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 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; } |
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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); } } |
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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 |
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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. |
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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 ?> |
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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)).
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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