# All possible strings of any length that can be formed from a given string

Given a string of distinct characters, print all possible strings of any length that can be formed from given string characters.

Examples:

Input: abc Output: a b c abc ab ac bc bac bca cb ca ba cab cba acb Input: abcd Output: a b ab ba c ac ca bc cb abc acb bac bca cab cba d ad da bd db abd adb bad bda dab dba cd dc acd adc cad cda dac dca bcd bdc cbd cdb dbc dcb abcd abdc acbd acdb adbc adcb bacd badc bcad bcda bdac bdca cabd cadb cbad cbda cdab cdba dabc dacb dbac dbca dcab dcba

The generation of all strings include the following steps.

1) Generate all subsequences of given string.

2) For every subsequence ‘subs’, print all permutations of ‘subs’

Below is C++ implementation. It uses next_permutation function in C++.

## C/C++

`/* C++ code to generate all possible strings ` ` ` `that can be formed from given string */` `#include<bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `void` `printAll(string str) ` `{ ` ` ` `/* Number of subsequences is (2**n -1)*/` ` ` `int` `n = str.length(); ` ` ` `unsigned ` `int` `opsize = ` `pow` `(2, n); ` ` ` ` ` `/* Generate all subsequences of a given string. ` ` ` `using counter 000..1 to 111..1*/` ` ` `for` `(` `int` `counter = 1; counter < opsize; counter++) ` ` ` `{ ` ` ` `string subs = ` `""` `; ` ` ` `for` `(` `int` `j = 0; j < n; j++) ` ` ` `{ ` ` ` `/* Check if jth bit in the counter is set ` ` ` `If set then print jth element from arr[] */` ` ` `if` `(counter & (1<<j)) ` ` ` `subs.push_back(str[j]); ` ` ` `} ` ` ` ` ` `/* Print all permutations of current subsequence */` ` ` `do` ` ` `{ ` ` ` `cout << subs << ` `" "` `; ` ` ` `} ` ` ` `while` `(next_permutation(subs.begin(), subs.end())); ` ` ` `} ` `} ` ` ` `// Driver program ` `int` `main() ` `{ ` ` ` `string str = ` `"abc"` `; ` ` ` `printSubsequences(str); ` ` ` `return` `0; ` `} ` |

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## Python3

`# Python3 code to generate all possible strings ` `# that can be formed from given string ` `from` `itertools ` `import` `permutations ` ` ` `def` `printAll( st): ` ` ` ` ` `# Number of subsequences is (2**n -1) ` ` ` `n ` `=` `len` `(st) ` ` ` `opsize ` `=` `pow` `(` `2` `, n) ` ` ` ` ` `# Generate all subsequences of a given string. ` ` ` `# using counter 000..1 to 111..1 ` ` ` `for` `counter ` `in` `range` `(` `1` `, opsize): ` ` ` ` ` `subs ` `=` `"" ` ` ` `for` `j ` `in` `range` `(n): ` ` ` ` ` `# Check if jth bit in the counter is set ` ` ` `# If set then print jth element from arr[] ` ` ` `if` `(counter & (` `1` `<<j)): ` ` ` `subs ` `+` `=` `(st[j]) ` ` ` ` ` `# Print all permutations of current subsequence ` ` ` `perm ` `=` `permutations(subs) ` ` ` ` ` `for` `i ` `in` `perm: ` ` ` `print` `(''.join(i),end` `=` `" "` `) ` ` ` `# Driver program ` `if` `__name__ ` `=` `=` `"__main__"` `: ` ` ` ` ` `st ` `=` `"abc"` ` ` `printAll((st)) ` ` ` `# This code is contributed by chitranayal ` |

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Output:

a b ab ba c ac ca bc cb abc acb bac bca cab cba

This article is contributed by **Hardik Gaur**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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