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Generate all distinct subsequences of array using backtracking
  • Last Updated : 14 Dec, 2020

Given an array arr[] consisting of N positive integers, the task is to generate all distinct subsequences of the array.

Examples:

Input: arr[] = {1, 2, 2}
Output: {} {1} {1, 2} {1, 2, 2} {2} {2, 2}
Explanation:
The total subsequences of the given array are {}, {1}, {2}, {2}, {1, 2}, {1, 2}, {2, 2}, {1, 2, 2}.
Since {2} and {1, 2} are repeated twice, print all the remaining subsequences of the array.

Input: arr[] = {1, 2, 3, 3}
Output: {} {1} {1, 2} {1, 2, 3} {1, 2, 3, 3} {1, 3} {1, 3, 3} {2} {2, 3} {2, 3, 3} {3} {3, 3}

 

Approach: Follow the steps below to solve the problem:

  1. Sort the given array.
  2. Initialize a vector of vectors to store all distinct subsequences.
  3. Traverse the array and considering two choices for each array element, to include it in a subsequence or not to include it.
  4. If duplicates are found, ignore them and check for the remaining elements. Otherwise, add the current array element to the current subsequence and traverse the remaining elements to generate subsequences.
  5. After generating the subsequences, remove the current array element.

Below is the implementation of the above approach:

C++14




// C++ program to implement
// the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to generate all distinct
// subsequences of the array using backtracking
void backtrack(vector<int>& nums, int start,
               vector<vector<int> >& resultset,
               vector<int> curr_set)
{
    resultset.push_back(curr_set);
 
    for (int i = start; i < nums.size(); i++) {
 
        // If the current element is repeating
        if (i > start
            && nums[i] == nums[i - 1]) {
            continue;
        }
 
        // Include current element
        // into the subsequence
        curr_set.push_back(nums[i]);
 
        // Proceed to the remaining array
        // to generate subsequences
        // including current array element
        backtrack(nums, i + 1, resultset,
                  curr_set);
 
        // Remove current element
        // from the subsequence
        curr_set.pop_back();
    }
}
 
// Function to sort the array and generate
// subsequences using Backtracking
vector<vector<int> > AllSubsets(
    vector<int> nums)
{
    // Stores the subsequences
    vector<vector<int> > resultset;
 
    // Stores the current
    // subsequence
    vector<int> curr_set;
 
    // Sort the vector
    sort(nums.begin(), nums.end());
 
    // Backtrack function to
    // generate subsequences
    backtrack(nums, 0, resultset,
              curr_set);
 
    // Return the result
    return resultset;
}
 
// Function to print all subsequences
void print(vector<vector<int> > result)
{
    for (int i = 0; i < result.size();
         i++) {
 
        cout << "{";
        for (int j = 0; j < result[i].size();
             j++) {
 
            cout << result[i][j];
            if (j < result[i].size() - 1) {
 
                cout << ", ";
            }
        }
        cout << "} ";
    }
}
 
// Driver Code
int main()
{
    vector<int> v{ 1, 2, 2 };
 
    // Function call
    vector<vector<int> > result
        = AllSubsets(v);
 
    // Print function
    print(result);
 
    return 0;
}

Java




// Java program to implement
// the above approach
import java.io.*;
import java.util.*;
 
class GFG{
     
// Function to generate all distinct
// subsequences of the array using backtracking
public static void backtrack(ArrayList<Integer> nums,
                             int start,
                             ArrayList<Integer> curr_set)
{
    System.out.print(curr_set + " ");
 
    for(int i = start; i < nums.size(); i++)
    {
         
        // If the current element is repeating
        if (i > start &&
            nums.get(i) == nums.get(i - 1))
        {
            continue;
        }
 
        // Include current element
        // into the subsequence
        curr_set.add(nums.get(i));
 
        // Proceed to the remaining array
        // to generate subsequences
        // including current array element
        backtrack(nums, i + 1, curr_set);
 
        // Remove current element
        // from the subsequence
        curr_set.remove(curr_set.size() - 1);
    }
}
 
// Function to sort the array and generate
// subsequences using Backtracking
public static void AllSubsets(ArrayList<Integer> nums)
{
 
    // Stores the current
    // subsequence
    ArrayList<Integer> curr_set = new ArrayList<>();
 
    // Sort the vector
    Collections.sort(nums);
 
    // Backtrack function to
    // generate subsequences
    backtrack(nums, 0, curr_set);
}
 
// Driver Code
public static void main(String[] args)
{
    ArrayList<Integer> v = new ArrayList<>();
    v.add(1);
    v.add(2);
    v.add(2);
     
    // Function call
    AllSubsets(v);
}
}
 
// This code is contributed by hemanthswarna1506

Python3




# Python3 program to implement
# the above approach
result = []
  
# Function to generate all distinct
# subsequences of the array
# using backtracking
def backtrack(nums, start, curr_set):
     
    # Global result
    result.append(list(curr_set))
     
    for i in range(start, len(nums)):
         
        # If the current element is repeating
        if (i > start and nums[i] == nums[i - 1]):
            continue
         
        # Include current element
        # into the subsequence
        curr_set.append(nums[i])
  
        # Proceed to the remaining array
        # to generate subsequences
        # including current array element
        backtrack(nums, i + 1, curr_set)
  
        # Remove current element
        # from the subsequence
        curr_set.pop()
     
# Function to sort the array and generate
# subsequences using Backtracking
def AllSubsets(nums):
 
    # Stores the current
    # subsequence
    curr_set = []
     
    # Sort the vector
    nums.sort()
     
    # Backtrack function to
    # generate subsequences
    backtrack(nums, 0, curr_set)
     
# Function to prints all subsequences
def prints():
     
    global result
 
    for i in range(len(result)):
        print('{', end = '')
         
        for j in range(len(result[i])):
            print(result[i][j], end = '')
 
            if (j < len(result[i]) - 1):
                print(',', end = ' ')
                 
        print('} ', end = '')
         
# Driver Code
if __name__=='__main__':
     
    v = [ 1, 2, 2 ]
      
    # Function call
    AllSubsets(v)
  
    # Print function
    prints()
 
# This code is contributed by rutvik_56

C#




// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG{
     
// Function to generate all distinct
// subsequences of the array using
// backtracking
public static void backtrack(List<int> nums,
                             int start,
                             List<int> curr_set)
{
  Console.Write(" {");
   
  foreach(int i in curr_set)
  {
    Console.Write(i);
    Console.Write(", ");
  }
 
  Console.Write("}");
   
  for(int i = start;
          i < nums.Count; i++)
  {
    // If the current element
    // is repeating
    if (i > start &&
        nums[i] == nums[i - 1])
    {
      continue;
    }
 
    // Include current element
    // into the subsequence
    curr_set.Add(nums[i]);
 
    // Proceed to the remaining array
    // to generate subsequences
    // including current array element
    backtrack(nums, i + 1, curr_set);
 
    // Remove current element
    // from the subsequence
    curr_set.Remove(curr_set.Count - 1);
  }
}
 
// Function to sort the array and generate
// subsequences using Backtracking
public static void AllSubsets(List<int> nums)
{
  // Stores the current
  // subsequence
  List<int> curr_set = new List<int>();
 
  // Sort the vector
  nums.Sort();
 
  // Backtrack function to
  // generate subsequences
  backtrack(nums, 0, curr_set);
}
 
// Driver Code
public static void Main(String[] args)
{
  List<int> v = new List<int>();
  v.Add(1);
  v.Add(2);
  v.Add(2);
 
  // Function call
  AllSubsets(v);
}
}
 
// This code is contributed by 29AjayKumar
Output
{} {1} {1, 2} {1, 2, 2} {2} {2, 2}

Time Complexity: O(2N)
Auxiliary Space: O(N)




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