Given an array arr[] consisting of positive integers that occur even number of times, except one number, which occurs odd number of times. The task is to find this odd number of times occurring number.
Examples :
Input : arr = {1, 2, 3, 2, 3, 1, 3}
Output : 3
Input : arr = {5, 7, 2, 7, 5, 2, 5}
Output : 5
Naive Approach: A Simple Solution is to run two nested loops. The outer loop picks all elements one by one and inner loop counts the number of occurrences of the element picked by outer loop.
Implementation:
C++
// C++ program to find the element
// occurring odd number of times
#include <bits/stdc++.h>
using namespace std;
// Function 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;
}
C
// C program to find the element
// occurring odd number of times
#include <stdio.h>
// Function 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
printf("%d",getOddOccurrence(arr, n));
return 0;
}
// This code is contributed by rbbansal
5
Time complexity: O(N2)
Auxiliary space: O(1)
Better Approach: 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.
Time complexity: O(N)
Auxiliary space: O(N)
Efficient Approach: 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++
// 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;
}
/* Driver 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;
}
C
// 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;
}
/* Driver 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;
}
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!
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