# Distinct permutations of the string | Set 2

Print all distinct permutation of a string having duplicates.

Examples:

```Input : ABCA
Output : AABC AACB ABAC ABCA ACBA
ACAB BAAC BACA BCAA CABA
CAAB CBAA
```

An algorithm to print all distinct permutations has already been discussed here. Here we’ll discuss one more approach to do the same. Recall first how we print permutations without any duplicates in the input string. It is given here. Let’s now take the case of the string “ABAC”. While generating permutations, let’s say we are at index = 0, swap it with all elements after it. When we reach at i=2, we see that in the string s[index…i-1], there was an index which is equal to s[i]. Thus, swapping it will produce repeated permutations. Thus, we don’t swap it. The below explains it better.

Illustration: Let us understand with below example.

```i = 0 1 2 3
A B A C
index = 0, s = A
Start swapping s[index] with s[i] following it:
i = index + 1 = 1

Since s[index] != s[i], swap and recur.

i = 2, s[index] == s[i], don't swap

i = 3,  s[index] != s[i], swap and recur.
```

Below code does the same.

## C++

 `// C++ program to distinct permutations of the string` `#include ` `using` `namespace` `std;`   `// Returns true if str[curr] does not matches with any of the` `// characters after str[start]` `bool` `shouldSwap(``char` `str[], ``int` `start, ``int` `curr)` `{` `    ``for` `(``int` `i = start; i < curr; i++) ` `        ``if` `(str[i] == str[curr])` `            ``return` `0;` `    ``return` `1;` `}`   `// Prints all distinct permutations in str[0..n-1]` `void` `findPermutations(``char` `str[], ``int` `index, ``int` `n)` `{` `    ``if` `(index >= n) {` `        ``cout << str << endl;` `        ``return``;` `    ``}`   `    ``for` `(``int` `i = index; i < n; i++) {`   `        ``// Proceed further for str[i] only if it ` `        ``// doesn't match with any of the characters` `        ``// after str[index]` `        ``bool` `check = shouldSwap(str, index, i);` `        ``if` `(check) {` `            ``swap(str[index], str[i]);` `            ``findPermutations(str, index + 1, n);` `            ``swap(str[index], str[i]);` `        ``}` `    ``}` `}`   `// Driver code` `int` `main()` `{` `    ``char` `str[] = ``"ABCA"``;` `    ``int` `n = ``strlen``(str);` `    ``findPermutations(str, 0, n);` `    ``return` `0;` `}`

## Java

 `// Java program to distinct permutations of the string` `public` `class` `GFG {`   `// Returns true if str[curr] does not matches with any of the ` `// characters after str[start] ` `    ``static` `boolean` `shouldSwap(``char` `str[], ``int` `start, ``int` `curr) {` `        ``for` `(``int` `i = start; i < curr; i++) {` `            ``if` `(str[i] == str[curr]) {` `                ``return` `false``;` `            ``}` `        ``}` `        ``return` `true``;` `    ``}`   `// Prints all distinct permutations in str[0..n-1] ` `    ``static` `void` `findPermutations(``char` `str[], ``int` `index, ``int` `n) {` `        ``if` `(index >= n) {` `            ``System.out.println(str);` `            ``return``;` `        ``}`   `        ``for` `(``int` `i = index; i < n; i++) {`   `            ``// Proceed further for str[i] only if it ` `            ``// doesn't match with any of the characters ` `            ``// after str[index] ` `            ``boolean` `check = shouldSwap(str, index, i);` `            ``if` `(check) {` `                ``swap(str, index, i);` `                ``findPermutations(str, index + ``1``, n);` `                ``swap(str, index, i);` `            ``}` `        ``}` `    ``}`   `    ``static` `void` `swap(``char``[] str, ``int` `i, ``int` `j) {` `        ``char` `c = str[i];` `        ``str[i] = str[j];` `        ``str[j] = c;` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `main(String[] args) {`   `        ``char` `str[] = {``'A'``, ``'B'``, ``'C'``, ``'A'``};` `        ``int` `n = str.length;` `        ``findPermutations(str, ``0``, n);` `    ``}`   `}`

## Python3

 `# Python3 program to distinct` `# permutations of the string `   `# Returns true if str[curr] does not ` `# matches with any of the characters` `# after str[start] ` `def` `shouldSwap(string, start, curr): `   `    ``for` `i ``in` `range``(start, curr): ` `        ``if` `string[i] ``=``=` `string[curr]: ` `            ``return` `0` `    ``return` `1`   `# Prints all distinct permutations` `# in str[0..n-1] ` `def` `findPermutations(string, index, n): `   `    ``if` `index >``=` `n: ` `        ``print``(''.join(string)) ` `        ``return`   `    ``for` `i ``in` `range``(index, n): `   `        ``# Proceed further for str[i] only ` `        ``# if it doesn't match with any of ` `        ``# the characters after str[index] ` `        ``check ``=` `shouldSwap(string, index, i) ` `        ``if` `check: ` `            ``string[index], string[i] ``=` `string[i], string[index] ` `            ``findPermutations(string, index ``+` `1``, n) ` `            ``string[index], string[i] ``=` `string[i], string[index] `   `# Driver code ` `if` `__name__ ``=``=` `"__main__"``:`   `    ``string ``=` `list``(``"ABCA"``) ` `    ``n ``=` `len``(string) ` `    ``findPermutations(string, ``0``, n) ` `    `  `# This code is contributed by Rituraj Jain`

## C#

 `// C# program to distinct permutations` `// of the string` `using` `System;`   `class` `GFG` `{ `   `// Returns true if str[curr] does ` `// not matches with any of the ` `// characters after str[start] ` `static` `bool` `shouldSwap(``char` `[]str, ` `                       ``int` `start, ``int` `curr) ` `{ ` `    ``for` `(``int` `i = start; i < curr; i++) ` `    ``{ ` `        ``if` `(str[i] == str[curr]) ` `        ``{ ` `            ``return` `false``; ` `        ``} ` `    ``} ` `    ``return` `true``; ` `} `   `// Prints all distinct permutations ` `// in str[0..n-1] ` `static` `void` `findPermutations(``char` `[]str, ` `                             ``int` `index, ``int` `n) ` `{ ` `    ``if` `(index >= n) ` `    ``{ ` `        ``Console.WriteLine(str); ` `        ``return``; ` `    ``} `   `    ``for` `(``int` `i = index; i < n; i++) ` `    ``{ `   `        ``// Proceed further for str[i] only ` `        ``// if it doesn't match with any of ` `        ``// the characters after str[index] ` `        ``bool` `check = shouldSwap(str, index, i); ` `        ``if` `(check) ` `        ``{ ` `            ``swap(str, index, i); ` `            ``findPermutations(str, index + 1, n); ` `            ``swap(str, index, i); ` `        ``} ` `    ``} ` `} `   `static` `void` `swap(``char``[] str, ``int` `i, ``int` `j)` `{ ` `    ``char` `c = str[i]; ` `    ``str[i] = str[j]; ` `    ``str[j] = c; ` `} `   `// Driver code ` `public` `static` `void` `Main() ` `{ ` `    ``char` `[]str = {``'A'``, ``'B'``, ``'C'``, ``'A'``}; ` `    ``int` `n = str.Length; ` `    ``findPermutations(str, 0, n); ` `} ` `} `   `// This code is contributed ` `// by 29AjayKumar`

Output

```ABCA
ABAC
ACBA
ACAB
AACB
AABC
BACA
BAAC
BCAA
CBAA
CABA
CAAB

```

Better Approach:

Generate all distinct strings is to simply use some if conditions. The technique above uses an extra loop inside the recursion which causes a major time complexity cost. Instead, we can improve it by little pre-processing. We first sort the given string and then apply the below code.

Below is the implementation of the above idea:

## C++

 `// C++ program to print all the permutation` `// of the given string.` `#include ` `#include ` `#include ` `using` `namespace` `std;`   `// count of total permutations` `int` `total = 0;`   `void` `permute(``int` `i, string s)` `{` `    ``// base case` `    ``if` `(i == (s.length() - 1)) {` `        ``cout << s << endl;` `        ``total++;` `        ``return``;` `    ``}` `  `  `    ``char` `prev = ``'*'``;` `  `  `    ``// loop from j = 1 to length of string` `    ``for` `(``int` `j = i; j < s.length(); j++) {` `        ``string temp = s;` `        ``if` `(j > i && temp[i] == temp[j])` `            ``continue``;` `        ``if` `(prev != ``'*'` `&& prev == s[j]) {` `            ``continue``;` `        ``}` `      `  `        ``// swap the elements` `        ``swap(temp[i], temp[j]);` `        ``prev = s[j];` `      `  `        ``// recursion call` `        ``permute(i + 1, temp);` `    ``}` `}`   `// Driver code` `int` `main()` `{` `    ``string s = ``"abca"``;` `    `  `    ``// sort` `    ``sort(s.begin(), s.end());` `  `  `    ``// Function call` `    ``permute(0, s);` `    ``cout << ``"Total distinct permutations = "` `<< total` `         ``<< endl;` `    ``return` `0;` `}`   `// This code is contributed by risingStark.`

Output

```aabc
aacb
abac
abca
acba
acab
baac
baca
bcaa
caba
caab
cbaa
Total distinct permutations = 12
```

Time complexity: If we take the length of string to be N, then the complexity of my code will be O(N log N) for sorting and O(N*N!) for the permutation. Total time complexity would be O(N log N + N*N!) which is effectively only O(N*N!).

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