Skip to main content

Bubble Sort in JAVA

What is bubble sort?

It is a sorting technique that is based on the comparison.Here we compare adjacent element, if the first element is larger than the second we swap each other. We do the same procedure again and again until array do not sort completely.

Example:-
5 1 4 2 8



 5 1 4 2 8 

here pass is nothing but iterating the loops equal to number of elements in the array but if it already sorted before then we can break the loop anddo exist
PASS1
case0 1 //check 0 and first element
 1 5 4 2 8 
case1 2 //check 1 and 2 element
 1 4 5 2 8 
case2 3
 1 4 2 5 8 
case3 4
 1 4 2 5 8 
swap istrue
PASS2 //first pass completed now do second pass
case0 1
 1 4 2 5 8 
case1 2
 1 2 4 5 8 
case2 3
 1 2 4 5 8 
case3 4
 1 2 4 5 8 
swap istrue
PASS3 // third pass

case0 1
 1 2 4 5 8 
case1 2
 1 2 4 5 8 
case2 3
 1 2 4 5 8 
case3 4
 1 2 4 5 8 
swap isfalse

since swap is false we break from the loop and do not go for fourth and fifth pass

Program:-

package sorting;

public class BubbleSort {
 
 //function to print array
 public static void print(int[] arr)
 {
  for(int i=0;i<=arr.length-1;i++)
   System.out.print(" "+arr[i]);
  System.out.println(" ");
 }
 
 //function to swap elements
 public static int[] swap(int[] arr,int i, int j)
 {
  int temp;
  temp=arr[i];
  arr[i]=arr[j];
  arr[j]=temp;
  
  return arr;
 }
 
public static void main(String[] args)
{
 int[] arr={5,1,4,2,8};
 print(arr);
 //declared to check the number of passes
 int pass=1;

 while(pass!=arr.length)
 {
  /*declared to check if there is no swap then we are 
  working on already sorted array and can break the loop */
 boolean swap=false;
 
 System.out.println("PASS"+pass);
 
 for(int i=0;i<arr.length-1;i++)
 { 
   if(arr[i]>arr[i+1])
   {
    arr=swap(arr,i,i+1);
    swap=true;
   }
   
   System.out.println("case"+i+" "+(i+1));
   
   print(arr);
  }


System.out.println("swap is"+swap);
if(swap==false)
{
 break;
}
pass++;
 }

}
 
}

Comments

.

Popular posts from this blog

Best LeetCode Lists for Interviews

Here is a list of some of the best questions asked in interviews Must do Top interview questions Top 100 liked Must do 75  Must do 60  Must do medium Data structures Tree Graph  Dynamic Programming Company Interviews FaceBook interviews Amazon Interviews Google Interviews Github master List

EDIT DISTANCE PROBLEM - LEETCODE 72

  Given two words  word1  and  word2 , find the minimum number of operations required to convert  word1  to  word2 . You have the following 3 operations permitted on a word: Insert a character Delete a character Replace a character Example 1: Input: word1 = "horse", word2 = "ros" Output: 3 Explanation: horse -> rorse (replace 'h' with 'r') rorse -> rose (remove 'r') rose -> ros (remove 'e') Solution:- class Solution { public int minDistance(String word1, String word2) { if(word1.length()==0){ return word2.length(); } if(word2.length()==0){ return word1.length(); } if(word1.length()==0 && word2.length()==0){ return 0; } int[][] result=new int[word1.length()+1][word2.length()+1]; for(int i=0;i<=word1.length();i++){ result[i][0]=i; } ...