Generate all permutations of a string that follow given constraints

Given a string, generate all permutations of it that do not contain ‘B’ after ‘A’, i.e., the string should not contain “AB” as a substring.

Examples:

Input : str = “ABC”
Output : ACB, BAC, BCA, CBA
Out of 6 permutations of “ABC”, 4 follow the given constraint and 2 (“ABC” and “CAB”) do not follow.



Input : str = “BCD”
Output : BCD, BDC, CDB, CBD, DCB, DBC

A simple solution is to generate all permutations. For every permutation, check if it follows the given constraint.

C++

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// Simple CPP program to print all permutations 
// of a string that follow given constraint
#include <bits/stdc++.h>
using namespace std;
  
void permute(string& str, int l, int r)
{
  
    // Check if current permutation is
    // valid
    if (l == r) {
        if (str.find("AB") == string::npos)
            cout << str << " ";
        return;
    }
  
    // Recursively generate all permutation
    for (int i = l; i <= r; i++) {
        swap(str[l], str[i]);
        permute(str, l + 1, r);
        swap(str[l], str[i]);
    }
}
  
// Driver Code
int main()
{
    string str = "ABC";
    permute(str, 0, str.length() - 1);
    return 0;
}

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Python3

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# Simple Python3 program to print all permutations 
# of a string that follow given constraint
def permute(str, l, r):
      
    # Check if current permutation is
    # valid
    if (l == r):
        if "AB" not in ''.join(str):
            print(''.join(str), end = " ")
        return
          
    # Recursively generate all permutation
    for i in range(l, r + 1):
        str[l], str[i] = str[i], str[l]
        permute(str, l + 1, r)
        str[l], str[i] = str[i], str[l]
          
# Driver Code
str = "ABC"
permute(list(str), 0, len(str) - 1)
  
# This code is contributed by SHUBHAMSINGH10

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Output:

ACB BAC BCA CBA

The above solution first generates all permutations, then for every permutation it checks if it follows given constraint or not.

NewPermutation

An efficient solution is to use Backtracking. We cut down the recursion tree whenever we see that substring “AB” is formed. How do we do this? we add a isSafe() function. Before doing a swap, we check if previous character is ‘A’ and current character is ‘B’.

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// Backtracking based CPP program to print all 
// permutations of a string that follow given 
// constraint
#include <bits/stdc++.h>
using namespace std;
  
bool isSafe(string &str, int l, int i, int r)
{
    // If previous character was 'A' and character
    // is 'B', then do not proceed and cut down
    // the recursion tree. 
    if (l != 0 && str[l-1] == 'A' && str[i] == 'B')
       return false;
  
    // This condition is explicitly required for
    // cases when last two characters are "BA". We
    // do not want them to swapped and become "AB"
    if (r == l+1 && str[i] == 'A' && str[l] == 'B')
      return false;
  
   return true;
}
  
void permute(string& str, int l, int r)
{
    // We reach here only when permutation
    // is valid
    if (l == r) {
        cout << str << " ";
        return;
    }
      
    // Fix all characters one by one
    for (int i = l; i <= r; i++) {
  
      // Fix str[i] only if it is a
      // valid move.
      if (isSafe(str, l, i, r)) {
        swap(str[l], str[i]);
        permute(str, l + 1, r);
        swap(str[l], str[i]);
      }
    }
}
  
// Driver Code
int main()
{
    string str = "ABC";
    permute(str, 0, str.length() - 1);
    return 0;
}

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Output:

ACB BAC BCA CBA


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Improved By : SHUBHAMSINGH10, nidhi_biet