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

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

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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++ 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 = "ABCDE";     printPattern(str);     return 0; }

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