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    <p>First of all, you have an out-of-bounds access:</p> <pre><code> for(int j=0; j&lt;a.length; j++) { if(a[j] &gt; a[j+1]) { </code></pre> <p>for <code>j == a.length-1</code>, so the loop condition should rather be <code>j &lt; a.length-1</code>.</p> <p>But, in Bubble sort, you know that after <code>k</code> passes, the largest <code>k</code> elements are sorted at the <code>k</code> last entries of the array, so the conventional Bubble sort uses</p> <pre><code>public static void bubblesort(int[] a) { for(int i=1; i&lt;a.length; i++) { boolean is_sorted = true; for(int j=0; j &lt; a.length - i; j++) { // skip the already sorted largest elements if(a[j] &gt; a[j+1]) { int temp = a[j]; a[j] = a[j+1]; a[j+1] = temp; is_sorted = false; } } if(is_sorted) return; } } </code></pre> <p>Now, that would still do a lot of unnecessary iterations when the array has a long sorted tail of largest elements, say you have <code>k,k-1,...,1</code> as the first <code>k</code> elements and <code>k+1</code> to <code>100000000</code> in order after that. The standard Bubble sort will pass <code>k</code> times through (almost) the entire array.</p> <p>But if you remember where you made your last swap, you know that after that index, there are the largest elements in order, so</p> <pre><code>public static void bubblesort(int[] a) { int lastSwap = a.length-1; for(int i=1; i&lt;a.length; i++) { boolean is_sorted = true; int currentSwap = -1; for(int j=0; j &lt; lastSwap; j++) { if(a[j] &gt; 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; } } </code></pre> <p>would sort the above example with only one pass through the entire array, and the remaining passes only through a (short) prefix.</p> <p>Of course, in general, that won't buy you much, but then optimising a Bubble sort is a rather futile exercise anyway.</p>
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    1. USDaniel Fischer
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    1. COappreciate your detailed explanation as well as spotting that jarring error of mine!
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      1. USkent
    2. COit is a bit cleaner/more clear to use a while loop for the outer loop and check the `currentSwap` variable.
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      1. USemployee-0
 

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