Given an array of integers, duplicates appear in it even number of times except two elements which appears odd number of times. Find both odd appearing element without using any extra memory.

Expected time complexity: O(n)

For example,

**Input: ** arr = [4, 3, 6, 2, 4, 2, 3, 4, 3, 3]

**Output: **Odd occurring element are 4 and 6

4 appears 3 times

3 appears 4 times

6 appears 1 times

2 appears 2 times

**Related Post:** Find odd occurring element in an array in single traversal

We can use hashing to solve this problem in O(n) time. The idea is to traverse the array and maintain frequency of each element in a hash table. Then after all array elements are processed, we return the elements with odd frequencies. The problem with this approach is that it requires O(n) extra space as well. Also it requires one traversal of array and one traversal of hash table. The advantage of this approach is its simplicity and the fact that it will work for any number of distinct odd elements present in the array.

We can solve this problem in O(1) space by using **XOR operator**. We know that if we XOR a number with itself odd number of times the result is number itself, otherwise if we XOR a number even number of times with itself, the result is 0. Also XOR with 0 is always the number itself.

XOR x with 0

x ^ 0 = x

XOR x with itself even number of times

x ^ x = 0

x ^ x ^ x ^ x = (x ^ x) ^ (x ^ x) = 0 ^ 0 = 0

XOR x with itself odd number of times

(x ^ x ^ x) = (x ^ (x ^ x)) = (x ^ 0) = x

(x ^ x ^ x ^ x ^ x) = (x ^ (x ^ x) ^ (x ^ x)) = (x ^ 0 ^ 0) = x

So, if we take XOR of all elements present in the array, even appearing elements will cancel out each other and we are left with XOR of x and y (x ^ y) where x and y are two odd appearing elements.

**How to find x and y?**

Let res = (x ^ y)

We know that any set bit in *res* will be either set in x or y (but not in both as a bit will only set in *res* when it is set in one number and unset in the other).

The idea is to consider the rightmost set bit in *res* (or any other set bit) and split the array into two sub-arrays –

- All elements that have this bit set.
- All elements that have this bit unset.

As this rightmost bit is set in one number and unset in the other, we will have one odd appearing element in each sub-array. Basically we have isolated trait of one number with other so that both x and y will go to different sub-array.

Now we iterate each sub-array once more, do XOR on each element of the sub-array and the result will be the odd appearing element present in the sub-array (since even appearing elements will cancel each other).

## C++

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#include <iostream> #include <math.h> using namespace std; pair<int, int> findOddOccuring(int arr[], int n) { int res = 0; // take XOR of all array elements for (int i = 0; i < n; i++) res = res ^ arr[i]; // find position of the rightmost set bit in res int k = log2(res & -res); // x and y are two odd appearing elements int x = 0, y = 0; // split the array into two sub-arrays for (int i = 0; i < n; i++) { // elements that have k'th bit 1 if (arr[i] & (1 << k)) x = x ^ arr[i]; // elements that have k'th bit 0 else y = y ^ arr[i]; } return make_pair(x, y); } // main function int main() { int arr[] = { 4, 3, 6, 2, 4, 2, 3, 4, 3, 3 }; int n = sizeof(arr) / sizeof(arr[0]); pair<int, int> p = findOddOccuring(arr, n); cout << "Odd occurring element are " << p.first << " and " << p.second; return 0; } |

`Output:`

Odd occurring element are 6 and 4

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## Leave a Reply

Umm slight error in the comments – code still works but

k= log2(X&-X)gives you the k offset of the most significant bit (the left most), not the least significant bit (or right most).what if I am finding it inside {1, 2, 3, 4, 5, 6, 7}

they will cancel each other

You can find lowest set bit with below

int lsb = x & (~x + 1);

`x = 1010000 (2)`

~x = 0101111 (2)

(~x+1) = 0110000 (2)

x & (~x+1) = 0010000 (2)