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

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```

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

The idea is to use recursion and create a buffer that one by one contains all output strings having spaces. We keep updating buffer in every recursive call. If the length of given string is ‘n’ our updated string can have maximum length of n + (n-1) i.e. 2n-1. So we create 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.

## C/C++

```// C++ program to print permutations of a given string with spaces.
#include <iostream>
#include <cstring>
using namespace std;

/* Function recursively prints the strings having space pattern.
i and j are indices in 'str[]' and 'buff[]' respectively */
void printPatternUtil(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(char *str)
{
int n = strlen(str);

// Buffer to hold the string containing spaces
char buf[2*n]; // 2n-1 characters and 1 string terminator

// Copy the first character as it is, since it will be always
// at first position
buf[0] = str[0];

printPatternUtil(str, buf, 1, 1, n);
}

// Driver program to test above functions
int main()
{
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 == '\0':
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] = '\0'
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
buff = [0] * (2*n) # 2n-1 characters and 1 string terminator

# 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
```

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

Time Complexity: Since number of Gaps are 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)).

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