A permutation also called an “arrangement number” or “order,” is a rearrangement of the elements of an ordered list S into a one-to-one correspondence with S itself. A string of length n has n! permutation.
Source: Mathword(http://mathworld.wolfram.com/Permutation.html)
Below are the permutations of string ABC.
ABC ACB BAC BCA CBA CAB
Here is a solution that is used as a basis in backtracking.
C++
// C++ program to print all permutations // with duplicates allowed #include <bits/stdc++.h> using namespace std; // Function to print permutations // of string // This function takes three parameters: // 1. String // 2. Starting index of the string // 3. Ending index of the string. void permute(string a, int l, int r) { // Base case if (l == r) cout<<a<<endl; else { // Permutations made for (int i = l; i <= r; i++) { // Swapping done swap(a[l], a[i]); // Recursion called permute(a, l+1, r); //backtrack swap(a[l], a[i]); } } } // Driver Code int main() { string str = "ABC"; int n = str.size(); permute(str, 0, n-1); return 0; } // This is code is contributed by rathbhupendra |
Output:
ABC ACB BAC BCA CBA CAB
Algorithm Paradigm: Backtracking
Time Complexity: O(n*n!) Note that there are n! permutations and it requires O(n) time to print a permutation.
Auxiliary Space: O(r – l)
Note: The above solution prints duplicate permutations if there are repeating characters in the input string. Please see the below link for a solution that prints only distinct permutations even if there are duplicates in input.
Print all distinct permutations of a given string with duplicates.
Permutations of a given string using STL
Another approach:
C++
// C++ program to implement // the above approach #include <bits/stdc++.h> #include <string> using namespace std; void permute(string s, string answer) { if(s.length() == 0) { cout << answer << " "; return; } for(int i = 0; i < s.length(); i++) { char ch = s[i]; string left_substr = s.substr(0, i); string right_substr = s.substr(i + 1); string rest = left_substr + right_substr; permute(rest , answer+ch); } } // Driver code int main() { string s; string answer = ""; cout << "Enter the string : "; cin >> s; cout << "All possible strings are : "; permute(s, answer); return 0; } |
Output:
Enter the string : abc All possible strings are : abc acb bac bca cab cba
Time Complexity: O(n*n!) The time complexity is the same as the above approach, i.e. there are n! permutations and it requires O(n) time to print a permutation.
Auxiliary Space: O(|s|)
Please refer complete article on Write a program to print all permutations of a given string for more details!
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