Bubble sort using two Stacks
Prerequisite : Bubble Sort
Write a function that sort an array of integers using stacks and also uses bubble sort paradigm.
Algorithm:
1. Push all elements of array in 1st stack
2. Run a loop for 'n' times(n is size of array)
having the following :
2.a. Keep on pushing elements in the 2nd
stack till the top of second stack is
smaller than element being pushed from
1st stack.
2.b. If the element being pushed is smaller
than top of 2nd stack then swap them
(as in bubble sort)
*Do above steps alternatively
TRICKY STEP: Once a stack is empty, then the
top of the next stack will be the largest
number so keep it at its position in array
i.e arr[len-1-i] and then pop it from that
stack.
Implementation:
C++
#include <bits/stdc++.h>
using namespace std;
void bubbleSortStack( int a[], int n)
{
stack< int > s1;
for ( int i = 0; i < n; i++)
{
s1.push(a[i]);
}
stack< int > s2;
for ( int i = 0; i < n; i++)
{
if (i % 2 == 0)
{
while (!s1.empty())
{
int t = s1.top();
s1.pop();
if (s2.empty())
{
s2.push(t);
}
else
{
if (s2.top() > t)
{
int temp = s2.top();
s2.pop();
s2.push(t);
s2.push(temp);
}
else
{
s2.push(t);
}
}
}
a[n - 1 - i] = s2.top();
s2.pop();
}
else
{
while (!s2.empty())
{
int t = s2.top();
s2.pop();
if (s1.empty())
{
s1.push(t);
}
else
{
if (s1.top() > t)
{
int temp = s1.top();
s1.pop();
s1.push(t);
s1.push(temp);
}
else
{
s1.push(t);
}
}
}
a[n - 1 - i] = s1.top();
s1.pop();
}
}
cout << "[" ;
for ( int i = 0; i < n; i++)
{
cout << a[i] << ", " ;
}
cout << "]" ;
}
int main()
{
int a[] = { 15, 12, 44, 2, 5, 10 };
int n = sizeof (a) / sizeof (a[0]);
bubbleSortStack(a, n);
return 0;
}
|
Java
import java.util.Arrays;
import java.util.Stack;
public class Test
{
static void bubbleSortStack( int arr[], int n)
{
Stack<Integer> s1 = new Stack<>();
for ( int num : arr)
s1.push(num);
Stack<Integer> s2 = new Stack<>();
for ( int i = 0 ; i < n; i++)
{
if (i % 2 == 0 )
{
while (!s1.isEmpty())
{
int t = s1.pop();
if (s2.isEmpty())
s2.push(t);
else
{
if (s2.peek() > t)
{
int temp = s2.pop();
s2.push(t);
s2.push(temp);
}
else
{
s2.push(t);
}
}
}
arr[n- 1 -i] = s2.pop();
}
else
{
while (!s2.isEmpty())
{
int t = s2.pop();
if (s1.isEmpty())
s1.push(t);
else
{
if (s1.peek() > t)
{
int temp = s1.pop();
s1.push(t);
s1.push(temp);
}
else
s1.push(t);
}
}
arr[n- 1 -i] = s1.pop();
}
}
System.out.println(Arrays.toString(arr));
}
public static void main(String[] args)
{
int arr[] = { 15 , 12 , 44 , 2 , 5 , 10 };
bubbleSortStack(arr, arr.length);
}
}
|
Python3
def bubbleSortStack(a, n):
s1 = []
for i in range (n):
s1.append(a[i]);
s2 = []
for i in range (n):
if (i % 2 = = 0 ):
while ( len (s1) ! = 0 ):
t = s1[ - 1 ]
s1.pop();
if ( len (s2) = = 0 ):
s2.append(t);
else :
if (s2[ - 1 ] > t):
temp = s2[ - 1 ]
s2.pop();
s2.append(t);
s2.append(temp);
else :
s2.append(t);
a[n - 1 - i] = s2[ - 1 ]
s2.pop();
else :
while ( len (s2) ! = 0 ):
t = s2[ - 1 ]
s2.pop();
if ( len (s1) = = 0 ):
s1.append(t);
else :
if (s1[ - 1 ] > t):
temp = s1[ - 1 ]
s1.pop();
s1.append(t);
s1.append(temp);
else :
s1.append(t);
a[n - 1 - i] = s1[ - 1 ]
s1.pop();
print ( "[" , end = '')
for i in range (n):
print (a[i], end = ', ' )
print ( ']' , end = '')
if __name__ = = '__main__' :
a = [ 15 , 12 , 44 , 2 , 5 , 10 ]
n = len (a)
bubbleSortStack(a, n);
|
C#
using System;
using System.Collections.Generic;
class GFG
{
static void bubbleSortStack( int []arr, int n)
{
Stack< int > s1 = new Stack< int >();
foreach ( int num in arr)
s1.Push(num);
Stack< int > s2 = new Stack< int >();
for ( int i = 0; i < n; i++)
{
if (i % 2 == 0)
{
while (s1.Count != 0)
{
int t = s1.Pop();
if (s2.Count == 0)
s2.Push(t);
else
{
if (s2.Peek() > t)
{
int temp = s2.Pop();
s2.Push(t);
s2.Push(temp);
}
else
{
s2.Push(t);
}
}
}
arr[n - 1 - i] = s2.Pop();
}
else
{
while (s2.Count != 0)
{
int t = s2.Pop();
if (s1.Count == 0)
s1.Push(t);
else
{
if (s1.Peek() > t)
{
int temp = s1.Pop();
s1.Push(t);
s1.Push(temp);
}
else
s1.Push(t);
}
}
arr[n - 1 - i] = s1.Pop();
}
}
Console.WriteLine( "[" + String.Join( ", " , arr) + "]" );
}
public static void Main(String[] args)
{
int []arr = {15, 12, 44, 2, 5, 10};
bubbleSortStack(arr, arr.Length);
}
}
|
Javascript
<script>
function bubbleSortStack(arr, n)
{
let s1 = [];
for (let num = 0; num < arr.length; num++)
s1.push(arr[num]);
let s2 = [];
for (let i = 0; i < n; i++)
{
if (i % 2 == 0)
{
while (s1.length != 0)
{
let t = s1.pop();
if (s2.length == 0)
s2.push(t);
else
{
if (s2[s2.length - 1] > t)
{
let temp = s2.pop();
s2.push(t);
s2.push(temp);
}
else
{
s2.push(t);
}
}
}
arr[n - 1 - i] = s2.pop();
}
else
{
while (s2.length != 0)
{
let t = s2.pop();
if (s1.length == 0)
s1.push(t);
else
{
if (s1[s1.length - 1] > t)
{
let temp = s1.pop();
s1.push(t);
s1.push(temp);
}
else
s1.push(t);
}
}
arr[n - 1 - i] = s1.pop();
}
}
document.write((arr).join( " " ));
}
let arr = [ 15, 12, 44, 2, 5, 10 ];
bubbleSortStack(arr, arr.length);
</script>
|
Output
[2, 5, 10, 12, 15, 44, ]
Time Complexity: O(n2) // we are using a for loop from 0 to n-1 and inside it we are traversing the stack elements hence the complexity is quadratic
Auxiliary Space: O(n) // since we are using a stack to store all the elements hence takes up the space
Last Updated :
14 Feb, 2023
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