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Posts tagged as “greedy”

花花酱 LeetCode 2087. Minimum Cost Homecoming of a Robot in a Grid

By zxi on November 27, 2021

There is an m x n grid, where (0, 0) is the top-left cell and (m - 1, n - 1) is the bottom-right cell. You are given an integer array startPos where startPos = [startrow, startcol] indicates that initially, a robot is at the cell (startrow, startcol). You are also given an integer array homePos where homePos = [homerow, homecol] indicates that its home is at the cell (homerow, homecol).

The robot needs to go to its home. It can move one cell in four directions: left, right, up, or down, and it can not move outside the boundary. Every move incurs some cost. You are further given two 0-indexed integer arrays: rowCosts of length m and colCosts of length n.

  • If the robot moves up or down into a cell whose row is r, then this move costs rowCosts[r].
  • If the robot moves left or right into a cell whose column is c, then this move costs colCosts[c].

Return the minimum total cost for this robot to return home.

Example 1:

Input: startPos = [1, 0], homePos = [2, 3], rowCosts = [5, 4, 3], colCosts = [8, 2, 6, 7]
Output: 18
Explanation: One optimal path is that:
Starting from (1, 0)
-> It goes down to (2, 0). This move costs rowCosts[2] = 3.
-> It goes right to (2, 1). This move costs colCosts[1] = 2.
-> It goes right to (2, 2). This move costs colCosts[2] = 6.
-> It goes right to (2, 3). This move costs colCosts[3] = 7.
The total cost is 3 + 2 + 6 + 7 = 18

Example 2:

Input: startPos = [0, 0], homePos = [0, 0], rowCosts = [5], colCosts = [26]
Output: 0
Explanation: The robot is already at its home. Since no moves occur, the total cost is 0.

Constraints:

  • m == rowCosts.length
  • n == colCosts.length
  • 1 <= m, n <= 105
  • 0 <= rowCosts[r], colCosts[c] <= 104
  • startPos.length == 2
  • homePos.length == 2
  • 0 <= startrow, homerow < m
  • 0 <= startcol, homecol < n

Solution: Manhattan distance

Move directly to the goal, no back and forth. Cost will be the same no matter which path you choose.

ans = sum(rowCosts[y1+1~y2]) + sum(colCosts[x1+1~x2])

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

C++

花花酱 LeetCode 2086. Minimum Number of Buckets Required to Collect Rainwater from Houses

By zxi on November 27, 2021

You are given a 0-indexed string street. Each character in street is either 'H' representing a house or '.' representing an empty space.

You can place buckets on the empty spaces to collect rainwater that falls from the adjacent houses. The rainwater from a house at index i is collected if a bucket is placed at index i - 1 and/or index i + 1. A single bucket, if placed adjacent to two houses, can collect the rainwater from both houses.

Return the minimum number of buckets needed so that for every house, there is at least one bucket collecting rainwater from it, or -1 if it is impossible.

Example 1:

Input: street = "H..H"
Output: 2
Explanation:
We can put buckets at index 1 and index 2.
"H..H" -> "HBBH" ('B' denotes where a bucket is placed).
The house at index 0 has a bucket to its right, and the house at index 3 has a bucket to its left.
Thus, for every house, there is at least one bucket collecting rainwater from it.

Example 2:

Input: street = ".H.H."
Output: 1
Explanation:
We can put a bucket at index 2.
".H.H." -> ".HBH." ('B' denotes where a bucket is placed).
The house at index 1 has a bucket to its right, and the house at index 3 has a bucket to its left.
Thus, for every house, there is at least one bucket collecting rainwater from it.

Example 3:

Input: street = ".HHH."
Output: -1
Explanation:
There is no empty space to place a bucket to collect the rainwater from the house at index 2.
Thus, it is impossible to collect the rainwater from all the houses.

Example 4:

Input: street = "H"
Output: -1
Explanation:
There is no empty space to place a bucket.
Thus, it is impossible to collect the rainwater from the house.

Example 5:

Constraints:

  • 1 <= street.length <= 105
  • street[i] is either'H' or '.'.

Solution: Greedy

Try to put a bucket after a house if possible, otherwise put it before the house, or impossible.

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

C++

花花酱 LeetCode 2078. Two Furthest Houses With Different Colors

By zxi on November 20, 2021

There are n houses evenly lined up on the street, and each house is beautifully painted. You are given a 0-indexed integer array colors of length n, where colors[i] represents the color of the ith house.

Return the maximum distance between two houses with different colors.

The distance between the ith and jth houses is abs(i - j), where abs(x) is the absolute value of x.

Example 1:

Input: colors = [1,1,1,6,1,1,1]
Output: 3
Explanation: In the above image, color 1 is blue, and color 6 is red.
The furthest two houses with different colors are house 0 and house 3.
House 0 has color 1, and house 3 has color 6. The distance between them is abs(0 - 3) = 3.
Note that houses 3 and 6 can also produce the optimal answer.

Example 2:

Input: colors = [1,8,3,8,3]
Output: 4
Explanation: In the above image, color 1 is blue, color 8 is yellow, and color 3 is green.
The furthest two houses with different colors are house 0 and house 4.
House 0 has color 1, and house 4 has color 3. The distance between them is abs(0 - 4) = 4.

Example 3:

Input: colors = [0,1]
Output: 1
Explanation: The furthest two houses with different colors are house 0 and house 1.
House 0 has color 0, and house 1 has color 1. The distance between them is abs(0 - 1) = 1.

Constraints:

  • n == colors.length
  • 2 <= n <= 100
  • 0 <= colors[i] <= 100
  • Test data are generated such that at least two houses have different colors.

Solution 1: Brute Force

Try all pairs.
Time complexity: O(n2)
Space complexity: O(1)

C++

Solution 2: Greedy / One pass

First house or last house must be involved in the ans.

Scan the house and check with first and last house.

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

C++

花花酱 LeetCode 2037. Minimum Number of Moves to Seat Everyone

By zxi on October 18, 2021

There are n seats and n students in a room. You are given an array seats of length n, where seats[i] is the position of the ith seat. You are also given the array students of length n, where students[j] is the position of the jth student.

You may perform the following move any number of times:

  • Increase or decrease the position of the ith student by 1 (i.e., moving the ith student from position x to x + 1 or x - 1)

Return the minimum number of moves required to move each student to a seat such that no two students are in the same seat.

Note that there may be multiple seats or students in the same position at the beginning.

Example 1:

Input: seats = [3,1,5], students = [2,7,4]
Output: 4
Explanation: The students are moved as follows:
- The first student is moved from from position 2 to position 1 using 1 move.
- The second student is moved from from position 7 to position 5 using 2 moves.
- The third student is moved from from position 4 to position 3 using 1 move.
In total, 1 + 2 + 1 = 4 moves were used.

Example 2:

Input: seats = [4,1,5,9], students = [1,3,2,6]
Output: 7
Explanation: The students are moved as follows:
- The first student is not moved.
- The second student is moved from from position 3 to position 4 using 1 move.
- The third student is moved from from position 2 to position 5 using 3 moves.
- The fourth student is moved from from position 6 to position 9 using 3 moves.
In total, 0 + 1 + 3 + 3 = 7 moves were used.

Example 3:

Input: seats = [2,2,6,6], students = [1,3,2,6]
Output: 4
Explanation: The students are moved as follows:
- The first student is moved from from position 1 to position 2 using 1 move.
- The second student is moved from from position 3 to position 6 using 3 moves.
- The third student is not moved.
- The fourth student is not moved.
In total, 1 + 3 + 0 + 0 = 4 moves were used.

Constraints:

  • n == seats.length == students.length
  • 1 <= n <= 100
  • 1 <= seats[i], students[j] <= 100

Solution: Greedy

Sort both arrays, move students[i] to seats[i].

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

C++

花花酱 LeetCode 1899. Merge Triplets to Form Target Triplet

By zxi on August 11, 2021

A triplet is an array of three integers. You are given a 2D integer array triplets, where triplets[i] = [ai, bi, ci] describes the ith triplet. You are also given an integer array target = [x, y, z] that describes the triplet you want to obtain.

To obtain target, you may apply the following operation on triplets any number of times (possibly zero):

  • Choose two indices (0-indexed) i and j (i != j) and update triplets[j] to become [max(ai, aj), max(bi, bj), max(ci, cj)].
    • For example, if triplets[i] = [2, 5, 3] and triplets[j] = [1, 7, 5], triplets[j] will be updated to [max(2, 1), max(5, 7), max(3, 5)] = [2, 7, 5].

Return true if it is possible to obtain the target triplet [x, y, z] as an element of triplets, or false otherwise.

Example 1:

Input: triplets = [[2,5,3],[1,8,4],[1,7,5]], target = [2,7,5]
Output: true
Explanation: Perform the following operations:
- Choose the first and last triplets [[2,5,3],[1,8,4],[1,7,5]]. Update the last triplet to be [max(2,1), max(5,7), max(3,5)] = [2,7,5]. triplets = [[2,5,3],[1,8,4],[2,7,5]]
The target triplet [2,7,5] is now an element of triplets.

Example 2:

Input: triplets = [[1,3,4],[2,5,8]], target = [2,5,8]
Output: true
Explanation: The target triplet [2,5,8] is already an element of triplets.

Example 3:

Input: triplets = [[2,5,3],[2,3,4],[1,2,5],[5,2,3]], target = [5,5,5]
Output: true
Explanation: Perform the following operations:
- Choose the first and third triplets [[2,5,3],[2,3,4],[1,2,5],[5,2,3]]. Update the third triplet to be [max(2,1), max(5,2), max(3,5)] = [2,5,5]. triplets = [[2,5,3],[2,3,4],[2,5,5],[5,2,3]].
- Choose the third and fourth triplets [[2,5,3],[2,3,4],[2,5,5],[5,2,3]]. Update the fourth triplet to be [max(2,5), max(5,2), max(5,3)] = [5,5,5]. triplets = [[2,5,3],[2,3,4],[2,5,5],[5,5,5]].
The target triplet [5,5,5] is now an element of triplets.

Example 4:

Input: triplets = [[3,4,5],[4,5,6]], target = [3,2,5]
Output: false
Explanation: It is impossible to have [3,2,5] as an element because there is no 2 in any of the triplets.

Constraints:

  • 1 <= triplets.length <= 105
  • triplets[i].length == target.length == 3
  • 1 <= ai, bi, ci, x, y, z <= 1000

Solution: Greedy

Exclude those bad ones (whose values are greater than x, y, z), check the max value for each dimension or whether there is x, y, z for each dimension.

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

C++