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Posts tagged as “hard”

花花酱 LeetCode 2197. Replace Non-Coprime Numbers in Array

By zxi on March 8, 2022

You are given an array of integers nums. Perform the following steps:

  1. Find any two adjacent numbers in nums that are non-coprime.
  2. If no such numbers are found, stop the process.
  3. Otherwise, delete the two numbers and replace them with their LCM (Least Common Multiple).
  4. Repeat this process as long as you keep finding two adjacent non-coprime numbers.

Return the final modified array. It can be shown that replacing adjacent non-coprime numbers in any arbitrary order will lead to the same result.

The test cases are generated such that the values in the final array are less than or equal to 108.

Two values x and y are non-coprime if GCD(x, y) > 1 where GCD(x, y) is the Greatest Common Divisor of x and y.

Example 1:

Input: nums = [6,4,3,2,7,6,2]
Output: [12,7,6]
Explanation: 
- (6, 4) are non-coprime with LCM(6, 4) = 12. Now, nums = [12,3,2,7,6,2].
- (12, 3) are non-coprime with LCM(12, 3) = 12. Now, nums = [12,2,7,6,2].
- (12, 2) are non-coprime with LCM(12, 2) = 12. Now, nums = [12,7,6,2].
- (6, 2) are non-coprime with LCM(6, 2) = 6. Now, nums = [12,7,6].
There are no more adjacent non-coprime numbers in nums.
Thus, the final modified array is [12,7,6].
Note that there are other ways to obtain the same resultant array.

Example 2:

Input: nums = [2,2,1,1,3,3,3]
Output: [2,1,1,3]
Explanation: 
- (3, 3) are non-coprime with LCM(3, 3) = 3. Now, nums = [2,2,1,1,3,3].
- (3, 3) are non-coprime with LCM(3, 3) = 3. Now, nums = [2,2,1,1,3].
- (2, 2) are non-coprime with LCM(2, 2) = 2. Now, nums = [2,1,1,3].
There are no more adjacent non-coprime numbers in nums.
Thus, the final modified array is [2,1,1,3].
Note that there are other ways to obtain the same resultant array.

Constraints:

  • 1 <= nums.length <= 105
  • 1 <= nums[i] <= 105
  • The test cases are generated such that the values in the final array are less than or equal to 108.

Solution: Stack

“””It can be shown that replacing adjacent non-coprime numbers in any arbitrary order will lead to the same result.”””

So that we can do it in one pass from left to right using a stack/vector.

Push the current number onto stack, and merge top two if they are not co-prime.

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

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 2188. Minimum Time to Finish the Race

By zxi on February 28, 2022

You are given a 0-indexed 2D integer array tires where tires[i] = [fi, ri] indicates that the ith tire can finish its xth successive lap in fi * ri(x-1) seconds.

  • For example, if fi = 3 and ri = 2, then the tire would finish its 1st lap in 3 seconds, its 2nd lap in 3 * 2 = 6 seconds, its 3rd lap in 3 * 22 = 12 seconds, etc.

You are also given an integer changeTime and an integer numLaps.

The race consists of numLaps laps and you may start the race with any tire. You have an unlimited supply of each tire and after every lap, you may change to any given tire (including the current tire type) if you wait changeTime seconds.

Return the minimum time to finish the race.

Example 1:

Input: tires = [[2,3],[3,4]], changeTime = 5, numLaps = 4
Output: 21
Explanation: 
Lap 1: Start with tire 0 and finish the lap in 2 seconds.
Lap 2: Continue with tire 0 and finish the lap in 2 * 3 = 6 seconds.
Lap 3: Change tires to a new tire 0 for 5 seconds and then finish the lap in another 2 seconds.
Lap 4: Continue with tire 0 and finish the lap in 2 * 3 = 6 seconds.
Total time = 2 + 6 + 5 + 2 + 6 = 21 seconds.
The minimum time to complete the race is 21 seconds.

Example 2:

Input: tires = [[1,10],[2,2],[3,4]], changeTime = 6, numLaps = 5
Output: 25
Explanation: 
Lap 1: Start with tire 1 and finish the lap in 2 seconds.
Lap 2: Continue with tire 1 and finish the lap in 2 * 2 = 4 seconds.
Lap 3: Change tires to a new tire 1 for 6 seconds and then finish the lap in another 2 seconds.
Lap 4: Continue with tire 1 and finish the lap in 2 * 2 = 4 seconds.
Lap 5: Change tires to tire 0 for 6 seconds then finish the lap in another 1 second.
Total time = 2 + 4 + 6 + 2 + 4 + 6 + 1 = 25 seconds.
The minimum time to complete the race is 25 seconds.

Constraints:

  • 1 <= tires.length <= 105
  • tires[i].length == 2
  • 1 <= fi, changeTime <= 105
  • 2 <= ri <= 105
  • 1 <= numLaps <= 1000

Solution: DP

Observation: since ri >= 2, we must change tire within 20 laps, otherwise it will be slower.

pre-compute the time to finish k laps using each type of tire (k < 20), find min for each lap.

dp[i] = best[i], i < 20,
dp[i] = min{dp[i – j] + changeTime + best[j]}, i > 20

Time complexity: O(n)
Space complexity: O(n) -> O(20)

C++

花花酱 LeetCode 2151. Maximum Good People Based on Statements

By zxi on February 4, 2022

There are two types of persons:

  • The good person: The person who always tells the truth.
  • The bad person: The person who might tell the truth and might lie.

You are given a 0-indexed 2D integer array statements of size n x n that represents the statements made by n people about each other. More specifically, statements[i][j] could be one of the following:

  • 0 which represents a statement made by person i that person j is a bad person.
  • 1 which represents a statement made by person i that person j is a good person.
  • 2 represents that no statement is made by person i about person j.

Additionally, no person ever makes a statement about themselves. Formally, we have that statements[i][i] = 2 for all 0 <= i < n.

Return the maximum number of people who can be good based on the statements made by the n people.

Example 1:

Input: statements = [[2,1,2],[1,2,2],[2,0,2]]
Output: 2
Explanation: Each person makes a single statement.
- Person 0 states that person 1 is good.
- Person 1 states that person 0 is good.
- Person 2 states that person 1 is bad.
Let's take person 2 as the key.
- Assuming that person 2 is a good person:
    - Based on the statement made by person 2, person 1 is a bad person.
    - Now we know for sure that person 1 is bad and person 2 is good.
    - Based on the statement made by person 1, and since person 1 is bad, they could be:
        - telling the truth. There will be a contradiction in this case and this assumption is invalid.
        - lying. In this case, person 0 is also a bad person and lied in their statement.
    - Following that person 2 is a good person, there will be only one good person in the group.
- Assuming that person 2 is a bad person:
    - Based on the statement made by person 2, and since person 2 is bad, they could be:
        - telling the truth. Following this scenario, person 0 and 1 are both bad as explained before.
            - Following that person 2 is bad but told the truth, there will be no good persons in the group.
        - lying. In this case person 1 is a good person.
            - Since person 1 is a good person, person 0 is also a good person.
            - Following that person 2 is bad and lied, there will be two good persons in the group.
We can see that at most 2 persons are good in the best case, so we return 2.
Note that there is more than one way to arrive at this conclusion.

Example 2:

Input: statements = [[2,0],[0,2]]
Output: 1
Explanation: Each person makes a single statement.
- Person 0 states that person 1 is bad.
- Person 1 states that person 0 is bad.
Let's take person 0 as the key.
- Assuming that person 0 is a good person:
    - Based on the statement made by person 0, person 1 is a bad person and was lying.
    - Following that person 0 is a good person, there will be only one good person in the group.
- Assuming that person 0 is a bad person:
    - Based on the statement made by person 0, and since person 0 is bad, they could be:
        - telling the truth. Following this scenario, person 0 and 1 are both bad.
            - Following that person 0 is bad but told the truth, there will be no good persons in the group.
        - lying. In this case person 1 is a good person.
            - Following that person 0 is bad and lied, there will be only one good person in the group.
We can see that at most, one person is good in the best case, so we return 1.
Note that there is more than one way to arrive at this conclusion.

Constraints:

  • n == statements.length == statements[i].length
  • 2 <= n <= 15
  • statements[i][j] is either 0, 1, or 2.
  • statements[i][i] == 2

Solution: Combination / Bitmask

Enumerate all subsets of n people and assume they are good people. Check whether their statements have any conflicts. We can ignore the statements from bad people since those can be either true or false and does not affect our checks.

Time complexity: O(n22n)
Space complexity: O(1)

C++

花花酱 LeetCode 2141. Maximum Running Time of N Computers

By zxi on January 30, 2022

You have n computers. You are given the integer n and a 0-indexed integer array batteries where the ith battery can run a computer for batteries[i] minutes. You are interested in running all n computers simultaneously using the given batteries.

Initially, you can insert at most one battery into each computer. After that and at any integer time moment, you can remove a battery from a computer and insert another battery any number of times. The inserted battery can be a totally new battery or a battery from another computer. You may assume that the removing and inserting processes take no time.

Note that the batteries cannot be recharged.

Return the maximum number of minutes you can run all the n computers simultaneously.

Example 1:

Input: n = 2, batteries = [3,3,3]
Output: 4
Explanation: 
Initially, insert battery 0 into the first computer and battery 1 into the second computer.
After two minutes, remove battery 1 from the second computer and insert battery 2 instead. Note that battery 1 can still run for one minute.
At the end of the third minute, battery 0 is drained, and you need to remove it from the first computer and insert battery 1 instead.
By the end of the fourth minute, battery 1 is also drained, and the first computer is no longer running.
We can run the two computers simultaneously for at most 4 minutes, so we return 4.

Example 2:

Input: n = 2, batteries = [1,1,1,1]
Output: 2
Explanation: 
Initially, insert battery 0 into the first computer and battery 2 into the second computer. 
After one minute, battery 0 and battery 2 are drained so you need to remove them and insert battery 1 into the first computer and battery 3 into the second computer. 
After another minute, battery 1 and battery 3 are also drained so the first and second computers are no longer running.
We can run the two computers simultaneously for at most 2 minutes, so we return 2.

Constraints:

  • 1 <= n <= batteries.length <= 105
  • 1 <= batteries[i] <= 109

Solution: Binary Search

Find the smallest L that we can not run, ans = L – 1.

For a guessing m, we check the total battery powers T = sum(min(m, batteries[i])), if T >= m * n, it means there is a way (doesn’t need to figure out how) to run n computers for m minutes by fully unitize those batteries.

Proof: If T >= m*n holds, there are two cases:

  1. There are only n batteries, can not swap, but each of them has power >= m.
  2. At least one of the batteries have power less than m, but there are more than n batteries and total power is sufficient, we can swap them with others.

Time complexity: O(Slogn) where S = sum(batteries)
Space complexity: O(1)

C++