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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
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One Erlang is equal to (A) 3600 CCS (B) 36 CCS (C) 60 CCS (D) 24 CCS Ans: One Erlang is equivalent to 3600 CCS.
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Prepare at least 3 sets of input data (Test data) along with expected output for testing your program.
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