Bubble sort is a rather futile exercise anyway.