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Posts published in “Math”

花花酱 LeetCode 2240. Number of Ways to Buy Pens and Pencils

You are given an integer total indicating the amount of money you have. You are also given two integers cost1 and cost2 indicating the price of a pen and pencil respectively. You can spend part or all of your money to buy multiple quantities (or none) of each kind of writing utensil.

Return the number of distinct ways you can buy some number of pens and pencils.

Example 1:

Input: total = 20, cost1 = 10, cost2 = 5
Output: 9
Explanation: The price of a pen is 10 and the price of a pencil is 5.
- If you buy 0 pens, you can buy 0, 1, 2, 3, or 4 pencils.
- If you buy 1 pen, you can buy 0, 1, or 2 pencils.
- If you buy 2 pens, you cannot buy any pencils.
The total number of ways to buy pens and pencils is 5 + 3 + 1 = 9.

Example 2:

Input: total = 5, cost1 = 10, cost2 = 10
Output: 1
Explanation: The price of both pens and pencils are 10, which cost more than total, so you cannot buy any writing utensils. Therefore, there is only 1 way: buy 0 pens and 0 pencils.

Constraints:

  • 1 <= total, cost1, cost2 <= 106

Solution:

Enumerate all possible ways to buy k pens, e.g. 0 pen, 1 pen, …, total / cost1.
The way to buy pencils are (total – k * cost1) / cost2 + 1.
ans = sum((total – k * cost1) / cost2 + 1)) for k = 0 to total / cost1.

Time complexity: O(total / cost1)
Space complexity: O(1)

C++

花花酱 LeetCode 2235. Add Two Integers

Given two integers num1 and num2, return the sum of the two integers.

Example 1:

Input: num1 = 12, num2 = 5
Output: 17
Explanation: num1 is 12, num2 is 5, and their sum is 12 + 5 = 17, so 17 is returned.

Example 2:

Input: num1 = -10, num2 = 4
Output: -6
Explanation: num1 + num2 = -6, so -6 is returned.

Constraints:

  • -100 <= num1, num2 <= 100

Solution: Just sum them up

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

C++

花花酱 LeetCode 2177. Find Three Consecutive Integers That Sum to a Given Number

Given an integer num, return three consecutive integers (as a sorted array) that sum to num. If num cannot be expressed as the sum of three consecutive integers, return an empty array.

Example 1:

Input: num = 33
Output: [10,11,12]
Explanation: 33 can be expressed as 10 + 11 + 12 = 33.
10, 11, 12 are 3 consecutive integers, so we return [10, 11, 12].

Example 2:

Input: num = 4
Output: []
Explanation: There is no way to express 4 as the sum of 3 consecutive integers.

Constraints:

  • 0 <= num <= 1015

Solution: Math

(x / 3 – 1) + (x / 3) + (x / 3 + 1) == 3x == num, num must be divisible by 3.

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

C++

花花酱 LeetCode 2183. Count Array Pairs Divisible by K

Given a 0-indexed integer array nums of length n and an integer k, return the number of pairs (i, j) such that:

  • 0 <= i < j <= n - 1 and
  • nums[i] * nums[j] is divisible by k.

Example 1:

Input: nums = [1,2,3,4,5], k = 2
Output: 7
Explanation: 
The 7 pairs of indices whose corresponding products are divisible by 2 are
(0, 1), (0, 3), (1, 2), (1, 3), (1, 4), (2, 3), and (3, 4).
Their products are 2, 4, 6, 8, 10, 12, and 20 respectively.
Other pairs such as (0, 2) and (2, 4) have products 3 and 15 respectively, which are not divisible by 2.    

Example 2:

Input: nums = [1,2,3,4], k = 5
Output: 0
Explanation: There does not exist any pair of indices whose corresponding product is divisible by 5.

Constraints:

  • 1 <= nums.length <= 105
  • 1 <= nums[i], k <= 105

Solution: Math

a * b % k == 0 <=> gcd(a, k) * gcd(b, k) == 0

Use a counter of gcd(x, k) so far to compute the number of pairs.

Time complexity: O(n*f), where f is the number of gcds, f <= 128 for x <= 1e5
Space complexity: O(f)

C++

花花酱 LeetCode 2147. Number of Ways to Divide a Long Corridor

Along a long library corridor, there is a line of seats and decorative plants. You are given a 0-indexed string corridor of length n consisting of letters 'S' and 'P' where each 'S' represents a seat and each 'P' represents a plant.

One room divider has already been installed to the left of index 0, and another to the right of index n - 1. Additional room dividers can be installed. For each position between indices i - 1 and i (1 <= i <= n - 1), at most one divider can be installed.

Divide the corridor into non-overlapping sections, where each section has exactly two seats with any number of plants. There may be multiple ways to perform the division. Two ways are different if there is a position with a room divider installed in the first way but not in the second way.

Return the number of ways to divide the corridor. Since the answer may be very large, return it modulo 109 + 7. If there is no way, return 0.

Example 1:

Input: corridor = "SSPPSPS"
Output: 3
Explanation: There are 3 different ways to divide the corridor.
The black bars in the above image indicate the two room dividers already installed.
Note that in each of the ways, each section has exactly two seats.

Example 2:

Input: corridor = "PPSPSP"
Output: 1
Explanation: There is only 1 way to divide the corridor, by not installing any additional dividers.
Installing any would create some section that does not have exactly two seats.

Example 3:

Input: corridor = "S"
Output: 0
Explanation: There is no way to divide the corridor because there will always be a section that does not have exactly two seats.

Constraints:

  • n == corridor.length
  • 1 <= n <= 105
  • corridor[i] is either 'S' or 'P'.

Solution: Combination

If the 2k-th seat is positioned at j, and the 2k+1-th seat is at i. There are (i – j) ways to split between these two groups.

ans = prod{ik – jk}

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

C++