C++ Program to Generate All Possible Subsets using Lexicographic Order

This is a C++ program to generate all subsets of a given set in the lexicographic order.

Problem Description

1. This algorithm print all the possible combination of each length from the given array in increasing order.
2. The time complexity of this algorithm is O(n*(2^n)).

Problem Solution

1. This algorithm takes the input of ‘n’ data element and prints all possible combination.
2. For that, it sorts the input array and maintains a boolean array ‘n’.
3. If the corresponding boolean value is true, then it prints that element.
4. Exit.

Program/Source Code

C++ Program to generate all subsets of a given set in the lexicographic order.
This program is successfully run on Dev-C++ using TDM-GCC 4.9.2 MinGW compiler on a Windows system.

#include<iostream>
 
using namespace std;
 
// A function to sort the data set.
void Sort(int a[], int n)
{
	int i, j, temp;
 
	for(i = 0; i < n; i++)
	{
		for(j = i+1; j < n; j++)
		{
			if(a[i] > a[j])
			{
				temp = a[i];
				a[i] = a[j];
				a[j] = temp;
			}
		}
	}
}
 
// A function to print all combination of a given length from the given array.
void GenSubset(int a[], int reqLen, int start, int currLen, bool check[], int len) 
{
	// Return if the currLen is more than the required length.
	if(currLen > reqLen)
	return;
	// If currLen is equal to required length then print the sequence.
	else if (currLen == reqLen) 
	{
		cout<<"\t";
		cout<<"{ ";
		for (int i = 0; i < len; i++) 
		{
			if (check[i] == true) 
			{
				cout<<a[i]<<" ";
			}
		}
		cout<<"}\n";
		return;
	}
	// If start equals to len then return since no further element left.
	if (start == len) 
	{
		return;
	}
	// For every index we have two options.
	// First is, we select it, means put true in check[] and increment currLen and start.
	check[start] = true;
	GenSubset(a, reqLen, start + 1, currLen + 1, check, len);
	// Second is, we don't select it, means put false in check[] and only start incremented.
	check[start] = false;
	GenSubset(a, reqLen, start + 1, currLen, check, len);
}
}
 
int main()
{
	int i, n;
	bool check[n];
	cout<<"Enter the number of element array have: ";
	cin>>n;
 
	int arr[n];
	cout<<"\n";
 
	// Take the input of the array.
	for(i = 0; i < n; i++)
	{
		cout<<"Enter "<<i+1<<" element: ";
		cin>>arr[i];
		check[i] = false;
	}
 
	// Sort the array.
	Sort(arr, n);
 
	cout<<"\nThe subset in the lexicographic order: \n";
	cout<<"\t{ }\n";
	for(i = 1; i <= n; i++)
	{
		GenSubset(arr, i, 0, 0, check, n);
	}
	return 0;
}

Program Explanation

1. Take the input of an array of ‘n’ data element.
2. Assign the boolean array to false initially.
3. Sort the array.
4. For each length ‘i’ call the Combination() function.
5. In Combination(), if currLen is more than the reqLen then return.
6. Otherwise, if currLen is equal to reqLen then there will be a new sequence generated, print it.
7. There will be two option either proceed with a start as ‘true’ and recursively call Combination() with incremented value of ‘currLen’ and ‘start’.
8. Or proceed with a start as ‘false’ and recursively call Combination() with incremented value of ‘start’.
9. Exit.

Runtime Test Cases

 
Case 1:
Enter the number of element array have: 5
 
Enter 1 element: 6
Enter 2 element: 8
Enter 3 element: 3
Enter 4 element: 4
Enter 5 element: 9
 
The subset in the lexicographic order:
        { }
        { 3 }
        { 4 }
        { 6 }
        { 8 }
        { 9 }
        { 3 4 }
        { 3 6 }
        { 3 8 }
        { 3 9 }
        { 4 6 }
        { 4 8 }
        { 4 9 }
        { 6 8 }
        { 6 9 }
        { 8 9 }
        { 3 4 6 }
        { 3 4 8 }
        { 3 4 9 }
        { 3 6 8 }
        { 3 6 9 }
        { 3 8 9 }
        { 4 6 8 }
        { 4 6 9 }
        { 4 8 9 }
        { 6 8 9 }
        { 3 4 6 8 }
        { 3 4 6 9 }
        { 3 4 8 9 }
        { 3 6 8 9 }
        { 4 6 8 9 }
        { 3 4 6 8 9 }

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