Related Articles

# Print all possible strings that can be made by placing spaces

• Difficulty Level : Medium
• Last Updated : 22 Jul, 2021

Given a string you need to print all possible strings that can be made by placing spaces (zero or one) in between them.

```Input:  str[] = "ABC"
Output: ABC
AB C
A BC
A B C```

The idea is to use recursion and create a buffer that one by one contains all output strings having spaces. We keep updating the buffer in every recursive call. If the length of the given string is ‘n’ our updated string can have a maximum length of n + (n-1) i.e. 2n-1. So we create a buffer size of 2n (one extra character for string termination).
We leave 1st character as it is, starting from the 2nd character, we can either fill a space or a character. Thus, one can write a recursive function like below.
Below is the implementation of the above approach:

## C++

 `// C++ program to print permutations``// of a given string with spaces.``#include ``#include ``using` `namespace` `std;` `/* Function recursively prints``   ``the strings having space pattern.``   ``i and j are indices in 'str[]' and``   ``'buff[]' respectively */``void` `printPatternUtil(``const` `char` `str[],``                      ``char` `buff[], ``int` `i,``                            ``int` `j, ``int` `n)``{``    ``if` `(i == n)``    ``{``        ``buff[j] = ``'\0'``;``        ``cout << buff << endl;``        ``return``;``    ``}` `    ``// Either put the character``    ``buff[j] = str[i];``    ``printPatternUtil(str, buff, i + 1, j + 1, n);` `    ``// Or put a space followed by next character``    ``buff[j] = ``' '``;``    ``buff[j + 1] = str[i];` `    ``printPatternUtil(str, buff, i + 1, j + 2, n);``}` `// This function creates buf[] to``// store individual output string and uses``// printPatternUtil() to print all permutations.``void` `printPattern(``const` `char``* str)``{``    ``int` `n = ``strlen``(str);` `    ``// Buffer to hold the string``    ``// containing spaces``    ``// 2n - 1 characters and 1 string terminator``    ``char` `buf[2 * n];` `    ``// Copy the first character as``    ``// it is, since it will be always``    ``// at first position``    ``buf = str;` `    ``printPatternUtil(str, buf, 1, 1, n);``}` `// Driver program to test above functions``int` `main()``{``    ``const` `char``* str = ``"ABCD"``;``    ``printPattern(str);``    ``return` `0;``}`

## Java

 `// Java program to print permutations``// of a given string with``// spaces``import` `java.io.*;` `class` `Permutation``{``    ` `    ``// Function recursively prints``    ``// the strings having space``    ``// pattern i and j are indices in 'String str' and``    ``// 'buf[]' respectively``    ``static` `void` `printPatternUtil(String str, ``char` `buf[],``                                 ``int` `i, ``int` `j, ``int` `n)``    ``{``        ``if` `(i == n)``        ``{``            ``buf[j] = ``'\0'``;``            ``System.out.println(buf);``            ``return``;``        ``}` `        ``// Either put the character``        ``buf[j] = str.charAt(i);``        ``printPatternUtil(str, buf, i + ``1``,``                               ``j + ``1``, n);` `        ``// Or put a space followed``        ``// by next character``        ``buf[j] = ``' '``;``        ``buf[j + ``1``] = str.charAt(i);` `        ``printPatternUtil(str, buf, i + ``1``,``                              ``j + ``2``, n);``    ``}` `    ``// Function creates buf[] to``    ``// store individual output``    ``// string and uses printPatternUtil()``    ``// to print all``    ``// permutations``    ``static` `void` `printPattern(String str)``    ``{``        ``int` `len = str.length();` `        ``// Buffer to hold the string``        ``// containing spaces``        ``// 2n-1 characters and 1``        ``// string terminator``        ``char``[] buf = ``new` `char``[``2` `* len];` `        ``// Copy the first character as it is, since it will``        ``// be always at first position``        ``buf[``0``] = str.charAt(``0``);``        ``printPatternUtil(str, buf, ``1``, ``1``, len);``    ``}` `    ``// Driver program``    ``public` `static` `void` `main(String[] args)``    ``{``        ``String str = ``"ABCD"``;``        ``printPattern(str);``    ``}``}`

## Python

 `# Python program to print permutations``# of a given string with``# spaces.` `# Utility function``def` `toString(``List``):``    ``s ``=` `""``    ``for` `x ``in` `List``:``        ``if` `x ``=``=` `'&# 092;&# 048;'``:``            ``break``        ``s ``+``=` `x``    ``return` `s` `# Function recursively prints the``# strings having space pattern.``# i and j are indices in 'str[]'``# and 'buff[]' respectively``def` `printPatternUtil(string, buff, i, j, n):``    ` `    ``if` `i ``=``=` `n:``        ``buff[j] ``=` `'&# 092;&# 048;'``        ``print` `toString(buff)``        ``return` `    ``# Either put the character``    ``buff[j] ``=` `string[i]``    ``printPatternUtil(string, buff, i ``+` `1``,``                                 ``j ``+` `1``, n)` `    ``# Or put a space followed by next character``    ``buff[j] ``=` `' '``    ``buff[j ``+` `1``] ``=` `string[i]` `    ``printPatternUtil(string, buff, i ``+` `1``,``                                 ``j ``+` `2``, n)` `# This function creates buf[] to``# store individual output string``# and uses printPatternUtil() to``# print all permutations.``def` `printPattern(string):``    ``n ``=` `len``(string)` `    ``# Buffer to hold the string``    ``# containing spaces``    ` `    ``# 2n - 1 characters and 1 string terminator``    ``buff ``=` `[``0``] ``*` `(``2` `*` `n)` `    ``# Copy the first character as it is,``    ``# since it will be always``    ``# at first position``    ``buff[``0``] ``=` `string[``0``]` `    ``printPatternUtil(string, buff, ``1``, ``1``, n)` `# Driver program``string ``=` `"ABCD"``printPattern(string)` `# This code is contributed by BHAVYA JAIN`

## C#

 `// C# program to print permutations of a``// given string with spaces``using` `System;` `class` `GFG``{` `    ``// Function recursively prints the``    ``// strings having space pattern``    ``// i and j are indices in 'String``    ``// str' and 'buf[]' respectively``    ``static` `void` `printPatternUtil(``string` `str,``                                 ``char``[] buf, ``int` `i,``                                 ``int` `j, ``int` `n)``    ``{``        ``if` `(i == n)``        ``{``            ``buf[j] = ``'\0'``;``            ``Console.WriteLine(buf);``            ``return``;``        ``}` `        ``// Either put the character``        ``buf[j] = str[i];``        ``printPatternUtil(str, buf, i + 1,``                               ``j + 1, n);` `        ``// Or put a space followed by next``        ``// character``        ``buf[j] = ``' '``;``        ``buf[j + 1] = str[i];` `        ``printPatternUtil(str, buf, i + 1,``                               ``j + 2, n);``    ``}` `    ``// Function creates buf[] to store``    ``// individual output string and uses``    ``// printPatternUtil() to print all``    ``// permutations``    ``static` `void` `printPattern(``string` `str)``    ``{``        ``int` `len = str.Length;` `        ``// Buffer to hold the string containing``        ``// spaces 2n-1 characters and 1 string``        ``// terminator``        ``char``[] buf = ``new` `char``[2 * len];` `        ``// Copy the first character as it is,``        ``// since it will be always at first``        ``// position``        ``buf = str;``        ``printPatternUtil(str, buf, 1, 1, len);``    ``}` `    ``// Driver program``    ``public` `static` `void` `Main()``    ``{``        ``string` `str = ``"ABCD"``;``        ``printPattern(str);``    ``}``}` `// This code is contributed by nitin mittal.`

## PHP

 ``

## Javascript

 ``
Output
```ABCD
ABC D
AB CD
AB C D
A BCD
A BC D
A B CD
A B C D```

Time Complexity: Since the number of Gaps is n-1, there are total 2^(n-1) patters each having length ranging from n to 2n-1. Thus overall complexity would be O(n*(2^n)).
Recursive Java Solution:

Steps:

1) Take the first character, and append space up the rest of the string;

2) First character+”space”+Rest of the spaced up string;

2) First character+Rest of the spaced up string;

## Java

 `// Java program for above approach``import` `java.util.*;` `public` `class` `GFG``{``    ``private` `static` `ArrayList``                         ``spaceString(String str)``    ``{` `        ``ArrayList strs = ``new``                           ``ArrayList();` `        ``// Check if str.length() is 1``        ``if` `(str.length() == ``1``)``        ``{``            ``strs.add(str);``            ``return` `strs;``        ``}` `        ``ArrayList strsTemp``            ``= spaceString(str.substring(``1``,``                             ``str.length()));` `        ``// Iterate over strsTemp``        ``for` `(``int` `i = ``0``; i < strsTemp.size(); i++)``        ``{` `            ``strs.add(str.charAt(``0``) +``                            ``strsTemp.get(i));``            ``strs.add(str.charAt(``0``) + ``" "` `+``                             ``strsTemp.get(i));``        ``}` `        ``// Return strs``        ``return` `strs;``    ``}``  ` `    ``// Driver Code``    ``public` `static` `void` `main(String args[])``    ``{``        ``ArrayList patterns``            ``= ``new` `ArrayList();` `        ``// Function Call``        ``patterns = spaceString(``"ABCD"``);` `        ``// Print patterns``        ``for` `(String s : patterns)``        ``{``            ``System.out.println(s);``        ``}``    ``}``}`

## Python3

 `# Python program for above approach``def` `spaceString(``str``):``    ``strs ``=` `[];``    ` `    ``# Check if str.length() is 1``    ``if``(``len``(``str``) ``=``=` `1``):``        ``strs.append(``str``)``        ``return` `strs``    ` `    ``strsTemp``=``spaceString(``str``[``1``:])``    ` `    ``# Iterate over strsTemp``    ``for` `i ``in` `range``(``len``(strsTemp)):``        ``strs.append(``str``[``0``] ``+` `strsTemp[i])``        ``strs.append(``str``[``0``] ``+` `" "` `+` `strsTemp[i])``    ` `    ``# Return strs``    ``return` `strs` `# Driver Code``patterns``=``[]` `# Function Call``patterns ``=` `spaceString(``"ABCD"``)` `# Print patterns``for` `s ``in` `patterns:``    ``print``(s)``    ` `# This code is contributed by rag2127`

## Javascript

 ``
Output
```ABCD
A BCD
AB CD
A B CD
ABC D
A BC D
AB C D
A B C D```