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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

Source: Amazon Interview Experience | Set 158, Round 1, Q 1. 

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 ā€˜nā€™ 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 <cstring>
#include <iostream>
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[0] = str[0];
 
    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[0] = str[0];
        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




<?php
// PHP program to print permutations
// of a given string with spaces.
 
/* Function recursively prints the strings
having space pattern. i and j are indices
in 'str[]' and 'buff[]' respectively */
function printPatternUtil($str, $buff,
                           $i, $j, $n)
{
    if ($i == $n)
    {
        $buff[$j] = '';
        echo str_replace(', ', '',
                  implode(', ', $buff))."\n";
        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.
function printPattern($str)
{
    $n = strlen($str);
 
    // Buffer to hold the string containing spaces
    // 2n-1 characters and 1 string terminator
    $buf = array_fill(0, 2 * $n, null);
 
    // 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 code
$str = "ABCD";
printPattern($str);
     
// This code is contributed by chandan_jnu
?>

Javascript




<script>
 
      // JavaScript program to print permutations of a
      // given string with spaces
       
      // Function recursively prints the
      // strings having space pattern
      // i and j are indices in 'String
      // str' and 'buf[]' respectively
      function printPatternUtil(str, buf, i, j, n) {
        if (i === n) {
          buf[j] = "\0";
          document.write(buf.join("") + "<br>");
          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
      function printPattern(str) {
        var len = str.length;
 
        // Buffer to hold the string containing
        // spaces 2n-1 characters and 1 string
        // terminator
        var buf = new Array(2 * len);
 
        // Copy the first character as it is,
        // since it will be always at first
        // position
        buf[0] = str[0];
        printPatternUtil(str, buf, 1, 1, len);
      }
 
      // Driver program
      var str = "ABCD";
      printPattern(str);
       
</script>
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<String>
                         spaceString(String str)
    {
 
        ArrayList<String> strs = new
                           ArrayList<String>();
 
        // Check if str.length() is 1
        if (str.length() == 1)
        {
            strs.add(str);
            return strs;
        }
 
        ArrayList<String> 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<String> patterns
            = new ArrayList<String>();
 
        // 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




<script>
// Javascript program for above approach
 
function spaceString(str)
{
    let strs = [];
    // Check if str.length() is 1
        if (str.length == 1)
        {
            strs.push(str);
            return strs;
        }
  
        let strsTemp
            = spaceString(str.substring(1,
                             str.length));
  
        // Iterate over strsTemp
        for (let i = 0; i < strsTemp.length; i++)
        {
  
            strs.push(str[0] +
                            strsTemp[i]);
            strs.push(str[0] + " " +
                             strsTemp[i]);
        }
  
        // Return strs
        return strs;
}
 
// Driver Code
let patterns = spaceString("ABCD");
  
// Print patterns
for (let s of patterns.values())
{
    document.write(s+"<br>");
}
 
// This code is contributed by avanitrachhadiya2155
</script>
Output
ABCD
A BCD
AB CD
A B CD
ABC D
A BC D
AB C D
A B C D

This article is contributed by Gaurav Sharma. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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