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!

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

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])

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.

Hi everyone. I keep getting wrong answers even though my outputs are exactly the same as expected output. Does anyone has the same issue ? Here is my solution:

static void countSwaps(int[] a) {

    int n = a.length;

    int counter = 0;

for (int i = 0; i < n; i++) {

    for (int j = 0; j < n - 1; j++) {

    if (a[j] > a[j + 1]) {

        int temp = a[j];
        a[j] = a[j+1];
        a[j+1] = temp;
        counter++;

    }
}
}

System.out.println("Array is sorted in " + counter + " swaps.");

System.out.println("First element: " + a[0]);

System.out.println("Last element: " + a[n - 1]);

return;
}