Dada una string como entrada. Necesitamos escribir un programa que imprima todas las substrings no vacías de esa string dada.
Ejemplos:
Input : abcd
Output : a
b
c
d
ab
bc
cd
abc
bcd
abcd
Podemos ejecutar tres bucles anidados, el bucle más externo elige un carácter inicial, el bucle medio considera todos los caracteres a la derecha del carácter seleccionado como el carácter final de la substring. El bucle más interno imprime caracteres desde el punto de inicio seleccionado actualmente hasta el punto final seleccionado.
Implementación:
C++
// C++ program to print all possible
// substrings of a given string
#include<bits/stdc++.h>
using namespace std;
// Function to print all sub strings
void subString(char str[], int n)
{
// Pick starting point
for (int len = 1; len <= n; len++)
{
// Pick ending point
for (int i = 0; i <= n - len; i++)
{
// Print characters from current
// starting point to current ending
// point.
int j = i + len - 1;
for (int k = i; k <= j; k++)
cout << str[k];
cout << endl;
}
}
}
// Driver program to test above function
int main()
{
char str[] = "abc";
subString(str, strlen(str));
return 0;
}
Java
//Java program to print all possible
// substrings of a given string
class GFG {
// Function to print all sub strings
static void subString(char str[], int n) {
// Pick starting point
for (int len = 1; len <= n; len++) {
// Pick ending point
for (int i = 0; i <= n - len; i++) {
// Print characters from current
// starting point to current ending
// point.
int j = i + len - 1;
for (int k = i; k <= j; k++) {
System.out.print(str[k]);
}
System.out.println();
}
}
}
// Driver program to test above function
public static void main(String[] args) {
char str[] = {'a', 'b', 'c'};
subString(str, str.length);
}
}
// This code is contributed by PrinciRaj1992
Python
# Python3 program to print all possible # substrings of a given string # Function to print all sub strings def subString(Str,n): # Pick starting point for Len in range(1,n + 1): # Pick ending point for i in range(n - Len + 1): # Print characters from current # starting point to current ending # point. j = i + Len - 1 for k in range(i,j + 1): print(Str[k],end="") print() # Driver program to test above function Str = "abc" subString(Str,len(Str)) # This code is contributed by mohit kumar
C#
// C# program to print all possible
// substrings of a given string
using System;
public class GFG {
// Function to print all sub
// strings
static void subString(string str,
int n)
{
// Pick starting point
for (int len = 1; len <= n;
len++)
{
// Pick ending point
for (int i = 0;
i <= n - len; i++)
{
// Print characters
// from current
// starting point to
// current ending
// point.
int j = i + len - 1;
for (int k = i; k <= j;
k++)
Console.Write(str[k]);
Console.WriteLine();
}
}
}
// Driver program to test
// above function
static public void Main ()
{
string str = "abc";
subString(str, str.Length);
}
}
// This code is contributed by anuj_67.
PHP
<?php
// PHP program to print all possible
// substrings of a given string
// Function to print all sub strings
function subString($str, $n)
{
// Pick starting point
for ($len = 1; $len <= $n; $len++)
{
// Pick ending point
for ($i = 0; $i <= $n - $len; $i++)
{
// Print characters from current
// starting point to current ending
// point.
$j = $i + $len - 1;
for ($k = $i; $k <= $j; $k++)
echo $str[$k];
echo "\n";
}
}
}
// Driver Code
$str = "abc";
subString($str, strlen($str));
// This code is contributed by nitin mittal.
?>
Javascript
<script>
//Javascript program to print all possible
// substrings of a given string
// Function to print all sub strings
function subString(str,n)
{
// Pick starting point
for (let len = 1; len <= n; len++) {
// Pick ending point
for (let i = 0; i <= n - len; i++) {
// Print characters from current
// starting point to current ending
// point.
let j = i + len - 1;
for (let k = i; k <= j; k++) {
document.write(str[k]);
}
document.write("<br>");
}
}
}
// Driver program to test above function
let str=['a', 'b', 'c'];
subString(str, str.length);
// This code is contributed by patel2127
</script>
Producción
a b c ab bc abc
Complejidad temporal: O( n 3 )
Espacio auxiliar: O(1)
Método 2 (usando la función substr()): s.substr(i, len) imprime una substring de longitud ‘len’ a partir del índice i en la string s.
Implementación:
C++
// C++ program to print all possible
// substrings of a given string
#include<bits/stdc++.h>
using namespace std;
// Function to print all sub strings
void subString(string s, int n)
{
// Pick starting point in outer loop
// and lengths of different strings for
// a given starting point
for (int i = 0; i < n; i++)
for (int len = 1; len <= n - i; len++)
cout << s.substr(i, len) << endl;
}
// Driver program to test above function
int main()
{
string s = "abcd";
subString(s,s.length());
return 0;
}
Java
// Java program to print all substrings of a string
public class GFG {
// Function to print all substring
public static void SubString(String str, int n)
{
for (int i = 0; i < n; i++)
for (int j = i+1; j <= n; j++)
// Please refer below article for details
// of substr in Java
// https://www.geeksforgeeks.org/java-lang-string-substring-java/
System.out.println(str.substring(i, j));
}
public static void main(String[] args)
{
String str = "abcd";
SubString(str, str.length());
}
}
// This code is contributed by ASHISH KUMAR PATEL
Python3
# Python program to print all possible # substrings of a given string # Function to print all sub strings def subString(s, n): # Pick starting point in outer loop # and lengths of different strings for # a given starting point for i in range(n): for len in range(i+1,n+1): print(s[i: len]); # Driver program to test above function s = "abcd"; subString(s,len(s)); # This code is contributed by princiraj1992
C#
// C# program to print all substrings of a string
using System;
public class GFG {
// Function to print all substring
public static void SubString(String str, int n)
{
for (int i = 0; i < n; i++)
for (int j = 1; j <= n - i; j++)
// Please refer below article for details
// of substr in Java
// https://www.geeksforgeeks.org/java-lang-string-substring-java/
Console.WriteLine(str.Substring(i, j));
}
public static void Main()
{
String str = "abcd";
SubString(str, str.Length);
}
}
/*This code is contributed by PrinciRaj1992*/
Javascript
<script>
// javascript program to print all substrings of a string
// Function to print all substring
function SubString( str , n)
{
for (var i = 0; i < n; i++)
for (var j = i+1; j <= n; j++)
// Please refer below article for details
// of substr in Java
// https://www.geeksforgeeks.org/java-lang-string-substring-java/
document.write(str.substring(i, j)+"<br/>");
}
var str = "abcd";
SubString(str, str.length);
// This code is contributed by gauravrajput1
</script>
Producción
a ab abc abcd b bc bcd c cd d
Complejidad temporal: O( n 3 )
Espacio auxiliar: O(1)
Este método es aportado por Ravi Shankar Rai
Método 3 (Generar una substring usando la substring anterior):
Implementación:
C++
/*
* C++ program to print all possible
* substrings of a given string
* without checking for duplication.
*/
#include<bits/stdc++.h>
using namespace std;
/*
* Function to print all (n * (n + 1)) / 2
* substrings of a given string s of length n.
*/
void printAllSubstrings(string s, int n)
{
/*
* Fix start index in outer loop.
* Reveal new character in inner loop till end of string.
* Print till-now-formed string.
*/
for (int i = 0; i < n; i++)
{
char temp[n - i + 1];
int tempindex = 0;
for (int j = i; j < n; j++)
{
temp[tempindex++] = s[j];
temp[tempindex] = '\0';
printf("%s\n", temp);
}
}
}
// Driver program to test above function
int main()
{
string s = "Geeky";
printAllSubstrings(s, s.length());
return 0;
}
Java
// Java program to print all possible
// subStrings of a given String
// without checking for duplication.
import java.io.*;
class GFG{
// Function to print all (n * (n + 1)) / 2
// subStrings of a given String s of length n.
public static void printAllSubStrings(String s,
int n)
{
// Fix start index in outer loop.
// Reveal new character in inner
// loop till end of String.
// Print till-now-formed String.
for(int i = 0; i < n; i++)
{
char[] temp = new char[n - i + 1];
int tempindex = 0;
for(int j = i; j < n; j++)
{
temp[tempindex++] = s.charAt(j);
temp[tempindex] = '\0';
System.out.println(temp);
}
}
}
// Driver code
public static void main(String[] args)
{
String s = "Geeky";
printAllSubStrings(s, s.length());
}
}
// This code is contributed by avanitrachhadiya2155
Python3
''' * Python3 program to print all possible * substrings of a given string * without checking for duplication. ''' ''' * Function to print all (n * (n + 1)) / 2 * substrings of a given string s of length n. ''' def printAllSubstrings(s, n): # Fix start index in outer loop. # Reveal new character in inner loop till end of string. # Print till-now-formed string. for i in range(n): temp="" for j in range(i,n): temp+=s[j] print(temp) # Driver program to test above function s = "Geeky" printAllSubstrings(s, len(s)) # This code is contributed by shubhamsingh10
C#
// C# program to print all possible
// subStrings of a given String
// without checking for duplication.
using System;
class GFG{
// Function to print all (n * (n + 1)) / 2
// subStrings of a given String s of length n.
public static void printAllSubStrings(String s, int n)
{
// Fix start index in outer loop.
// Reveal new character in inner
// loop till end of String.
// Print till-now-formed String.
for(int i = 0; i < n; i++)
{
char[] temp = new char[n - i + 1];
int tempindex = 0;
for(int j = i; j < n; j++)
{
temp[tempindex++] = s[j];
temp[tempindex] = '\0';
Console.WriteLine(temp);
}
}
}
// Driver code
public static void Main()
{
String s = "Geeky";
printAllSubStrings(s, s.Length);
}
}
// This code is contributed by Shubhamsingh10
Javascript
<script>
// Javascript program to print all possible
// subStrings of a given String
// without checking for duplication.
// Function to print all (n * (n + 1)) / 2
// subStrings of a given String s of length n.
function printAllSubStrings(s, n)
{
// Fix start index in outer loop.
// Reveal new character in inner
// loop till end of String.
// Print till-now-formed String.
for(let i = 0; i < n; i++)
{
let temp = new Array(n - i + 1);
let tempindex = 0;
for(let j = i; j < n; j++)
{
temp[tempindex++] = s[j];
temp[tempindex] = '\0';
document.write(temp.join("") + "</br>");
}
}
}
let s = "Geeky";
printAllSubStrings(s, s.length);
</script>
Producción
G Ge Gee Geek Geeky e ee eek eeky e ek eky k ky y
Complejidad temporal: O( n 2 )
Espacio auxiliar: O(n)
Método 4 (usando tres bucles anidados):
Implementación:
C++
// CPP program for the above approach
#include <iostream>
using namespace std;
void printSubstrings(string str)
{
// finding the length of the string
int n = str.length();
// outermost for loop
// this is for the selection
// of starting point
for (int i = 0; i < n; i++) {
// 2nd for loop is for selection
// of ending point
for (int j = i; j < n; j++) {
// 3rd loop is for printing from
// starting point to ending point
for (int k = i; k <= j; k++) {
cout << str[k];
}
// changing the line after printing
// from starting point to ending point
cout << endl;
}
}
}
// Driver Code
int main()
{
string str = "abcd";
printSubstrings(str);
return 0;
}
C
// C program for the above approach
#include <stdio.h>
void printSubstrings(char str[])
{
// outermost for loop
// this is for the selection
// of starting point
for (int start = 0; str[start] != '\0'; start++) {
// 2nd for loop is for selection
// of ending point
for (int end = start; str[end] != '\0'; end++) {
// 3rd loop is for printing from
// starting point to ending point
for (int i = start; i <= end; i++) {
printf("%c", str[i]);
}
// changing the line after printing
// from starting point to ending point
printf("\n");
}
}
}
// Driver Code
int main()
{
// code
char str[] = { 'a', 'b', 'c', 'd', '\0' };
// calling the method to print the substring
printSubstrings(str);
return 0;
}
Java
// Java program for the above approach
import java.io.*;
class GFG {
public static void printSubstrings(String str)
{
// finding the length of the string
int n = str.length();
// outermost for loop
// this is for the selection
// of starting point
for (int i = 0; i < n; i++) {
// 2nd for loop is for selection
// of ending point
for (int j = i; j < n; j++) {
// 3rd loop is for printing from
// starting point to ending point
for (int k = i; k <= j; k++) {
System.out.print(str.charAt(k));
}
// changing the line after printing
// from starting point to ending point
System.out.println();
}
}
}
// Driver Code
public static void main(String[] args)
{
String str = "abcd";
// calling method for printing substring
printSubstrings(str);
}
}
Python3
# Python program for the above approach def printSubstrings(string, n): # this is for the selection # of starting point for i in range(n): # 2nd for loop is for selection # of ending point for j in range(i, n): # 3rd loop is for printing from # starting point to ending point for k in range(i, (j + 1)): print(string[k], end="") # changing the line after printing # from starting point to ending point print() # Driver Code string = "abcd" # calling the method to print the substring printSubstrings(string, len(string))
C#
// C# program for the above approach
using System;
public class GFG {
public static void printSubstrings(String str)
{
// finding the length of the string
int n = str.Length;
// outermost for loop
// this is for the selection
// of starting point
for (int i = 0; i < n; i++) {
// 2nd for loop is for selection
// of ending point
for (int j = i; j < n; j++) {
// 3rd loop is for printing from
// starting point to ending point
for (int k = i; k <= j; k++) {
Console.Write(str[k]);
}
// changing the line after printing
// from starting point to ending point
Console.WriteLine();
}
}
}
// Driver Code
public static void Main(String[] args)
{
String str = "abcd";
// calling method for printing substring
printSubstrings(str);
}
}
// This code is contributed by gauravrajput1
Javascript
<script>
// JavaScript program for the above approach
function printSubstrings(str)
{
// finding the length of the string
var n = str.length;
// outermost for loop
// this is for the selection
// of starting point
for (var i = 0; i < n; i++) {
// 2nd for loop is for selection
// of ending point
for (var j = i; j < n; j++) {
// 3rd loop is for printing from
// starting point to ending point
for (var k = i; k <= j; k++) {
document.write(str.charAt(k));
}
// changing the line after printing
// from starting point to ending point
document.write("<br>");
}
}
}
// Driver Code
var str = "abcd";
// calling method for printing substring
printSubstrings(str);
// This code is contributed by shivanisinghss2110
</script>
Producción
a ab abc abcd b bc bcd c cd d
Complejidad de tiempo: O(N 3 ), donde N es la longitud de la string de entrada
Espacio auxiliar: O(1)
Método 5 (usando dos bucles anidados):
Implementación:
C++
// CPP program for the above approach
#include <iostream>
using namespace std;
void printSubstrings(string str)
{
// First loop for starting index
for (int i = 0; i < str.length(); i++) {
string subStr;
// Second loop is generating sub-string
for (int j = i; j < str.length(); j++) {
subStr += str[j];
cout << subStr << endl;
}
}
}
// Driver Code
int main()
{
string str = "abcd";
printSubstrings(str);
return 0;
// this code is contributed by defcdr
}
Java
// JAVA program for the above approach
import java.util.*;
class GFG{
static void printSubStrings(String str)
{
// First loop for starting index
for (int i = 0; i < str.length(); i++) {
String subStr="";
// Second loop is generating sub-String
for (int j = i; j < str.length(); j++) {
subStr += str.charAt(j);
System.out.print(subStr +"\n");
}
}
}
// Driver Code
public static void main(String[] args)
{
String str = "abcd";
printSubStrings(str);
}
}
// This code is contributed by gauravrajput1
Python3
# Python program for the above approach def printSubStrings(str): # First loop for starting index for i in range(len(str)): subStr = ""; # Second loop is generating sub-String for j in range(i,len(str)): subStr += str[j]; print(subStr + ""); # Driver Code if __name__ == '__main__': str = "abcd"; printSubStrings(str); # This code is contributed by umadevi9616
C#
// C# program for the above approach
using System;
public class GFG{
static void printSubStrings(String str)
{
// First loop for starting index
for (int i = 0; i < str.Length; i++) {
String subStr="";
// Second loop is generating sub-String
for (int j = i; j < str.Length; j++) {
subStr += str[j];
Console.Write(subStr +"\n");
}
}
}
// Driver Code
public static void Main(String[] args)
{
String str = "abcd";
printSubStrings(str);
}
}
// This code is contributed by gauravrajput1
Javascript
<script>
// javascript program for the above approach
function printSubStrings( str) {
// First loop for starting index
for (i = 0; i < str.length; i++) {
var subStr = "";
// Second loop is generating sub-String
for (var j = i; j < str.length; j++) {
subStr += str.charAt(j);
document.write(subStr + "<br/>");
}
}
}
// Driver Code
var str = "abcd";
printSubStrings(str);
// This code is contributed by gauravrajput1
</script>
Producción
a ab abc abcd b bc bcd c cd d
Complejidad de tiempo: O(N 2 ) , donde N es la longitud de la string de entrada.
Espacio auxiliar : O(N), donde N es la longitud de la string de entrada.
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA