C/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 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++

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// C++ program to find the element
// occurring odd number of times
#include <bits/stdc++.h>
using namespace std;
  
// Funtion to find the element
// occurring odd number of times
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
int main()
{
    int arr[] = { 2, 3, 5, 4, 5, 2,
                  4, 3, 5, 2, 4, 4, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    // Function calling
    cout << getOddOccurrence(arr, n);
  
    return 0;
}

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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.

The Best Solution is to do bitwise XOR of all the elements. XOR of all elements gives us odd occurring element. Please note that XOR of two elements is 0 if both elements are same and XOR of a number x with 0 is x.

Below is the implementation of the above approach.

C++

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// C++ program to find the element
// occurring odd number of times
#include <bits/stdc++.h>
using namespace std;
  
// Function to find element occurring
// odd number of times
int getOddOccurrence(int ar[], int ar_size)
{
    int res = 0;
    for (int i = 0; i < ar_size; i++)
        res = res ^ ar[i];
  
    return res;
}
  
/* Diver function to test above function */
int main()
{
    int ar[] = { 2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2 };
    int n = sizeof(ar) / sizeof(ar[0]);
  
    // Function calling
    cout << getOddOccurrence(ar, n);
  
    return 0;
}

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Output:

5

C

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// C program to find the element
// occurring odd number of times
#include <stdio.h>
  
// Function to find element occurring
// odd number of times
int getOddOccurrence(int ar[], int ar_size)
{
    int res = 0;
    for (int i = 0; i < ar_size; i++)
        res = res ^ ar[i];
  
    return res;
}
  
/* Diver function to test above function */
int main()
{
    int ar[] = { 2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2 };
    int n = sizeof(ar) / sizeof(ar[0]);
  
    // Function calling
    printf("%d", getOddOccurrence(ar, n));
    return 0;
}

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Output:

5

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



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