# Posts published in “Math”

We have a wooden plank of the length n units. Some ants are walking on the plank, each ant moves with speed 1 unit per second. Some of the ants move to the left, the other move to the right.

When two ants moving in two different directions meet at some point, they change their directions and continue moving again. Assume changing directions doesn’t take any additional time.

When an ant reaches one end of the plank at a time t, it falls out of the plank imediately.

Given an integer n and two integer arrays left and right, the positions of the ants moving to the left and the right. Return the moment when the last ant(s) fall out of the plank.

Example 1:

Input: n = 4, left = [4,3], right = [0,1]
Output: 4
Explanation: In the image above:
-The ant at index 0 is named A and going to the right.
-The ant at index 1 is named B and going to the right.
-The ant at index 3 is named C and going to the left.
-The ant at index 4 is named D and going to the left.
Note that the last moment when an ant was on the plank is t = 4 second, after that it falls imediately out of the plank. (i.e. We can say that at t = 4.0000000001, there is no ants on the plank).


Example 2:

Input: n = 7, left = [], right = [0,1,2,3,4,5,6,7]
Output: 7
Explanation: All ants are going to the right, the ant at index 0 needs 7 seconds to fall.


Example 3:

Input: n = 7, left = [0,1,2,3,4,5,6,7], right = []
Output: 7
Explanation: All ants are going to the left, the ant at index 7 needs 7 seconds to fall.


Example 4:

Input: n = 9, left = , right = 
Output: 5
Explanation: At t = 1 second, both ants will be at the same intial position but with different direction.


Example 5:

Input: n = 6, left = , right = 
Output: 6


Constraints:

• 1 <= n <= 10^4
• 0 <= left.length <= n + 1
• 0 <= left[i] <= n
• 0 <= right.length <= n + 1
• 0 <= right[i] <= n
• 1 <= left.length + right.length <= n + 1
• All values of left and right are unique, and each value can appear only in one of the two arrays.

## Solution: Keep Walking

When two ants A –> and <– B meet at some point, they change directions <– A B –>, we can swap the ids of the ants as <– B A–>, so it’s the same as walking individually and passed by. Then we just need to find the max/min of the left/right arrays.

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

## Python3

Given an array of integers arr of even length n and an integer k.

We want to divide the array into exactly n / 2 pairs such that the sum of each pair is divisible by k.

Return True If you can find a way to do that or False otherwise.

Example 1:

Input: arr = [1,2,3,4,5,10,6,7,8,9], k = 5
Output: true
Explanation: Pairs are (1,9),(2,8),(3,7),(4,6) and (5,10).


Example 2:

Input: arr = [1,2,3,4,5,6], k = 7
Output: true
Explanation: Pairs are (1,6),(2,5) and(3,4).


Example 3:

Input: arr = [1,2,3,4,5,6], k = 10
Output: false
Explanation: You can try all possible pairs to see that there is no way to divide arr into 3 pairs each with sum divisible by 10.


Example 4:

Input: arr = [-10,10], k = 2
Output: true


Example 5:

Input: arr = [-1,1,-2,2,-3,3,-4,4], k = 3
Output: true


Constraints:

• arr.length == n
• 1 <= n <= 10^5
• n is even.
• -10^9 <= arr[i] <= 10^9
• 1 <= k <= 10^5

## Solution: Mod and Count

Count the frequency of (x % k + k) % k.
f should be even (zero is also even)
f = f[k -1] ((1 + k – 1) % k == 0)
f = f[k -2] ((2 + k – 2) % k == 0)

Time complexity: O(n)
Space complexity: O(k)

## C++

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++

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++

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