Imprimir todas las subsecuencias de una string – Part 1

Dada una string, tenemos que encontrar todas las subsecuencias de la misma. Una String es una subsecuencia de una String dada, que se genera eliminando algún carácter de una string dada sin cambiar su orden.

Ejemplos: 

Input : abc
Output : a, b, c, ab, bc, ac, abc

Input : aaa
Output : a, aa, aaa

Método 1 (Concepto de elegir y no elegir)  :

Implementación:

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Find all subsequences
void printSubsequence(string input, string output)
{
    // Base Case
    // if the input is empty print the output string
    if (input.empty()) {
        cout << output << endl;
        return;
    }
 
    // output is passed with including
    // the Ist character of
    // Input string
    printSubsequence(input.substr(1), output + input[0]);
 
    // output is passed without
    // including the Ist character
    // of Input string
    printSubsequence(input.substr(1), output);
}
 
// Driver code
int main()
{
    // output is set to null before passing in as a
    // parameter
    string output = "";
    string input = "abcd";
 
    printSubsequence(input, output);
 
    return 0;
}

// Java program for the above approach
import java.util.*;
class GFG {
 
    // Declare a global list
    static List<String> al = new ArrayList<>();
 
    // Creating a public static Arraylist such that
    // we can store values
    // IF there is any question of returning the
    // we can directly return too// public static
    // ArrayList<String> al = new ArrayList<String>();
    public static void main(String[] args)
    {
        String s = "abcd";
        findsubsequences(s, ""); // Calling a function
        System.out.println(al);
    }
 
    private static void findsubsequences(String s,
                                         String ans)
    {
        if (s.length() == 0) {
            al.add(ans);
            return;
        }
 
        // We add adding 1st character in string
        findsubsequences(s.substring(1), ans + s.charAt(0));
 
        // Not adding first character of the string
        // because the concept of subsequence either
        // character will present or not
        findsubsequences(s.substring(1), ans);
    }
}

# Below is the implementation of the above approach
def printSubsequence(input, output):
 
    # Base Case
    # if the input is empty print the output string
    if len(input) == 0:
        print(output, end=' ')
        return
 
    # output is passed with including the
    # 1st character of input string
    printSubsequence(input[1:], output+input[0])
 
    # output is passed without including the
    # 1st character of input string
    printSubsequence(input[1:], output)
 
 
# Driver code
# output is set to null before passing in
# as a parameter
output = ""
input = "abcd"
 
printSubsequence(input, output)
 
# This code is contributed by Tharun Reddy

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
static void printSubsequence(string input,
                             string output)
{
     
    // Base Case
    // If the input is empty print the output string
    if (input.Length == 0)
    {
        Console.WriteLine(output);
        return;
    }
 
    // Output is passed with including
    // the Ist character of
    // Input string
    printSubsequence(input.Substring(1),
                     output + input[0]);
 
    // Output is passed without
    // including the Ist character
    // of Input string
    printSubsequence(input.Substring(1),
                     output);
}
 
// Driver code
static void Main()
{
     
    // output is set to null before passing
    // in as a parameter
    string output = "";
    string input = "abcd";
     
    printSubsequence(input, output);
}
}
  
// This code is contributed by SoumikMondal

<script>
 
// JavaScript program for the above approach
 
// Find all subsequences
function printSubsequence(input, output)
{
    // Base Case
    // if the input is empty print the output string
    if (input.length==0) {
        document.write( output + "<br>");
        return;
    }
 
    // output is passed with including
    // the Ist character of
    // Input string
    printSubsequence(input.substring(1), output + input[0]);
 
    // output is passed without
    // including the Ist character
    // of Input string
    printSubsequence(input.substring(1), output);
}
 
// Driver code
// output is set to null before passing in as a
// parameter
var output = "";
var input = "abcd";
printSubsequence(input, output);
 
 
</script>

abcd
abc
abd
ab
acd
ac
ad
a
bcd
bc
bd
b
cd
c
d

Método 2 

Explicación : 

Step 1: Iterate over the entire String
Step 2: Iterate from the end of string 
        in order to generate different substring
        add the substring to the list
Step 3: Drop kth character from the substring obtained 
        from above to generate different subsequence.
Step 4: if the subsequence is not in the list then recur.

A continuación se muestra la implementación del enfoque. 

// CPP program to print all subsequence of a
// given string.
#include <bits/stdc++.h>
using namespace std;
 
// set to store all the subsequences
unordered_set<string> st;
 
// Function computes all the subsequence of an string
void subsequence(string str)
{
 
    // Iterate over the entire string
    for (int i = 0; i < str.length(); i++) {
 
        // Iterate from the end of the string
        // to generate substrings
        for (int j = str.length(); j > i; j--) {
            string sub_str = str.substr(i, j);
            st.insert(sub_str);
 
            // Drop kth character in the substring
            // and if its not in the set then recur
            for (int k = 1; k < sub_str.length(); k++) {
                string sb = sub_str;
 
                // Drop character from the string
                sb.erase(sb.begin() + k);
                subsequence(sb);
            }
        }
    }
}
 
// Driver Code
int main()
{
    string s = "aabc";
    subsequence(s);
    for (auto i : st)
        cout << i << " ";
    cout << endl;
 
    return 0;
}
 
// This code is contributed by
// sanjeev2552

// Java Program to print all subsequence of a
// given string.
import java.util.HashSet;
 
public class Subsequence {
 
    // Set to store all the subsequences
    static HashSet<String> st = new HashSet<>();
 
    // Function computes all the subsequence of an string
    static void subsequence(String str)
    {
 
        // Iterate over the entire string
        for (int i = 0; i < str.length(); i++) {
 
            // Iterate from the end of the string
            // to generate substrings
            for (int j = str.length(); j > i; j--) {
                String sub_str = str.substring(i, j);
 
                if (!st.contains(sub_str))
                    st.add(sub_str);
 
                // Drop kth character in the substring
                // and if its not in the set then recur
                for (int k = 1; k < sub_str.length() - 1;
                     k++) {
                    StringBuffer sb
                        = new StringBuffer(sub_str);
 
                    // Drop character from the string
                    sb.deleteCharAt(k);
                    if (!st.contains(sb))
                        ;
                    subsequence(sb.toString());
                }
            }
        }
    }
 
    // Driver code
    public static void main(String[] args)
    {
        String s = "aabc";
        subsequence(s);
        System.out.println(st);
    }
}

# Python program to print all subsequence of a
# given string.
 
# set to store all the subsequences
st = set()
 
# Function computes all the subsequence of an string
def subsequence(str):
 
    # Iterate over the entire string
    for i in range(len(str)):
 
        # Iterate from the end of the string
        # to generate substrings
        for j in range(len(str),i,-1):
            sub_str = str[i: i+j]
            st.add(sub_str)
 
            # Drop kth character in the substring
            # and if its not in the set then recur
            for k in range(1,len(sub_str)):
                sb = sub_str
 
                # Drop character from the string
                sb = sb.replace(sb[k],"")
                subsequence(sb)
 
# Driver Code
 
s = "aabc"
subsequence(s)
for i in st:
    print(i,end = " ")
print()
 
# This code is contributed by shinjanpatra

<script>
 
// JavaScript program to print all subsequence of a
// given string.
 
 
// set to store all the subsequences
let st = new Set();
 
// Function computes all the subsequence of an string
function subsequence(str)
{
 
    // Iterate over the entire string
    for (let i = 0; i < str.length; i++) {
 
        // Iterate from the end of the string
        // to generate substrings
        for (let j = str.length; j > i; j--) {
            let sub_str = str.substr(i, i+j);
            st.add(sub_str);
 
            // Drop kth character in the substring
            // and if its not in the set then recur
            for (let k = 1; k < sub_str.length; k++) {
                let sb = sub_str;
 
                // Drop character from the string
                sb =sb.replace(sb[k],"");
                subsequence(sb);
            }
        }
    }
}
 
// Driver Code
 
let s = "aabc";
subsequence(s);
for (let i of st)
    document.write(i," ");
document.write("</br>");
 
// This code is contributed by shinjanpatra
 
</script>

aab aa aac bc b abc aabc ab ac a c 

Método 3: uno por uno corrige los caracteres y genera recursivamente todos los subconjuntos a partir de ellos. Después de cada llamada recursiva, eliminamos el último carácter para que se pueda generar la siguiente permutación. 

Implementación:

// CPP program to generate power set in
// lexicographic order.
#include <bits/stdc++.h>
using namespace std;
 
// str : Stores input string
// n : Length of str.
// curr : Stores current permutation
// index : Index in current permutation, curr
void printSubSeqRec(string str, int n, int index = -1,
                    string curr = "")
{
    // base case
    if (index == n)
        return;
 
    if (!curr.empty()) {
        cout << curr << "\n";
    }
 
    for (int i = index + 1; i < n; i++) {
 
        curr += str[i];
        printSubSeqRec(str, n, i, curr);
 
        // backtracking
        curr = curr.erase(curr.size() - 1);
    }
    return;
}
 
// Generates power set in lexicographic
// order.
void printSubSeq(string str)
{
    printSubSeqRec(str, str.size());
}
 
// Driver code
int main()
{
    string str = "cab";
    printSubSeq(str);
    return 0;
}

// Java program to generate power set in
// lexicographic order.
class GFG {
 
    // str : Stores input string
    // n : Length of str.
    // curr : Stores current permutation
    // index : Index in current permutation, curr
    static void printSubSeqRec(String str, int n, int index,
                               String curr)
    {
        // base case
        if (index == n) {
            return;
        }
        if (curr != null && !curr.trim().isEmpty()) {
            System.out.println(curr);
        }
        for (int i = index + 1; i < n; i++) {
            curr += str.charAt(i);
            printSubSeqRec(str, n, i, curr);
 
            // backtracking
            curr = curr.substring(0, curr.length() - 1);
        }
    }
 
    // Generates power set in
    // lexicographic order.
    static void printSubSeq(String str)
    {
        int index = -1;
        String curr = "";
 
        printSubSeqRec(str, str.length(), index, curr);
    }
 
    // Driver code
    public static void main(String[] args)
    {
        String str = "cab";
        printSubSeq(str);
    }
}
 
// This code is contributed by PrinciRaj1992

# Python program to generate power set in lexicographic order.
 
 # str: Stores input string
 # n: Length of str.
 # curr: Stores current permutation
 # index: Index in current permutation, curr
def printSubSeqRec(str, n, index = -1, curr = ""):
   
  # base case
     if (index == n):
       return
     if (len(curr) > 0):
       print(curr)
 
     i = index + 1
 
     while(i < n):
        curr = curr + str[i]
        printSubSeqRec(str, n, i, curr)
        curr = curr[0:-1]
        i = i + 1
        
#  Generates power set in lexicographic order.
#  function
def printSubSeq(str):
   printSubSeqRec(str, len(str))
 
# // Driver code
str = "cab"
printSubSeq(str)
 
# This code is contributed by shinjanpatra

// Include namespace system
using System;
 
// C# program to generate power set in
// lexicographic order.
public class GFG
{
    // str : Stores input string
    // n : Length of str.
    // curr : Stores current permutation
    // index : Index in current permutation, curr
    public static void printSubSeqRec(String str, int n, int index, String curr)
    {
        // base case
        if (index == n)
        {
            return;
        }
        if (curr != null && !(curr.Trim().Length == 0))
        {
            Console.WriteLine(curr);
        }
        for (int i = index + 1; i < n; i++)
        {
            curr += str[i];
            GFG.printSubSeqRec(str, n, i, curr);
            // backtracking
            curr = curr.Substring(0,curr.Length - 1-0);
        }
    }
    // Generates power set in
    // lexicographic order.
    public static void printSubSeq(String str)
    {
        var index = -1;
        var curr = "";
        GFG.printSubSeqRec(str, str.Length, index, curr);
    }
    // Driver code
    public static void Main(String[] args)
    {
        var str = "cab";
        GFG.printSubSeq(str);
    }
}
// This code is contributed by mukulsomukesh

<script>
// JavaScript program to generate power set in
// lexicographic order.
  
// str : Stores input string
// n : Length of str.
// curr : Stores current permutation
// index : Index in current permutation, curr
 
function printSubSeqRec(str,n,index = -1,curr = "")
{
    // base case
    if (index == n)
        return;
  
    if (curr.length>0) {
        document.write(curr)
    }
  
    for (let i = index + 1; i < n; i++) {
  
        curr += str[i];
        printSubSeqRec(str, n, i, curr);
  
        // backtracking
        curr = curr.slice(0, - 1);
    }
    return;
}
  
// Generates power set in lexicographic
// order.
function printSubSeq(str)
{
    printSubSeqRec(str, str.length);
}
  
// Driver code
 
let str = "cab";
printSubSeq(str);
</script>

c
ca
cab
cb
a
ab
b

Complejidad de tiempo: O(n * 2 ⁿ ) , donde n es el tamaño de la string dada
Espacio auxiliar: O(1), ya que no se utiliza espacio adicional

Este artículo es una contribución de Sumit Ghosh . Si te gusta GeeksforGeeks y te gustaría contribuir, también puedes escribir un artículo usando write.geeksforgeeks.org o enviar tu artículo por correo a review-team@geeksforgeeks.org. Vea su artículo que aparece en la página principal de GeeksforGeeks y ayude a otros Geeks.