Print first n distinct permutations of string using itertools in Python

When working with permutations of strings that contain duplicate characters, we often need to find only the distinct permutations. Python's itertools.permutations() generates all possible arrangements, including duplicates, so we need additional logic to filter unique results.

Syntax

from itertools import permutations

def get_distinct_permutations(string, n):
    # Sort characters to group duplicates together
    sorted_chars = sorted(list(string))
    
    # Generate all permutations
    all_perms = permutations(sorted_chars)
    
    # Use set to store unique permutations
    unique_perms = set()
    count = 0
    
    # Iterate until we find n distinct permutations
    for perm in all_perms:
        if count >= n:
            break
        perm_str = ''.join(perm)
        if perm_str not in unique_perms:
            unique_perms.add(perm_str)
            print(perm_str)
            count += 1

Example

Let's find the first 8 distinct permutations of a string with duplicate characters ?

from itertools import permutations

def get_distinct_permutations(string, n):
    sorted_chars = sorted(list(string))
    all_perms = permutations(sorted_chars)
    
    unique_perms = set()
    count = 0
    
    for perm in all_perms:
        if count >= n:
            break
        perm_str = ''.join(perm)
        if perm_str not in unique_perms:
            unique_perms.add(perm_str)
            print(perm_str)
            count += 1

# Test with string containing duplicates
string = "xyxxz"
n = 8
print(f"First {n} distinct permutations of '{string}':")
get_distinct_permutations(string, n)

First 8 distinct permutations of 'xyxxz':
xxxyz
xxxzy
xxyxz
xxyzx
xxzxy
xxzyx
xyxxz
xyxzx

How It Works

The algorithm follows these key steps:

• Sorting: sorted(list(string)) arranges characters alphabetically, grouping duplicates together
• Permutation Generation: permutations() creates all possible arrangements
• Duplicate Filtering: A set() tracks already seen permutations
• Counting: Stops after finding the required number of distinct permutations

Alternative Approach Using List

Instead of printing directly, you can collect the results in a list ?

from itertools import permutations

def get_distinct_permutations_list(string, n):
    sorted_chars = sorted(list(string))
    all_perms = permutations(sorted_chars)
    
    unique_perms = set()
    result = []
    
    for perm in all_perms:
        if len(result) >= n:
            break
        perm_str = ''.join(perm)
        if perm_str not in unique_perms:
            unique_perms.add(perm_str)
            result.append(perm_str)
    
    return result

# Example usage
string = "aab"
n = 3
distinct_perms = get_distinct_permutations_list(string, n)
print(f"Distinct permutations: {distinct_perms}")

Distinct permutations: ['aab', 'aba', 'baa']

Performance Comparison

Conclusion

Use itertools.permutations() with a set to filter distinct permutations efficiently. Sort the input string first to improve performance, and use a counter to limit results to the first n distinct permutations.