Python Program to Find Element Occurring Odd Number of Times in a List

When it is required to find an element that occurs odd number of times in a list, several approaches can be used. The most common methods include nested loops to count occurrences, using Python's Counter from collections module, or using XOR operations for optimization.

Method 1: Using Nested Loops

This method iterates through the list and counts occurrences of each element using nested loops ?

def odd_occurrence(my_list, list_size):
    for i in range(0, list_size):
        count = 0
        for j in range(0, list_size):
            if my_list[i] == my_list[j]:
                count += 1
        
        if (count % 2 != 0):
            return my_list[i]
    
    return -1

my_list = [34, 56, 78, 99, 23, 34, 34, 56, 78, 99, 99, 99, 99, 34, 34, 56, 56]
print("The list is:")
print(my_list)

n = len(my_list)
print("The length is:", n)

result = odd_occurrence(my_list, n)
print("The element that occurs odd number of times is:", result)

The list is:
[34, 56, 78, 99, 23, 34, 34, 56, 78, 99, 99, 99, 99, 34, 34, 56, 56]
The length is: 17
The element that occurs odd number of times is: 34

Method 2: Using Counter

Python's Counter class provides a more efficient way to count occurrences ?

from collections import Counter

def find_odd_occurrence(my_list):
    count_dict = Counter(my_list)
    
    for element, count in count_dict.items():
        if count % 2 != 0:
            return element
    
    return -1

my_list = [34, 56, 78, 99, 23, 34, 34, 56, 78, 99, 99, 99, 99, 34, 34, 56, 56]
print("The list is:", my_list)

result = find_odd_occurrence(my_list)
print("Element with odd occurrence:", result)

The list is: [34, 56, 78, 99, 23, 34, 34, 56, 78, 99, 99, 99, 99, 34, 34, 56, 56]
Element with odd occurrence: 34

Method 3: Using XOR Operation

For lists where only one element occurs odd times, XOR operation provides an optimal solution ?

def find_odd_using_xor(my_list):
    result = 0
    for element in my_list:
        result ^= element
    return result

# Example with only one element having odd occurrence
my_list = [2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 2]
print("The list is:", my_list)

result = find_odd_using_xor(my_list)
print("Element with odd occurrence:", result)

The list is: [2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 2]
Element with odd occurrence: 5

Comparison

Conclusion

Use the Counter method for general cases with multiple elements having odd occurrences. The XOR method is optimal when only one element occurs odd times, while nested loops provide a straightforward but less efficient approach.