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Posts tagged as “greedy”

花花酱 LeetCode 1386. Cinema Seat Allocation

A cinema has n rows of seats, numbered from 1 to n and there are ten seats in each row, labelled from 1 to 10 as shown in the figure above.

Given the array reservedSeats containing the numbers of seats already reserved, for example, reservedSeats[i]=[3,8] means the seat located in row 3 and labelled with 8 is already reserved. 

Return the maximum number of four-person families you can allocate on the cinema seats. A four-person family occupies fours seats in one row, that are next to each other. Seats across an aisle (such as [3,3] and [3,4]) are not considered to be next to each other, however, It is permissible for the four-person family to be separated by an aisle, but in that case, exactly two people have to sit on each side of the aisle.

Example 1:

Input: n = 3, reservedSeats = [[1,2],[1,3],[1,8],[2,6],[3,1],[3,10]]
Output: 4
Explanation: The figure above shows the optimal allocation for four families, where seats mark with blue are already reserved and contiguous seats mark with orange are for one family. 

Example 2:

Input: n = 2, reservedSeats = [[2,1],[1,8],[2,6]]
Output: 2

Example 3:

Input: n = 4, reservedSeats = [[4,3],[1,4],[4,6],[1,7]]
Output: 4

Constraints:

  • 1 <= n <= 10^9
  • 1 <= reservedSeats.length <= min(10*n, 10^4)
  • reservedSeats[i].length == 2
  • 1 <= reservedSeats[i][0] <= n
  • 1 <= reservedSeats[i][1] <= 10
  • All reservedSeats[i] are distinct.

Solution: HashTable + Greedy

if both seat[2~5] seat[6~9] are empty, seat two groups.
if any of seat[2~5] seat[4~7] seat[6~9] is empty seat one group.
if there is no one sit in a row, seat two groups.

Time complexity: O(|reservedSeats|)
Space complexity: O(|rows|)

C++

花花酱 LeetCode 1383. Maximum Performance of a Team

There are n engineers numbered from 1 to n and two arrays: speed and efficiency, where speed[i] and efficiency[i] represent the speed and efficiency for the i-th engineer respectively. Return the maximum performance of a team composed of at most k engineers, since the answer can be a huge number, return this modulo 10^9 + 7.

The performance of a team is the sum of their engineers’ speeds multiplied by the minimum efficiency among their engineers. 

Example 1:

Input: n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 2
Output: 60
Explanation: 
We have the maximum performance of the team by selecting engineer 2 (with speed=10 and efficiency=4) and engineer 5 (with speed=5 and efficiency=7). That is, performance = (10 + 5) * min(4, 7) = 60.

Example 2:

Input: n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 3
Output: 68
Explanation:
This is the same example as the first but k = 3. We can select engineer 1, engineer 2 and engineer 5 to get the maximum performance of the team. That is, performance = (2 + 10 + 5) * min(5, 4, 7) = 68.

Example 3:

Input: n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 4
Output: 72

Constraints:

  • 1 <= n <= 10^5
  • speed.length == n
  • efficiency.length == n
  • 1 <= speed[i] <= 10^5
  • 1 <= efficiency[i] <= 10^8
  • 1 <= k <= n

Solution: Greedy + Sliding Window

  1. Sort engineers by their efficiency in descending order.
  2. For each window of K engineers (we can have less than K people in the first k-1 windows), ans is sum(speed) * min(efficiency).

Time complexity: O(nlogn) + O(nlogk)
Space complexity: O(n)

C++

Python3

花花酱 LeetCode 1363. Largest Multiple of Three

Given an integer array of digits, return the largest multiple of three that can be formed by concatenating some of the given digits in any order.

Since the answer may not fit in an integer data type, return the answer as a string.

If there is no answer return an empty string.

Example 1:

Input: digits = [8,1,9]
Output: "981"

Example 2:

Input: digits = [8,6,7,1,0]
Output: "8760"

Example 3:

Input: digits = [1]
Output: ""

Example 4:

Input: digits = [0,0,0,0,0,0]
Output: "0"

Constraints:

  • 1 <= digits.length <= 10^4
  • 0 <= digits[i] <= 9
  • The returning answer must not contain unnecessary leading zeros.

Solution: Greedy + Math + Counting sort

Count the numbers of each digit.
if sum % 3 == 0, we can use all digits.
if sum % 1 == 0, we can remove one digits among {1, 4, 7} => sum % 3 == 0
if sum % 2 == 0, we can remove one digits among {2, 5, 8} => sum % 3 == 0
if sum % 2 == 0, we have to remove two digits among {1, 4, 7} => sum % 3 == 0
if sum % 1 == 0, we have to remove two digits among {2, 5, 8} => sum % 3 == 0

Time complexity: O(n)
Space complexity: O(n) w/ output, O(1) w/o output

C++

Python3

花花酱 LeetCode 1353. Maximum Number of Events That Can Be Attended

Given an array of events where events[i] = [startDayi, endDayi]. Every event i starts at startDayiand ends at endDayi.

You can attend an event i at any day d where startTimei <= d <= endTimei. Notice that you can only attend one event at any time d.

Return the maximum number of events you can attend.

Example 1:

Input: events = [[1,2],[2,3],[3,4]]
Output: 3
Explanation: You can attend all the three events.
One way to attend them all is as shown.
Attend the first event on day 1.
Attend the second event on day 2.
Attend the third event on day 3.

Example 2:

Input: events= [[1,2],[2,3],[3,4],[1,2]]
Output: 4

Example 3:

Input: events = [[1,4],[4,4],[2,2],[3,4],[1,1]]
Output: 4

Example 4:

Input: events = [[1,100000]]
Output: 1

Example 5:

Input: events = [[1,1],[1,2],[1,3],[1,4],[1,5],[1,6],[1,7]]
Output: 7

Constraints:

  • 1 <= events.length <= 10^5
  • events[i].length == 2
  • 1 <= events[i][0] <= events[i][1] <= 10^5

Solution: Greedy

Sort events by end time, for each event find the first available day to attend.

Time complexity: O(sum(endtime – starttime)) = O(10^10)
Space complexity: O(max(endtime – starttime) = O(10^5)

C++

C++

Python

We can use a TreeSet to maintain the open days and do a binary search to find the first available day.

Time complexity: O(nlogd)
Space complexity: O(d)

C++

花花酱 LeetCode 1330. Reverse Subarray To Maximize Array Value

You are given an integer array nums. The value of this array is defined as the sum of |nums[i]-nums[i+1]| for all 0 <= i < nums.length-1.

You are allowed to select any subarray of the given array and reverse it. You can perform this operation only once.

Find maximum possible value of the final array.

Example 1:

Input: nums = [2,3,1,5,4]
Output: 10
Explanation: By reversing the subarray [3,1,5] the array becomes [2,5,1,3,4] whose value is 10.

Example 2:

Input: nums = [2,4,9,24,2,1,10]
Output: 68

Constraints:

  • 1 <= nums.length <= 3*10^4
  • -10^5 <= nums[i] <= 10^5

Solution: Greedy

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

C++