Print all subsequences of a string using ArrayList
Last Updated :
01 Nov, 2023
Given a string str, the task is to print all the sub-sequences of str.
A subsequence is a sequence that can be derived from another sequence by deleting some or no elements without changing the order of the remaining elements.
Examples:
Input: str = "abc"
Output: a b ab c ac bc abc
Input: str = "geek"
Output: g e ge e ge ee gee k gk ek gek ek gek eek geek
Approach: Write a recursive function that prints every sub-sequence of the sub-string starting from the second character str[1, n - 1] after appending the first character of the string str[0] in the beginning of every sub-sequence. Terminating condition will be when the passed string is empty, in that case the function will return an empty arraylist.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Utility function to print the contents
// of the List
void printList(vector<string> arrL)
{
arrL.erase(find(arrL.begin(), arrL.end(), ""));
for (int i = 0; i < arrL.size(); i++)
cout << arrL[i] << " ";
}
// Function to returns the arraylist which contains
// all the sub-sequences of str
vector<string> getSequence(string str)
{
// If string is empty
if (str.length() == 0)
{
// Return an empty arraylist
vector<string> empty;
empty.push_back("");
return empty;
}
// Take first character of str
char ch = str[0];
// Take sub-string starting from the
// second character
string subStr = str.substr(1);
// Recursive call for all the sub-sequences
// starting from the second character
vector<string> subSequences = getSequence(subStr);
// Add first character from str in the beginning
// of every character from the sub-sequences
// and then store it into the resultant arraylist
vector<string> res;
for(string val : subSequences)
{
res.push_back(val);
res.push_back(ch + val);
}
// Return the resultant arraylist
return res;
}
int main()
{
string str = "geek";
printList(getSequence(str));
return 0;
}
// This code is contributed by rameshtravel07.
// Java implementation of the approach
import java.util.ArrayList;
public class GFG {
// Utility function to print the contents
// of the ArrayList
static void printArrayList(ArrayList<String> arrL)
{
arrL.remove("");
for (int i = 0; i < arrL.size(); i++)
System.out.print(arrL.get(i) + " ");
}
// Function to returns the arraylist which contains
// all the sub-sequences of str
public static ArrayList<String> getSequence(String str)
{
// If string is empty
if (str.length() == 0) {
// Return an empty arraylist
ArrayList<String> empty = new ArrayList<>();
empty.add("");
return empty;
}
// Take first character of str
char ch = str.charAt(0);
// Take sub-string starting from the
// second character
String subStr = str.substring(1);
// Recursive call for all the sub-sequences
// starting from the second character
ArrayList<String> subSequences =
getSequence(subStr);
// Add first character from str in the beginning
// of every character from the sub-sequences
// and then store it into the resultant arraylist
ArrayList<String> res = new ArrayList<>();
for (String val : subSequences) {
res.add(val);
res.add(ch + val);
}
// Return the resultant arraylist
return res;
}
// Driver code
public static void main(String[] args)
{
String str = "geek";
printArrayList(getSequence(str));
// System.out.print(getSequence(str));
}
}
# Python implementation of the approach
# Utility function to print the contents
# of the ArrayList
def printArrayList(arrL):
arrL.remove("")
print(*arrL, sep = " ")
# Function to returns the arraylist which contains
# all the sub-sequences of str
def getSequence(Str):
# If string is empty
if(len(Str) == 0):
# Return an empty arraylist
empty = []
empty.append("")
return empty
# Take first character of str
ch = Str[0]
# Take sub-string starting from the
# second character
subStr = Str[1:]
# Recursive call for all the sub-sequences
# starting from the second character
subSequences = getSequence(subStr)
# Add first character from str in the beginning
# of every character from the sub-sequences
# and then store it into the resultant arraylist
res = []
for val in subSequences:
res.append(val)
res.append(ch + val)
# Return the resultant arraylist
return res
# Driver code
Str = "geek"
printArrayList(getSequence(Str))
# This code is contributed by avanitrachhadiya2155
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG{
// Utility function to print the contents
// of the List
static void printList(List<String> arrL)
{
arrL.Remove("");
for (int i = 0; i < arrL.Count; i++)
Console.Write(arrL[i] + " ");
}
// Function to returns the arraylist which contains
// all the sub-sequences of str
public static List<String> getSequence(String str)
{
// If string is empty
if (str.Length == 0)
{
// Return an empty arraylist
List<String> empty = new List<String>();
empty.Add("");
return empty;
}
// Take first character of str
char ch = str[0];
// Take sub-string starting from the
// second character
String subStr = str.Substring(1);
// Recursive call for all the sub-sequences
// starting from the second character
List<String> subSequences = getSequence(subStr);
// Add first character from str in the beginning
// of every character from the sub-sequences
// and then store it into the resultant arraylist
List<String> res = new List<String>();
foreach (String val in subSequences)
{
res.Add(val);
res.Add(ch + val);
}
// Return the resultant arraylist
return res;
}
// Driver code
public static void Main(String[] args)
{
String str = "geek";
printList(getSequence(str));
// Console.Write(getSequence(str));
}
}
// This code is contributed by Rohit_ranjan
<script>
// JavaScript implementation of the approach
// Utility function to print the contents
// of the ArrayList
function printArrayList(arrL)
{
arrL.splice(arrL.indexOf(""), 1);
for (let i = 0; i < arrL.length; i++)
document.write(arrL[i] + " ");
}
// Function to returns the arraylist which contains
// all the sub-sequences of str
function getSequence(str)
{
// If string is empty
if (str.length == 0) {
// Return an empty arraylist
let empty = [];
empty.push("");
return empty;
}
// Take first character of str
let ch = str[0];
// Take sub-string starting from the
// second character
let subStr = str.substring(1);
// Recursive call for all the sub-sequences
// starting from the second character
let subSequences = getSequence(subStr);
// Add first character from str in the beginning
// of every character from the sub-sequences
// and then store it into the resultant arraylist
let res = [];
for (let val = 0; val < subSequences.length; val++) {
res.push(subSequences[val]);
res.push(ch + subSequences[val]);
}
// Return the resultant arraylist
return res;
}
let str = "geek";
printArrayList(getSequence(str));
</script>
Outputg e ge e ge ee gee k gk ek gek ek gek eek geek
Time Complexity: O(2^n), where n is the length of the string.
Space Complexity: O(2^n), as the recursive stack grows until the length of the string.
Alternate Solution: One by one fix characters and recursively generate all subsets starting from them.
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to print all the sub-sequences
// of a string
void printSubSeq(string sub, string ans)
{
if (sub.length() == 0) {
cout << "" << ans << " ";
return;
}
// First character of sub
char ch = sub[0];
// Sub-string starting from second
// character of sub
string ros = sub.substr(1);
// Excluding first character
printSubSeq(ros, ans);
// Including first character
printSubSeq(ros, ans + ch);
}
// Driver code
int main()
{
string str = "abc";
printSubSeq(str, "");
return 0;
}
// This code is contributed by divyesh072019
// Java implementation of the approach
public class sub_sequence {
// Function to print all the sub-sequences
// of a string
public static void printSubSeq(String sub,
String ans)
{
if (sub.length() == 0) {
System.out.print("" + ans + " ");
return;
}
// First character of sub
char ch = sub.charAt(0);
// Sub-string starting from second
// character of sub
String ros = sub.substring(1);
// Excluding first character
printSubSeq(ros, ans);
// Including first character
printSubSeq(ros, ans + ch);
}
// Driver code
public static void main(String[] args)
{
String str = "abc";
printSubSeq(str, "");
}
}
# Python3 implementation of the approach
# Function to print all the sub-sequences
# of a string
def printSubSeq(sub, ans) :
if (len(sub) == 0) :
print(ans , end = " ")
return
# First character of sub
ch = sub[0]
# Sub-string starting from second
# character of sub
ros = sub[1 : ]
# Excluding first character
printSubSeq(ros, ans)
# Including first character
printSubSeq(ros, ans + ch)
Str = "abc"
printSubSeq(Str, "")
# This code iscontributed by divyeshrabadiya07
// C# implementation of the approach
using System;
class GFG
{
// Function to print all the
// sub-sequences of a string
public static void printSubSeq(string sub,
string ans)
{
if (sub.Length == 0)
{
Console.Write("" + ans + " ");
return;
}
// First character of sub
char ch = sub[0];
// Sub-string starting from second
// character of sub
string ros = sub.Substring(1);
// Excluding first character
printSubSeq(ros, ans);
// Including first character
printSubSeq(ros, ans + ch);
}
// Driver code
public static void Main()
{
string str = "abc";
printSubSeq(str, "") ;
}
}
// This code is contributed by Ryuga
<script>
// Javascript implementation of the approach
// Function to print all the
// sub-sequences of a string
function printSubSeq(sub, ans)
{
if (sub.length == 0)
{
document.write("" + ans + " ");
return;
}
// First character of sub
let ch = sub[0];
// Sub-string starting from second
// character of sub
let ros = sub.substring(1);
// Excluding first character
printSubSeq(ros, ans);
// Including first character
printSubSeq(ros, ans + ch);
}
let str = "abc";
printSubSeq(str, "") ;
// This code is contributed by decode2207.
</script>
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