# Posts tagged as “math”

Given an integer n, return a list of all simplified fractions between 0 and 1 (exclusive) such that the denominator is less-than-or-equal-to n. The fractions can be in any order.

Example 1:

Input: n = 2
Output: ["1/2"]
Explanation: "1/2" is the only unique fraction with a denominator less-than-or-equal-to 2.

Example 2:

Input: n = 3
Output: ["1/2","1/3","2/3"]


Example 3:

Input: n = 4
Output: ["1/2","1/3","1/4","2/3","3/4"]
Explanation: "2/4" is not a simplified fraction because it can be simplified to "1/2".

Example 4:

Input: n = 1
Output: []


Constraints:

• 1 <= n <= 100

## Solution: GCD

if gcd(a, b) == 1 then a/b is a simplified frication.

std::gcd is available since c++17.

Time complexity: O(n^2logn)
Space complexity: O(1)

## C++

Given an integer array nums, return the sum of divisors of the integers in that array that have exactly four divisors.

If there is no such integer in the array, return 0.

Example 1:

Input: nums = [21,4,7]
Output: 32
Explanation:
21 has 4 divisors: 1, 3, 7, 21
4 has 3 divisors: 1, 2, 4
7 has 2 divisors: 1, 7
The answer is the sum of divisors of 21 only.


Constraints:

• 1 <= nums.length <= 10^4
• 1 <= nums[i] <= 10^5

## Solution: Math

If a number is a perfect square (e.g. 9 = 3 * 3), it will have odd number of divisors. (9: 1, 3, 9).

Time complexity: O(sum(sqrt(num_i))
Space complexity: O(1)

## C++

There is a room with n lights which are turned on initially and 4 buttons on the wall. After performing exactly m unknown operations towards buttons, you need to return how many different kinds of status of the n lights could be.

Suppose n lights are labeled as number [1, 2, 3 …, n], function of these 4 buttons are given below:

1. Flip all the lights.
2. Flip lights with even numbers.
3. Flip lights with odd numbers.
4. Flip lights with (3k + 1) numbers, k = 0, 1, 2, …

Example 1:

Input: n = 1, m = 1.
Output: 2
Explanation: Status can be: [on], [off]


Example 2:

Input: n = 2, m = 1.
Output: 3
Explanation: Status can be: [on, off], [off, on], [off, off]


Example 3:

Input: n = 3, m = 1.
Output: 4
Explanation: Status can be: [off, on, off], [on, off, on], [off, off, off], [off, on, on].


Note: n and m both fit in range [0, 1000].

The light pattern will be repeated if we have more than 6 lights, so n = n % 6, n = 6 if n == 0.

Time complexity: O(m*2^6)
Space complexity: O(2^6)

## C++

You are given two jugs with capacities x and y litres. There is an infinite amount of water supply available. You need to determine whether it is possible to measure exactly z litres using these two jugs.

If z liters of water is measurable, you must have z liters of water contained within one or both buckets by the end.

Operations allowed:

• Fill any of the jugs completely with water.
• Empty any of the jugs.
• Pour water from one jug into another till the other jug is completely full or the first jug itself is empty.

Example 1: (From the famous “Die Hard” example)

Input: x = 3, y = 5, z = 4
Output: True


Example 2:

Input: x = 2, y = 6, z = 5
Output: False

## Solution: Math

special case 1: x == z or y == z or x + y == z: return True
special case 2: x + y < z: return False
normal case: z must be a factor of gcd(x, y)

Time complexity: O(log(min(x, y))
Space complexity: O(1)

## C++

Given a non-negative integer num, repeatedly add all its digits until the result has only one digit.

Example:

Input: 38
Output: 2
Explanation: The process is like: 3 + 8 = 11, 1 + 1 = 2.
Since 2 has only one digit, return it.


Could you do it without any loop/recursion in O(1) runtime?

## Solution 1: Simulation

Time complexity: O(logn)
Space complexity: O(1)

## Solution 2: Math

https://en.wikipedia.org/wiki/Digital_root#Congruence_formula

Digit root = num % 9 if num % 9 != 0 else min(num, 9) e.g. 0 or 9

Time complexity: O(1)
Space complexity: O(1)

## C++

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