Prime factors of a number

Given a number n, find all prime factors of n.
Note : Prime number is a natural number greater than 1 that has exactly two factors:1 and itself.

Examples:

Input: n = 18
Output: [2, 3, 3]
Explanation: The prime factorization of 18 is 2×3².

Input: n = 25
Output: [5, 5]
Explanation: The prime factorization of 25 is 5².

[Naive Approach] Iterative Time O(n) and space O(1)

Step by Step implementation:

• Start a loop from i = 2 up to n.
• For each i, check if n is divisible by i (i.e n % i == 0).
• While n is divisible by i, store i as a factor and divide n by i.
• Continue this process for all i up to n to extract all prime factors.

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[Expected Approach] Prime Factorisation Time O(sqrt(n)) and Space (1)

Every composite number has at least one prime factor less than or equal to its square root. This property can be proved using a counter statement -

• Let a and b be two factors of n such that a*b = n.
• If both are greater than sqrt(n), a*b > sqrt(n)*sqrt(n), which contradicts the expression a * b = n

Step by Step implementation

• While n is divisible by 2, print 2 and divide n by 2.
• After step 1, n must be odd. Now start a loop from i = 3 to the square root of n. While i divides n, print i, and divide n by i. After i fails to divide n, increment i by 2 because next prime factor will also be odd since we have already taken care of all the 2 and continue.
• If n greater than 2, then n is the last prime factor.

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Related article: 

• Prime Factorization using Sieve O(log n) for multiple queries