C# 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 the 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.
C#
// C# program to find the element// occurring odd number of timesusing System;class GFG { // Function to find the element // occurring odd number of times static int getOddOccurrence(int[] arr, int arr_size) { for (int i = 0; i < arr_size; i++) { int count = 0; for (int j = 0; j < arr_size; j++) { if (arr[i] == arr[j]) count++; } if (count % 2 != 0) return arr[i]; } return -1; } // Driver code public static void Main() { int[] arr = { 2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2 }; int n = arr.Length; Console.Write(getOddOccurrence(arr, n)); }}// This code is contributed by Sam007 |
Output:
5
Time Complexity : O(n^2)
Auxiliary Space: O(1)
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 :
C#
// C# program to find the element// occurring odd number of timesusing System;class GFG { // Function to find the element // occurring odd number of times static int getOddOccurrence(int[] arr, int arr_size) { int res = 0; for (int i = 0; i < arr_size; i++) { res = res ^ arr[i]; } return res; } // Driver code public static void Main() { int[] arr = { 2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2 }; int n = arr.Length; Console.Write(getOddOccurrence(arr, n)); }}// This code is contributed by Sam007 |
Output:
5
Please refer complete article on Find the Number Occurring Odd Number of Times for more details!
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