Home Practice Programming Print all permutations of a string with unique characters

# Print all permutations of a string with unique characters

A permutation is an act of rearranging or reordering elements of a set or string etc.
For n elements, n! (n factorial) permutations are possible.
Ex-> Possible permutations of abc are abc, acb, bac, bca, cab, cba.

Here, given a string with n elements, we have to generate all possible permutation of this string.

#### Algorithm

1. Start with the original string str and call the function find_permuation() with parameters original string, start index(0) and end index(str.size()-1).
2. Iterate through every element of the string and perform the following operation i.e, for(i in range start to end-1).
3. For every ith element of the string swap it with the 0th element, i.e, swap(str[start],str[i])
4. Fix the current element, and again call the find_permuatation() function with parameters: the updated string, start index = current start index + 1 and original end index i.e, find_permuatation(str, start+1, end);
5. After the current recursion terminates, re-swap the ith index with the 0th index i.e., swap(str[i],str[start])
6. If start ==  end, we have reached the end of the recursion and found the permutation, print it and return
7. Repeat step 2 to step 6 for every string element.

```find_permutation(str, start, end){
if(start == end)
print(str)
for i in range [start to end){
swap(str[start], str[i]);
//we fixed the ith index and now in next recursion
//we will work with the remaining string elements
find_permutation(str, start+1, end);
swap(str[i], str[start]);
}
return 0;
}

```

#### Implementation of the above Algorithm in CPP

```#include <bits/stdc++.h>
using namespace std;

int find_permutation(string str, int start, int end){

if(start==end)
cout<<str<<endl;

for(int i=start; i<end; i++){
swap(str[start],str[i]);
find_permutation(str, start+1,end);
swap(str[start],str[i]);
}
return 0;
}

int main()
{
string str = "abc";
find_permutation(str, 0, str.size());

return 0;
}

```

```abc
acb
bac
bca
cba
cab```

#### Time Complexity

The time complexity of the above algorithm is O(N*N!) where N is the size of the string.