An alternate way associated with putting the actual largest element at the highest index in the array makes use of an algorithm known as bubble sort.

## Algorithm:

BUBBLESORT(A) for i ▭ 1 to length[A] do for j ▭ length[A] downto i + 1 do if A[j] < A[j -1] then exchange A[j] ▭ A[j-1]

## Bubble-Sort Running Time:

BUBBLESORT(A) | |

for i <---- 1 to length[A] | c_{1} |

do for j <---- length[A] downto i + 1 | c_{2} |

Comparisons: ≈ n2/2, do if A[j] < A[j -1] | c_{3} |

Exchanges: ≈ n2/2, then exchange A[j] « A[j-1] | c_{3} |

### Example1:

## Example2: consider the array:

45 | 67 | 12 | 34 | 25 | 39 |

In the first stage, the particular focus is actually about the first two elements, that are in comparison as well as swapped, if necessary. In this instance, because the element at index 1 is actually larger than the one at index 0, no swap takes place.

45 | 67 | 12 | 34 | 25 | 39 |

Then this focus move to the elements at index 1 and also 2 that is in comparison as well as swapped, if necessary. In example, 67 is actually bigger than 12 therefore the two elements are swapped. The result is that the largest of the first three elements is now at index 2.

45 | 67 | 12 | 34 | 25 | 39 |

The procedure is actually repeated before the focus moves to the end of the array, at which point the largest of all the elements ends up at the highest possible index. The rest of the steps as well as result are:

45 | 12 | 34 | 67 | 25 | 39 |

45 | 12 | 34 | 25 | 67 | 39 |

45 | 12 | 34 | 25 | 39 | 67 |

### Example 3:

#### Example 4:

1st move:

( 5 1 4 2 8 ) | -> | ( 1 5 4 2 8 ) | Here, algorithm compares the first two elements, as well as swaps since 5 > 1. |

( 1 5 4 2 8 ) | -> | ( 1 4 5 2 8 ) | Swap since 5 > 4 |

( 1 4 5 2 8 ) | -> | ( 1 4 2 5 8 ) | Swap since 5 > 2 |

( 1 4 2 5 8 ) | -> | ( 1 4 2 5 8 ) | Right now, because these types of elements tend to be already so as (8 > 5), algorithm doesn’t swap them. |

2^{nd} move:

( 1 4 2 5 8 ) | -> | ( 1 4 2 5 8 ) | |

( 1 4 2 5 8 ) | -> | ( 1 2 4 5 8 ) | Swap since 4 > 2 |

( 1 2 4 5 8 ) | -> | ( 1 2 4 5 8 ) | |

( 1 2 4 5 8 ) | -> | ( 1 2 4 5 8 ) |

Now, the particular array has already been sorted, however the algorithm doesn't know if it is completed. The actual algorithm needs one whole move with no exchange to understand it's sorted.

3^{rd} move:

( 1 2 4 5 8) | -> | ( 1 2 4 5 8 ) |

( 1 2 4 5 8) | -> | ( 1 2 4 5 8 ) |

( 1 2 4 5 8) | -> | ( 1 2 4 5 8 ) |

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