# Posts published in “Greedy”

Given the number kreturn the minimum number of Fibonacci numbers whose sum is equal to k, whether a Fibonacci number could be used multiple times.

The Fibonacci numbers are defined as:

• F1 = 1
• F2 = 1
• Fn = Fn-1 + Fn-2 , for n > 2.

It is guaranteed that for the given constraints we can always find such fibonacci numbers that sum k.

Example 1:

Input: k = 7
Output: 2
Explanation: The Fibonacci numbers are: 1, 1, 2, 3, 5, 8, 13, ...
For k = 7 we can use 2 + 5 = 7.

Example 2:

Input: k = 10
Output: 2
Explanation: For k = 10 we can use 2 + 8 = 10.


Example 3:

Input: k = 19
Output: 3
Explanation: For k = 19 we can use 1 + 5 + 13 = 19.


Constraints:

• 1 <= k <= 10^9

## Solution: Greedy

Find the largest fibonacci numbers x that x <= k, ans = 1 + find(k – x)

Time complexity: O(logk^2) -> O(logk)
Space complexity: O(logk) -> O(1)

Recursive

Iterative

## C++

A string is called happy if it does not have any of the strings 'aaa''bbb' or 'ccc' as a substring.

Given three integers ab and c, return any string s, which satisfies following conditions:

• s is happy and longest possible.
• s contains at most a occurrences of the letter 'a'at most b occurrences of the letter 'b' and at most c occurrences of the letter 'c'.
• s will only contain 'a''b' and 'c' letters.

If there is no such string s return the empty string "".

Example 1:

Input: a = 1, b = 1, c = 7
Output: "ccaccbcc"
Explanation: "ccbccacc" would also be a correct answer.


Example 2:

Input: a = 2, b = 2, c = 1
Output: "aabbc"


Example 3:

Input: a = 7, b = 1, c = 0
Output: "aabaa"
Explanation: It's the only correct answer in this case.


Constraints:

• 0 <= a, b, c <= 100
• a + b + c > 0

## Solution: Greedy

Put the char with highest frequency first if its consecutive length of that char is < 2
or put one char if any of other two chars has consecutive length of 2.

increase the consecutive length of itself and reset that for other two chars.

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

## C++

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 = [6]
Output: [6]


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][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)

## 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++

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