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Suppose we have an array A. In this array there are different numbers that occurs twice. But there is only one number that occurs once. We have to find that element from that array.

Suppose A = [1, 1, 5, 3, 2, 5, 2], then the output will be 3. As there are each number twice, we can perform XOR to cancel out that element. because we know y XOR y = 0

To solve this, we will follow these steps.

Take one variable res = 0

for each element e in array A, preform res := res XOR e

return res

Let us see the following implementation to get better understanding −

class Solution(object): def singleNumber(self, nums): ans = nums[0] for i in range(1,len(nums)): ans ^=nums[i] return ans ob1 = Solution() print(ob1.singleNumber([1,1,5,3,2,5,2]))

[1,1,5,3,2,5,2]

3

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