# Posts published in “Greedy”

Given the array nums, obtain a subsequence of the array whose sum of elements is strictly greater than the sum of the non included elements in such subsequence.

If there are multiple solutions, return the subsequence with minimum size and if there still exist multiple solutions, return the subsequence with the maximum total sum of all its elements. A subsequence of an array can be obtained by erasing some (possibly zero) elements from the array.

Note that the solution with the given constraints is guaranteed to be unique. Also return the answer sorted in non-increasing order.

Example 1:

Input: nums = [4,3,10,9,8]
Output: [10,9]
Explanation: The subsequences [10,9] and [10,8] are minimal such that the sum of their elements is strictly greater than the sum of elements not included, however, the subsequence [10,9] has the maximum total sum of its elements.


Example 2:

Input: nums = [4,4,7,6,7]
Output: [7,7,6]
Explanation: The subsequence [7,7] has the sum of its elements equal to 14 which is not strictly greater than the sum of elements not included (14 = 4 + 4 + 6). Therefore, the subsequence [7,6,7] is the minimal satisfying the conditions. Note the subsequence has to returned in non-decreasing order.


Example 3:

Input: nums = 
Output: 


Constraints:

• 1 <= nums.length <= 500
• 1 <= nums[i] <= 100

## Solution: Greedy

Sort the elements in reverse order, pick the largest elements until their sum is greater than the sum of the left elements or total / 2.

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

## C++

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)

## Python3

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] <= events[i] <= 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)

## 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++

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++

Given a palindromic string palindrome, replace exactly one character by any lowercase English letter so that the string becomes the lexicographically smallest possible string that isn’t a palindrome.

After doing so, return the final string.  If there is no way to do so, return the empty string.

Example 1:

Input: palindrome = "abccba"
Output: "aaccba"


Example 2:

Input: palindrome = "a"
Output: ""


Constraints:

• 1 <= palindrome.length <= 1000
• palindrome consists of only lowercase English letters.

## Solution: Greedy

For the first half of the string, replace the first non ‘a’ character to ‘a’.

e.g. abcdcba => aacdcba

If not found which means the the entire string is ‘a’ expect the middle one if the length is odd, like aa or aba, replace the last character to ‘b’.

aa => ab
aba => abb

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

## C++

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