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Problem Statement
You're given a notepad
that initially displays
a single character
'A'. You have two actions
available to perform on this notepad for each step:
- Copy All: It allows you to
copy everything
on the screen. - Paste: You can
paste
the characters which arecopied
last time.
Given an integer n
, return the minimum
number of operations
to print the character 'A' exactly n
times on the screen.
Examples
-
Example 1:
- Input:
n = 4
- Expected Output:
4
- Justification: Start with 'A', copy it (
1
operation), and paste it(1
operations). Now, you have 'AA'. Copy it, and paste it. Total =4
operations.
- Input:
-
Example 2:
- Input:
n = 5
- Expected Output:
5
- Justification: Start with 'A', copy it, and paste it four times. Total =
5
operations.
- Input:
-
Example 3:
- Input:
n = 9
- Expected Output:
6
- Justification: Start with 'A', copy and paste twice (3 operations) to get 'AAA'. Copy 'AAA' and paste it twice (3 operations) to get 'AAAAAAAAA'. Total =
6
operations.
- Input:
Solution
To solve this problem, we'll focus on breaking down the target number n
into its prime factors, because the minimum number of operations needed to achieve n
'A's on the screen can be thought of in terms of multiplying the counts of 'A's by repeatedly copying and pasting. This strategy works because copying and pasting a larger block of 'A's, which is a multiple of smaller blocks, mirrors the process of constructing a number by multiplying its factors. Thus, our approach will be to find the sum of the prime factors of n
as this sum represents the minimum number of steps needed to get n
'A's on the screen.
This approach is effective because it leverages the mathematical property that any composite number can be expressed as a product of prime numbers, and the operations to achieve the target n
can be seen as the process of building up to n
through these prime multiplicands. Therefore, by analyzing n
in terms of its prime components, we can efficiently determine the least number of steps required.
Step-by-step algorithm
- Initialize
totalOperations
to 0, which will hold our final count of operations. - Iterate from
i = 2
ton
:- While
n % i == 0
(meaningi
is a factor ofn
):- Add
i
tototalOperations
(representing the operation count needed to achievei
copies). - Divide
n
byi
to reduce our problem size and reflect that we've accounted for this factor.
- Add
- While
- Return
totalOperations
as the answer.
Algorithm Walkthrough
Let's consider the input: n = 9
.
- Initialize
operations
to 0. This variable tracks the total number of operations required. - Start iterating
i
from 2 ton
(inclusive).- For
n = 9
, start withi = 2
.
- For
- Check if
n % i == 0
(ifn
is divisible byi
without a remainder).- For
i = 2
,9 % 2 != 0
. So, move to the nexti
. - For
i = 3
,9 % 3 == 0
. This is true, so enter the while loop.
- For
- Inside the while loop, add
i
tooperations
. For the first iteration, add 3 tooperations
. - Divide
n
byi
, settingn
ton / i
. For the first iteration, setn = 9 / 3 = 3
. - Check again if
n % i == 0
. Sincen
is now 3 andi
is also 3, the condition is true.- Add
i
(3) tooperations
again. Now,operations = 6
. - Divide
n
byi
again, nown = 3 / 3 = 1
.
- Add
- Since
n
is now 1, the loop fori = 3
finishes. - There are no more divisors of
n
(now 1) greater than 1, so the outer loop terminates. - The value of
operations
is now 6, which is the expected minimum number of operations to get the character 'A' exactlyn
times on the screen forn = 9
.
Code
Complexity Analysis
Time Complexity
The time complexity of the provided code is O(n sqrt(n)) in the worst case, primarily because it iterates through all numbers up to n
and performs division operations that, in the worst case, can be as costly as iterating up to the square root of n
for each number.
Space Complexity
The space complexity is O(1), as it uses a fixed amount of space regardless of input size.
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