Explain me the given solution that what approach it is and how this approach came in his mind

#1

/*
Time Complexity: O(Q)
Space Complexity: O(Q),

Where ‘Q’ is the number of queries
*/

vector xorQuery(vector<vector> &queries)
{
// Create an empty array ans
vectorans;

// Create an array xorArray of the size of 10^5+1 initialized with 0
vector<int> xorArray(100001, 0);

// Iterate over all the queries
for (int i = 0; i < queries.size(); i++)
{
	if (queries[i][0] == 1)
	{
		ans.push_back(queries[i][1]);
	}
	else
	{
		xorArray[0] ^= queries[i][1];
		xorArray[ans.size()] ^= queries[i][1];
	}

}

// Computing cumulative prefix XOR and evaluating the answer
for (int i = 0; i < ans.size(); i++)
{
	if (i == 0)
	{
		ans[i] = ans[i] ^ xorArray[i];
	}
	else
	{
		xorArray[i] ^= xorArray[i - 1];
		ans[i] ^= xorArray[i];
	}

}

return ans;

}

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#2

hi @rjhk6767_a9801e8bdca93439,
the overall idea is not to perform o(n) operation when a xor query comes (which you r doing here just once for all query in second for loop)
in first for loop you are maintaining what values you have to perform operation (type 1 query), in between any type 2 query comes you are just keeping that information in second vector initially all 0.
Now why we only xor first and element after last (i.e 0 and ans.size() )element in xorArray?
lets understand with simple eg {1,2},{1,3}, {1,4}, {2,2}
2,3,4 comes store in ans array as they are of type 1
now type 2 query with 2 comes we store it in xorArray first and after last index (to nullify effect) which becomes 2 0 0 2.
we have now type 1 info (which elements) + type 2 info (to xor with)
so use for loop to and take cumulative xor.
when it reaches after last index in general it will be for here 2^2 i.e 0 which nullifies the effect of this xor for further array if there as it says to xor elements till that time.

hope this clears the doubt

#3

@rjhk6767_a9801e8bdca93439 regarding how it came in mind is range questions are mostly easily done by cumulative sum or cumulative xor here. Hope u have solved question asking range sum in array

#4

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