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Python Program to Find the Number Occurring Odd Number of Times

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Write a Python program for a given array of positive integers. All numbers occur an even number of times except one number which occurs an odd number of times. Find the number in O(n) time & constant space.

Examples :

Input: arr = {1, 2, 3, 2, 3, 1, 3}
Output : 3

Input: arr = {5, 7, 2, 7, 5, 2, 5}
Output: 5

Python Program to Find the Number Occurring Odd Number of Times using Nested Loop:

A Simple Solution is to run two nested loops. The outer loop picks all elements one by one and the inner loop counts the number of occurrences of the element picked by the outer loop.

Below is the implementation of the brute force approach : 

Python3




# Python program to find the element occurring
# odd number of times
     
# function to find the element occurring odd
# number of times
def getOddOccurrence(arr, arr_size):
     
    for i in range(0,arr_size):
        count = 0
        for j in range(0, arr_size):
            if arr[i] == arr[j]:
                count+=1
             
        if (count % 2 != 0):
            return arr[i]
         
    return -1
     
     
# driver code
arr = [2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2 ]
n = len(arr)
print(getOddOccurrence(arr, n))
 
# This code has been contributed by
# Smitha Dinesh Semwal


Output

5

Time Complexity: O(n^2)
Auxiliary Space: O(1)

Python Program to Find the Number Occurring Odd Number of Times using Hashing:

A Better Solution is to use Hashing. Use array elements as a key and their counts as values. Create an empty hash table. One by one traverse the given array elements and store counts.

Below is the implementation of the above approach:

Python3




# Python3 program to find the element 
# occurring odd number of times
  
# function to find the element
# occurring odd number of times
def getOddOccurrence(arr,size):
      
    # Defining HashMap in C++
    Hash=dict()
  
    # Putting all elements into the HashMap
    for i in range(size):
        Hash[arr[i]]=Hash.get(arr[i],0) + 1;
     
    # Iterate through HashMap to check an element
    # occurring odd number of times and return it
    for i in Hash:
 
        if(Hash[i]% 2 != 0):
            return i
    return -1
 
  
# Driver code
arr=[2, 3, 5, 4, 5, 2, 4,3, 5, 2, 4, 4, 2]
n = len(arr)
  
# Function calling
print(getOddOccurrence(arr, n))
 
# This code is contributed by mohit kumar


Output

5

Time Complexity: O(n)
Auxiliary Space: O(n)

Python Program to Find the Number Occurring Odd Number of Times using Bit Manipulation:

The Best Solution is to do bitwise XOR of all the elements. XOR of all elements gives us odd occurring elements. 

Here ^ is the XOR operators;
Note :
x^0 = x
x^y=y^x (Commutative property holds)
(x^y)^z = x^(y^z) (Distributive property holds)
x^x=0

Below is the implementation of the above approach.  

Python3




# Python program to find the element occurring odd number of times
 
def getOddOccurrence(arr):
 
    # Initialize result
    res = 0
     
    # Traverse the array
    for element in arr:
        # XOR with the result
        res = res ^ element
 
    return res
 
# Test array
arr = [ 2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2]
 
print("%d" % getOddOccurrence(arr))


Output

5

Time Complexity: O(n)
Auxiliary Space: O(1)

Please refer complete article on Find the Number Occurring Odd Number of Times for more details!



Last Updated : 24 Oct, 2023
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