Find the number which occurs odd number of times

In an integer array, except for one number which occurs odd number of times, all other numbers occur even number of times. Find the number.
Example:
Input: 2 3 4 3 1 4 5 1 4 2 5
Output: 4
Algorithm:
Initialize result = 0.
Iterate over the array and XOR result with each element of the input array.
Once iteration over the array is done, print result as the output.
Order of the Algorithm:
Time Complexity: O(n)
Space Complexity: O(1)
Code and Algorithm Visualization:
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[@mamathagollavilli9]

Tnx

[@umangbarbhaya1605]

//java code class Test { public static void main(String args[]) { // x is stored using 32 bit 2's complement form. // Binary representation of -1 is all 1s (111..1) int a[] = {2,3,4,3,1,4,5,1,2,4,5}; int ans=a[0]; for(int i=1;i

[@srinidhiii]

awesome

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Thanks Srinidhi for your kind words :) We would really appreciate if you could spread the word about IDeserve in your college and to your colleagues. Also please check out our website at: www.ideserve.co.in It has features like Algorithm Visualization, Learn Together and many more coming soon. Please check it out and leave us a comment there! Thanks, -Team IDeserve.

[@ashishgupta8123]

but for serching pairs it will take more than o(n)

[@StockDC2]

I was thinking the same thing. If you need to find the same number in the array, wouldn't the time complexity be n^2?

[@stalinsoft]

since xor is commutative, you really no need to find the pair element. you just simply start 0 xor i'th element and take the result and xor with i+1'the element the final result value will be the answer.. this can be done in o(n). In this video he demonstrated picking the same element for the intuition reason.

[@varun-bhandari]

Code Does not work for this input : {2, 7, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2, 7, 4, 4}; Returns '6' which is not even in the array.

[@Horhew]

That is because both 3 and 5 appear an odd number of times in your array.