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First Run a Sieve Algorithm and find all the Prime Numbers Up to Square root of N . Then Start dividing the Given Number N by the Prime factors less than sqrt(N) Untill N become Undivisible . This will take sqrt(N)/ln(sqrt(N)) time .Store the Prime Number which divides N and How many times They divide . This will take log2(N) time . So overall , Inside the Factorisation Function Complexity will be O(Sqrt(N)/ln(sqrt(N))xlog2(N))
But Overall Complexity will be O(Nln(N)) . Because for the Sieve Algorithm the Complexity will be O(Nln(N)) and Inside the Factorisation Function Complexity will be O(Sqrt(N)/ln(sqrt(N))xlog2(N)) . But O(NlnN) > O(Sqrt(N)/ln(sqrt(N))xlog2(N)) That's Why overall complexity will be But O(Nln(N)) . But if there is Multiple Test cases and O(Tx(Sqrt(N)/ln(sqrt(N))xlog2(N)) > O(Nln(N)) then the complexity will be O(Tx(Sqrt(N)/ln(sqrt(N)) + log2(N))
Prime Factorisation of a Number
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