CareerCup

bucket sort. Count frequency of each characters and then output in order.

merge from mergesort

Yes. Merge sort will serve the purpose.

bogo sort.

Sort both the arrays A1 and A2 separately using any sorting which guarantees nlogn time
Create an output array A3 of size A1.length+A2.length
Have 2 pointers at the end of each array, compare them (may be using ASCII) and copy the elements to the new array A3, decrement the pointer based on the comparison

sample comparison part:

if (A1[end1] > A2[end2])
{
A3[end3] = A1[end1];
end1--;
}
else
{
A3[end3] = A2[end2];
end2--;
}
end3--;

#include <iostream>
#include <string>
#include <algorithm>

using namespace std;

bool comp(char c1, char c2)
{
	return tolower(c1) < tolower(c2);
}

int main()
{
	std::string str1 = "ab";
	std::string str2 = "badg";
	std::string str3;
	str3 += str1 + str2;
	std::sort(str3.begin(), str3.end(), comp);
	cout << str3 << endl;

	system("pause");
	return 0;
}

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