Hackerrank - Sorting: Bubble Sort Solution
Consider the following version of Bubble Sort:
for (int i = 0; i < n; i++) {
for (int j = 0; j < n - 1; j++) {
// Swap adjacent elements if they are in decreasing order
if (a[j] > a[j + 1]) {
swap(a[j], a[j + 1]);
}
}
}
Given an array of integers, sort the array in ascending order using the Bubble Sort algorithm above. Once sorted, print the following three lines:
Array is sorted in numSwaps swaps.
, where is the number of swaps that took place.First Element: firstElement
, where is the first element in the sorted array.Last Element: lastElement
, where is the last element in the sorted array.
Hint: To complete this challenge, you must add a variable that keeps a running tally of all swaps that occur during execution.
For example, given a worst-case but small array to sort: we go through the following steps:
swap a
0 [6,4,1]
1 [4,6,1]
2 [4,1,6]
3 [1,4,6]
It took swaps to sort the array. Output would beArray is sorted in 3 swaps. First Element: 1 Last Element: 6
Function Description
Complete the function countSwaps in the editor below. It should print the three lines required, then return.
countSwaps has the following parameter(s):
- a: an array of integers .
Input Format
The first line contains an integer, , the size of the array .
The second line contains space-separated integers .
Constraints
Output Format
You must print the following three lines of output:
Array is sorted in numSwaps swaps.
, where is the number of swaps that took place.First Element: firstElement
, where is the first element in the sorted array.Last Element: lastElement
, where is the last element in the sorted array.
Sample Input 0
3
1 2 3
Sample Output 0
Array is sorted in 0 swaps.
First Element: 1
Last Element: 3
Explanation 0
The array is already sorted, so swaps take place and we print the necessary three lines of output shown above.
Sample Input 1
3
3 2 1
Sample Output 1
Array is sorted in 3 swaps.
First Element: 1
Last Element: 3
Explanation 1
The array is not sorted, and its initial values are: . The following swaps take place:
At this point the array is sorted and we print the necessary three lines of output shown above.
Solution in Python
from itertools import product
def countSwaps(a):
swaps = 0
for i,j in product(range(len(a)),range(len(a)-1)):
if a[j]>a[j+1]:
a[j],a[j+1] = a[j+1],a[j]
swaps += 1
print("Array is sorted in %s swaps." %swaps)
print("First Element:", a[0])
print("Last Element:", a[-1])
n = int(input())
a = list(map(int,input().split()))
countSwaps(a)
Additional Info
The following two gives same output
for i in range(n):
for j in range(n-1):
print(i,j)
from itertools import product
for i,j in product(range(n),range(n-1)):
print(i,j)