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Posts tagged as “math”

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

By zxi on March 4, 2022

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

By zxi on March 4, 2022

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 2169. Count Operations to Obtain Zero

By zxi on February 13, 2022

You are given two non-negative integers num1 and num2.

In one operation, if num1 >= num2, you must subtract num2 from num1, otherwise subtract num1 from num2.

  • For example, if num1 = 5 and num2 = 4, subtract num2 from num1, thus obtaining num1 = 1 and num2 = 4. However, if num1 = 4 and num2 = 5, after one operation, num1 = 4 and num2 = 1.

Return the number of operations required to make either num1 = 0 or num2 = 0.

Example 1:

Input: num1 = 2, num2 = 3
Output: 3
Explanation: 
- Operation 1: num1 = 2, num2 = 3. Since num1 < num2, we subtract num1 from num2 and get num1 = 2, num2 = 3 - 2 = 1.
- Operation 2: num1 = 2, num2 = 1. Since num1 > num2, we subtract num2 from num1.
- Operation 3: num1 = 1, num2 = 1. Since num1 == num2, we subtract num2 from num1.
Now num1 = 0 and num2 = 1. Since num1 == 0, we do not need to perform any further operations.
So the total number of operations required is 3.

Example 2:

Input: num1 = 10, num2 = 10
Output: 1
Explanation: 
- Operation 1: num1 = 10, num2 = 10. Since num1 == num2, we subtract num2 from num1 and get num1 = 10 - 10 = 0.
Now num1 = 0 and num2 = 10. Since num1 == 0, we are done.
So the total number of operations required is 1.

Constraints:

  • 0 <= num1, num2 <= 105

Solution 1: Simulation

Time complexity: O(max(n,m) / min(n, m))
Space complexity: O(1)

No code

Solution 2: Simualtion + Math

For the case of 100, 3
100 – 3 = 97
97 – 3 = 94
…
4 – 3 = 1
Swap
3 – 1 = 2
2 – 1 = 1
1 – 1 = 0
It takes 36 steps.

We can do 100 / 3 to skip 33 steps
100 %= 3 = 1
3 / 1 = 3 to skip 3 steps
3 %= 1 = 0
total is 33 + 3 = 36.

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

C++

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

By zxi on February 4, 2022

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++

花花酱 LeetCode 2145. Count the Hidden Sequences

By zxi on January 31, 2022

You are given a 0-indexed array of n integers differences, which describes the differences between each pair of consecutive integers of a hidden sequence of length (n + 1). More formally, call the hidden sequence hidden, then we have that differences[i] = hidden[i + 1] - hidden[i].

You are further given two integers lower and upper that describe the inclusive range of values [lower, upper] that the hidden sequence can contain.

  • For example, given differences = [1, -3, 4], lower = 1, upper = 6, the hidden sequence is a sequence of length 4 whose elements are in between 1 and 6 (inclusive).
    • [3, 4, 1, 5] and [4, 5, 2, 6] are possible hidden sequences.
    • [5, 6, 3, 7] is not possible since it contains an element greater than 6.
    • [1, 2, 3, 4] is not possible since the differences are not correct.

Return the number of possible hidden sequences there are. If there are no possible sequences, return 0.

Example 1:

Input: differences = [1,-3,4], lower = 1, upper = 6
Output: 2
Explanation: The possible hidden sequences are:
- [3, 4, 1, 5]
- [4, 5, 2, 6]
Thus, we return 2.

Example 2:

Input: differences = [3,-4,5,1,-2], lower = -4, upper = 5
Output: 4
Explanation: The possible hidden sequences are:
- [-3, 0, -4, 1, 2, 0]
- [-2, 1, -3, 2, 3, 1]
- [-1, 2, -2, 3, 4, 2]
- [0, 3, -1, 4, 5, 3]
Thus, we return 4.

Example 3:

Input: differences = [4,-7,2], lower = 3, upper = 6
Output: 0
Explanation: There are no possible hidden sequences. Thus, we return 0.

Constraints:

  • n == differences.length
  • 1 <= n <= 105
  • -105 <= differences[i] <= 105
  • -105 <= lower <= upper <= 105

Solution: Math

Find the min and max of the cumulative sum of the differences.

Ans = max(0, upper – lower – (hi – lo) + 1)

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

C++