Python Program to Find the Number Occurring Odd Number of Times
Given an array of positive integers. All numbers occur even number of times except one number which occurs 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
Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.
Python3
# Python program to find the element occurring# odd number of times # function to find the element occurring odd# number of timesdef 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
A Better Solution is to use Hashing. Use array elements as key and their counts as value. Create an empty hash table. One by one traverse the given array elements and store counts. Time complexity of this solution is O(n). But it requires extra space for hashing.
Program :
Python
# 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 arrayarr = [ 2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2] print "% d" % getOddOccurrence(arr) |
Output:
5
Please refer complete article on Find the Number Occurring Odd Number of Times for more details!
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