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  1. Sorting: Bubble Sort
  2. Discussions

Sorting: Bubble Sort

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  • thugwaffle
     + 5 comments

    I think this is a bit too easy to classify as "Medium", especially since you not only give most of the code in the question, but the solution too!

  • amrkhaledccd
     + 1 comment

    This should be classified as easy. The fact it is marked as Medium made me think 100 times before submitting my solution, i thought there is something i didn't understand

  • X1011
     + 2 comments

    fun fact: you don't need to actually perform the bubble sort to find the total number of swaps necessary; you can just count the inversions, as in the problem Merge Sort: Counting Inversions. not that it would be easier to code, but it would have lower time complexity.

  • MaxNRG
     + 7 comments

    Python 3 solution:

    n = int(input().strip())
a = list(map(int, input().strip().split(' ')))

numSwaps = 0

while True:
    SwapsFlag = False
    for i in range(len(a)-1):
        if a[i] > a[i+1]:
            a[i], a[i+1] = a[i+1], a[i]
            numSwaps += 1
            SwapsFlag = True
    if not SwapsFlag:
        break


print('Array is sorted in', numSwaps, 'swaps.')
print('First Element:', a[0])
print('Last Element:', a[-1])

  • bullshit
     + 2 comments

    JAVA 8 BubbleSort Solution:

    static void countSwaps(int[] a) {

        int count = 0;
        boolean flag = false;
        while (!flag) {
            flag = true;
            for (int i = 0; i < a.length - 1; i++) {
                if (a[i] > a[i + 1]) {
                    int temp = a[i];
                    a[i] = a[i + 1];
                    a[i + 1] = temp;
                    flag = false;
                    count++;
                }
            }
        }

        System.out.println(String.format(
                "Array is sorted in %d swaps.%n" +
                        "First Element: %d%n" +
                        "Last Element: %d%n",
                count,
                a[0],
                a[a.length - 1]));
    }
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