Optimized Bubble Sort (Java)

Question

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         


        

Answer 1

you should use a variable "size" for the inner loop and change it to the latest swapped element in each cycle.This way your inner loop goes up to the latest "swapped" element and passes the rest that are unswapped (aka in their correctplace). i.e

do {
        int newsize =0;
        for (int i = 1; i < size; i++) {
            if (a[i - 1] > a[i]) {
                int temp;
                temp = a[i - 1];
                a[i - 1] = a[i];
                a[i] = temp;
                newsize =i;
            }
        }
        size = newsize;
   } while (size > 0);

Answer 2

    public static void BubbleSorter(params int[] input){
        int newSize = input.Length-1, size = 0;
        bool swap;
        do
        {
            swap = false;
            for (int j = 0; j < newSize; j++)
            {
                if (input[j] > input[j + 1])
                {
                    int temp = input[j + 1];
                    input[j + 1] = input[j];
                    input[j] = temp;
                    swap = true;
                    size = j;
                }
            } newSize = size;
        } while (swap);

        DisplayArrayElements(input);
    }

Answer 3

Here is the simplest, best and optimal Bubble Sort Algorithm using a while loop. It sorts the numbers in the given array form left to right in ascending order. It is very simple to understand and easy to implement.

private static int[] bubbleSort(int[] array) {

        int length = array.length - 1;
        int index = 0;

        while ( index < length) {

            if (array[index] > array[index + 1]) {
                swap(array, index, index + 1);
            }
            index++;

            if (index == length) {
                index = 0;
                length--;
            }
        }
        return array;
    }

    private static void swap(int[] array, int index1, int index2) {

        int temp = array[index1];
        array[index1] = array[index2];
        array[index2] = temp;
    }

Answer 4

In the above example, the array got sorted after 3rd pass, but we will still continue with the 4th, 5th pass. Suppose if the array is already sorted, then there will be no swapping (because adjacent elements are always in order), but still we will continue with the passes and there will still be (n-1) passes.

If we can identify, that the array is sorted, then we should stop execution of further passes. This is the optimization over the original bubble sort algorithm.

If there is no swapping in a particular pass, it means the array has become sorted, so we should not perform the further passes. For this we can have a flag variable which is set to true before each pass and is made false when a swapping is performed.

void bubbleSort(int *arr, int n){
for(int i=0; i<n; i++)
{  
  bool flag = false;
   for(int j=0; j<n-i-1; j++)
   {
      if(array[j]>array[j+1])
      {
        flag = true;
         int temp = array[j+1];
         array[j+1] = array[j];
         array[j] = temp;
      }
   }
  // No Swapping happened, array is sorted
  if(!flag){ 
     return; 
  }}}

Answer 5

Optimized bubble sort with just 1 for Loop

/*Advanced BUBBLE SORT with ONE PASS*/
/*Authored by :: Brooks Tare  AAU*/

public class Bubble {

    public int[] bubble(int b[]){ 
    int temp,temp1; 

    for(int i=0;i<b.length-1;i++){

            if(b[i]>b[i+1] ){
                ///swap(b[i],b[i+1]);

                temp=b[i];
                b[i]=b[i+1];
                b[i+1]=temp;

    /*Checking if there is any number(s) greater than 
      the current number. If there is swap them.*/
                while(i>0){


                    if(b[i]<b[i-1]){
                    ///swap(b[i]<b[i-1])

                        temp1=b[i];
                        b[i]=b[i-1];
                        b[i-1]=temp1;
                        i--;
                    }
                    else if(b[i]>b[i-1]){i--;}
                }
            }
            else{continue;}

        }



        return b;
    }
///the following is a function to display the Array 
        public void see(int []a){
            for(int j=0;j<a.length;j++){
                System.out.print(a[j]+",");
            }
        }



    public static void main(String []args){
        ///You can change the Array to your preference.. u can even make it dynamic 

        int b[]={5,1,4,2,0,3}; 
        int v[]=new int[100]; 
        Bubble br=new Bubble();
        v=br.bubble(b);
        br.see(v);

    }
}

Answer 6

I devised a method that reduces the number of iterations by excluding parts at the beginning and end of the array that have been ordered in previous loops.

static int[] BubbleSortOptimized(int arr[]) {
    int start = 0, stop = arr.length - 1, control = 0;
    boolean ordered, nsCaught;
    while (true){
        ordered = true;
        nsCaught = false;
        for (int i = start; i < stop; i++) {
            if (i > 1) {
                if (!nsCaught && arr[i-2] > arr[i-1]){
                    ordered = false;
                    start = i-2;
                    nsCaught = true;
                }
            }
            if (arr[i] > arr[i+1]){
                int hold = arr[i];
                arr[i] = arr[i+1];
                arr[i+1] = hold;
                control = i;
            }
        }
        System.out.println(Arrays.toString(arr));
        if (ordered) return arr;
        stop = control;
    }
}

But as @Daniel Fischer said in an earlier answer, it doesn't do a lot with larger arrays.