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Q. Explain the odd-even transposition algorithm?
The algorithm needs one 'for loop' beginning from I=1 to N it implies that N times and for every value of I, one 'for loop' of J is executed concurrently. So the time complexity of the algorithm is O(n) because there are total n phases and every phase executes either even or odd transposition in O(1) time.
Figure: Example
* Public, protected and private are 3 access specifier in C++. * Public data members and member functions are accessible outside the class. * Protected data members and memb
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