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Posts tagged as “greedy”

花花酱 LeetCode 2656. Maximum Sum With Exactly K Elements

By zxi on April 30, 2023

You are given a 0-indexed integer array nums and an integer k. Your task is to perform the following operation exactly k times in order to maximize your score:

  1. Select an element m from nums.
  2. Remove the selected element m from the array.
  3. Add a new element with a value of m + 1 to the array.
  4. Increase your score by m.

Return the maximum score you can achieve after performing the operation exactly k times.

Example 1:

Input: nums = [1,2,3,4,5], k = 3
Output: 18
Explanation: We need to choose exactly 3 elements from nums to maximize the sum.
For the first iteration, we choose 5. Then sum is 5 and nums = [1,2,3,4,6]
For the second iteration, we choose 6. Then sum is 5 + 6 and nums = [1,2,3,4,7]
For the third iteration, we choose 7. Then sum is 5 + 6 + 7 = 18 and nums = [1,2,3,4,8]
So, we will return 18.
It can be proven, that 18 is the maximum answer that we can achieve.

Example 2:

Input: nums = [5,5,5], k = 2
Output: 11
Explanation: We need to choose exactly 2 elements from nums to maximize the sum.
For the first iteration, we choose 5. Then sum is 5 and nums = [5,5,6]
For the second iteration, we choose 6. Then sum is 5 + 6 = 11 and nums = [5,5,7]
So, we will return 11.
It can be proven, that 11 is the maximum answer that we can achieve.

Constraints:

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

Solution: Greedy

Always to chose the largest element from the array.

We can find the largest element of the array m, then the total score will be
m + (m + 1) + (m + 2) + … + (m + k – 1),
We can use summation formula of arithmetic sequence to compute that in O(1)
ans = (m + (m + k – 1)) * k / 2

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

C++

花花酱 LeetCode 2587. Rearrange Array to Maximize Prefix Score

By zxi on March 11, 2023

You are given a 0-indexed integer array nums. You can rearrange the elements of nums to any order (including the given order).

Let prefix be the array containing the prefix sums of nums after rearranging it. In other words, prefix[i] is the sum of the elements from 0 to i in nums after rearranging it. The score of nums is the number of positive integers in the array prefix.

Return the maximum score you can achieve.

Example 1:

Input: nums = [2,-1,0,1,-3,3,-3]
Output: 6
Explanation: We can rearrange the array into nums = [2,3,1,-1,-3,0,-3].
prefix = [2,5,6,5,2,2,-1], so the score is 6.
It can be shown that 6 is the maximum score we can obtain.

Example 2:

Input: nums = [-2,-3,0]
Output: 0
Explanation: Any rearrangement of the array will result in a score of 0.

Constraints:

  • 1 <= nums.length <= 105
  • -106 <= nums[i] <= 106

Solution: Greedy

Sort the numbers in descending order.

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

C++

花花酱 LeetCode 2554. Maximum Number of Integers to Choose From a Range I

By zxi on February 13, 2023

You are given an integer array banned and two integers n and maxSum. You are choosing some number of integers following the below rules:

  • The chosen integers have to be in the range [1, n].
  • Each integer can be chosen at most once.
  • The chosen integers should not be in the array banned.
  • The sum of the chosen integers should not exceed maxSum.

Return the maximum number of integers you can choose following the mentioned rules.

Example 1:

Input: banned = [1,6,5], n = 5, maxSum = 6
Output: 2
Explanation: You can choose the integers 2 and 4.
2 and 4 are from the range [1, 5], both did not appear in banned, and their sum is 6, which did not exceed maxSum.

Example 2:

Input: banned = [1,2,3,4,5,6,7], n = 8, maxSum = 1
Output: 0
Explanation: You cannot choose any integer while following the mentioned conditions.

Example 3:

Input: banned = [11], n = 7, maxSum = 50
Output: 7
Explanation: You can choose the integers 1, 2, 3, 4, 5, 6, and 7.
They are from the range [1, 7], all did not appear in banned, and their sum is 28, which did not exceed maxSum.

Constraints:

  • 1 <= banned.length <= 104
  • 1 <= banned[i], n <= 104
  • 1 <= maxSum <= 109

Solution 1: Greedy + HashSet

We would like to use the smallest numbers possible. Store all the banned numbers into a hashset, and enumerate numbers from 1 to n and check whether we can use that number.

Time complexity: O(m + n)
Space complexity: O(m)

C++

Solution 2: Two Pointers

Sort the banned numbers. Use one pointer j and compare with the current number i.

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

C++

花花酱 LeetCode 2477. Minimum Fuel Cost to Report to the Capital

By zxi on November 26, 2022

There is a tree (i.e., a connected, undirected graph with no cycles) structure country network consisting of n cities numbered from 0 to n - 1 and exactly n - 1 roads. The capital city is city 0. You are given a 2D integer array roads where roads[i] = [ai, bi] denotes that there exists a bidirectional road connecting cities ai and bi.

There is a meeting for the representatives of each city. The meeting is in the capital city.

There is a car in each city. You are given an integer seats that indicates the number of seats in each car.

A representative can use the car in their city to travel or change the car and ride with another representative. The cost of traveling between two cities is one liter of fuel.

Return the minimum number of liters of fuel to reach the capital city.

Example 1:

Input: roads = [[0,1],[0,2],[0,3]], seats = 5
Output: 3
Explanation: 
- Representative1 goes directly to the capital with 1 liter of fuel.
- Representative2 goes directly to the capital with 1 liter of fuel.
- Representative3 goes directly to the capital with 1 liter of fuel.
It costs 3 liters of fuel at minimum. 
It can be proven that 3 is the minimum number of liters of fuel needed.

Example 2:

Input: roads = [[3,1],[3,2],[1,0],[0,4],[0,5],[4,6]], seats = 2
Output: 7
Explanation: 
- Representative2 goes directly to city 3 with 1 liter of fuel.
- Representative2 and representative3 go together to city 1 with 1 liter of fuel.
- Representative2 and representative3 go together to the capital with 1 liter of fuel.
- Representative1 goes directly to the capital with 1 liter of fuel.
- Representative5 goes directly to the capital with 1 liter of fuel.
- Representative6 goes directly to city 4 with 1 liter of fuel.
- Representative4 and representative6 go together to the capital with 1 liter of fuel.
It costs 7 liters of fuel at minimum. 
It can be proven that 7 is the minimum number of liters of fuel needed.

Example 3:

Input: roads = [], seats = 1
Output: 0
Explanation: No representatives need to travel to the capital city.

Constraints:

  • 1 <= n <= 105
  • roads.length == n - 1
  • roads[i].length == 2
  • 0 <= ai, bi < n
  • ai != bi
  • roads represents a valid tree.
  • 1 <= seats <= 105

Solution: Greedy + DFS

To reach the minimum cost, we must share cars if possible, say X reps from children nodes to an intermediate node u on the way towards capital 0. Then they all changes cars at node u, and we need (X + 1) // seats cars/fuel from u to 0.

We use DFS to count # of reps at each node u while accumulating the total cost.

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

C++

花花酱 LeetCode 2419. Longest Subarray With Maximum Bitwise AND

By zxi on September 26, 2022

You are given an integer array nums of size n.

Consider a non-empty subarray from nums that has the maximum possible bitwise AND.

  • In other words, let k be the maximum value of the bitwise AND of any subarray of nums. Then, only subarrays with a bitwise AND equal to k should be considered.

Return the length of the longest such subarray.

The bitwise AND of an array is the bitwise AND of all the numbers in it.

A subarray is a contiguous sequence of elements within an array.

Example 1:

Input: nums = [1,2,3,3,2,2]
Output: 2
Explanation:
The maximum possible bitwise AND of a subarray is 3.
The longest subarray with that value is [3,3], so we return 2.

Example 2:

Input: nums = [1,2,3,4]
Output: 1
Explanation:
The maximum possible bitwise AND of a subarray is 4.
The longest subarray with that value is [4], so we return 1.

Constraints:

  • 1 <= nums.length <= 105
  • 1 <= nums[i] <= 106

Solution: Find the largest number

a & b <= a
a & b <= b
if b > a, a & b < b, we choose to start a new sequence of “b” instead of continuing with “ab”

Basically, we find the largest number in the array and count the longest sequence of it. Note, there will be some tricky cases like.
b b b b a b
b a b b b b
We need to return 4 instead of 1.

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

C++