# Distinct permutations of a string containing duplicates using HashSet in Java

Given a string **str** that may contain duplicate characters, the task is to print all the distinct permutations of the given string such that no permutation is repeated in the output.

**Examples:**

Input:str = “ABA”Output:

ABA

AAB

BAA

Input:str = “ABC”Output:

ABC

ACB

BAC

BCA

CBA

CAB

**Approach:** An approach to generate all the permutations of a given string has been discussed in this article. All the permutations generated by this approach can be stored in a HashSet in order to avoid duplicates.

Below is the implementation of the above approach:

`// Java implementation of the approach` `import` `java.util.HashSet;` ` ` `public` `class` `GFG {` ` ` ` ` `// To store all the generated permutations` ` ` `public` `static` `HashSet<String> h = ` `new` `HashSet<String>();` ` ` ` ` `public` `static` `void` `permute(` `char` `s[], ` `int` `i, ` `int` `n)` ` ` `{` ` ` ` ` `// If the permutation is complete` ` ` `if` `(i == n) {` ` ` ` ` `// If set doesn't contain` ` ` `// the permutation already` ` ` `if` `(!(h.contains(String.copyValueOf(s)))) {` ` ` ` ` `h.add(String.copyValueOf(s));` ` ` ` ` `// Print the generated permutation` ` ` `System.out.println(s);` ` ` `}` ` ` `}` ` ` ` ` `else` `{` ` ` ` ` `// One by one swap the jth` ` ` `// character with the ith` ` ` `for` `(` `int` `j = i; j <= n; j++) {` ` ` ` ` `// Swapping a[i] and a[j];` ` ` `char` `temp = s[i];` ` ` `s[i] = s[j];` ` ` `s[j] = temp;` ` ` ` ` `// Revert the swapping` ` ` `permute(s, i + ` `1` `, n);` ` ` ` ` `temp = s[i];` ` ` `s[i] = s[j];` ` ` `s[j] = temp;` ` ` `}` ` ` `}` ` ` `}` ` ` ` ` `// Driver code` ` ` `public` `static` `void` `main(String args[])` ` ` `{` ` ` `char` `s[] = { ` `'A'` `, ` `'B'` `, ` `'A'` `};` ` ` `permute(s, ` `0` `, s.length - ` `1` `);` ` ` `}` `}` |

**Output:**

ABA AAB BAA

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