Sort an array using Bubble Sort without using loops
Last Updated :
22 Jun, 2021
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Given an array arr[] consisting of N integers, the task is to sort the given array by using Bubble Sort without using loops.
Examples:
Input: arr[] = {1, 3, 4, 2, 5}
Output: 1 2 3 4 5Input: arr[] = {1, 3, 4, 2}
Output: 1 2 3 4
Approach: The idea to implement Bubble Sort without using loops is based on the following observations:
- The sorting algorithm of Bubble Sort performs the following steps:
- The outer loop traverses the given array (N – 1) times.
- The inner loop traverses the array and swaps two adjacent elements if arr[i] > arr[i + 1], for every i over the range [0, N – 1].
- It can be observed that in every N – 1 iteration, the largest element over the range [0, N – 1 – i] shifts to the position (N – 1 – i) for every i over the range [0, N – 1].
- Therefore, the idea is to use recursion to avoid loops.
Follow the steps below to solve the problem:
- Consider the following base cases:
- If the array contains a single element, simply print the array.
- If the array contains two elements, swap the pair of elements (if required) to obtain a sorted sequence of array elements. Print the sorted array.
- Store the first two elements from the current array in variables, say a and b.
- Initialize an array, say bs[], to store the remaining array elements.
- Place the smaller value among a and b at the front of the current array.
- Recursively repeat the above steps with a new array formed by appending bs[] to the end of the larger value among a and b.
- Store the sorted list returned in a variable, say res[].
- Now, recursively repeat the above steps with a new array formed by appending res[] (excluding the last element) to the end of res[res.size() – 1] ( the last element ).
- After completing the above steps, print the returned list.
Below is the implementation of the above approach:
- C++
- Java
- Python3
- C#
- Javascript
C++
// C++ program for the above approach#include<bits/stdc++.h>using namespace std;// Function to implement bubble// sort without using loopsvector<int> bubble_sort(vector<int> ar){ // Base Case: If array // contains a single element if (ar.size() <= 1) return ar; // Base Case: If array // contains two elements if (ar.size() == 2){ if(ar[0] < ar[1]) return ar; else return {ar[1], ar[0]}; } // Store the first two elements // of the list in variables a and b int a = ar[0]; int b = ar[1]; // Store remaining elements // in the list bs vector<int> bs; for(int i = 2; i < ar.size(); i++) bs.push_back(ar[i]); // Store the list after // each recursive call vector<int> res; // If a < b if (a < b){ vector<int> temp1; temp1.push_back(b); for(int i = 0; i < bs.size(); i++) temp1.push_back(bs[i]); vector<int> v = bubble_sort(temp1); v.insert(v.begin(), a); res = v; } // Otherwise, if b >= a else{ vector<int> temp1; temp1.push_back(a); for(int i = 0; i < bs.size(); i++) temp1.push_back(bs[i]); vector<int> v = bubble_sort(temp1); v.insert(v.begin(), b); res = v; } // Recursively call for the list // less than the last element and // and return the newly formed list vector<int> pass; for(int i = 0; i < res.size() - 1; i++) pass.push_back(res[i]); vector<int> ans = bubble_sort(pass); ans.push_back(res[res.size() - 1]); return ans;}// Driver Codeint main(){ vector<int> arr{1, 3, 4, 5, 6, 2}; vector<int> res = bubble_sort(arr); // Print the array for(int i = 0; i < res.size(); i++) cout << res[i] << " ";}// This code is contributed by ipg2016107. |
Java
Python3
C#
Javascript
Output:
1 2 3 4 5 6
Time Complexity: O(N2)
Auxiliary Space: O(N)
