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Posts tagged as “medium”

花花酱 LeetCode 1334. Find the City With the Smallest Number of Neighbors at a Threshold Distance

There are n cities numbered from 0 to n-1. Given the array edges where edges[i] = [fromi, toi, weighti] represents a bidirectional and weighted edge between cities fromi and toi, and given the integer distanceThreshold.

Return the city with the smallest numberof cities that are reachable through some path and whose distance is at most distanceThreshold, If there are multiple such cities, return the city with the greatest number.

Notice that the distance of a path connecting cities i and j is equal to the sum of the edges’ weights along that path.

Example 1:

Input: n = 4, edges = [[0,1,3],[1,2,1],[1,3,4],[2,3,1]], distanceThreshold = 4
Output: 3
Explanation: The figure above describes the graph. 
The neighboring cities at a distanceThreshold = 4 for each city are:
City 0 -> [City 1, City 2] 
City 1 -> [City 0, City 2, City 3] 
City 2 -> [City 0, City 1, City 3] 
City 3 -> [City 1, City 2] 
Cities 0 and 3 have 2 neighboring cities at a distanceThreshold = 4, but we have to return city 3 since it has the greatest number.

Example 2:

Input: n = 5, edges = [[0,1,2],[0,4,8],[1,2,3],[1,4,2],[2,3,1],[3,4,1]], distanceThreshold = 2
Output: 0
Explanation: The figure above describes the graph. 
The neighboring cities at a distanceThreshold = 2 for each city are:
City 0 -> [City 1] 
City 1 -> [City 0, City 4] 
City 2 -> [City 3, City 4] 
City 3 -> [City 2, City 4]
City 4 -> [City 1, City 2, City 3] 
The city 0 has 1 neighboring city at a distanceThreshold = 2.

Constraints:

  • 2 <= n <= 100
  • 1 <= edges.length <= n * (n - 1) / 2
  • edges[i].length == 3
  • 0 <= fromi < toi < n
  • 1 <= weighti, distanceThreshold <= 10^4
  • All pairs (fromi, toi) are distinct.

Solution: Floyd-Warshall

All pair shortest path

Time complexity: O(n^3)
Space complexity: O(n^2)

C++

花花酱 LeetCode 1333. Filter Restaurants by Vegan-Friendly, Price and Distance

Given the array restaurants where  restaurants[i] = [idi, ratingi, veganFriendlyi, pricei, distancei]. You have to filter the restaurants using three filters.

The veganFriendly filter will be either true (meaning you should only include restaurants with veganFriendlyi set to true) or false (meaning you can include any restaurant). In addition, you have the filters maxPrice and maxDistance which are the maximum value for price and distance of restaurants you should consider respectively.

Return the array of restaurant IDs after filtering, ordered by rating from highest to lowest. For restaurants with the same rating, order them by id from highest to lowest. For simplicity veganFriendlyi and veganFriendly take value 1 when it is true, and 0 when it is false.

Example 1:

Input: restaurants = [[1,4,1,40,10],[2,8,0,50,5],[3,8,1,30,4],[4,10,0,10,3],[5,1,1,15,1]], veganFriendly = 1, maxPrice = 50, maxDistance = 10
Output: [3,1,5] 
Explanation: 
The restaurants are:
Restaurant 1 [id=1, rating=4, veganFriendly=1, price=40, distance=10]
Restaurant 2 [id=2, rating=8, veganFriendly=0, price=50, distance=5]
Restaurant 3 [id=3, rating=8, veganFriendly=1, price=30, distance=4]
Restaurant 4 [id=4, rating=10, veganFriendly=0, price=10, distance=3]
Restaurant 5 [id=5, rating=1, veganFriendly=1, price=15, distance=1] 
After filter restaurants with veganFriendly = 1, maxPrice = 50 and maxDistance = 10 we have restaurant 3, restaurant 1 and restaurant 5 (ordered by rating from highest to lowest). 

Example 2:

Input: restaurants = [[1,4,1,40,10],[2,8,0,50,5],[3,8,1,30,4],[4,10,0,10,3],[5,1,1,15,1]], veganFriendly = 0, maxPrice = 50, maxDistance = 10
Output: [4,3,2,1,5]
Explanation: The restaurants are the same as in example 1, but in this case the filter veganFriendly = 0, therefore all restaurants are considered.

Example 3:

Input: restaurants = [[1,4,1,40,10],[2,8,0,50,5],[3,8,1,30,4],[4,10,0,10,3],[5,1,1,15,1]], veganFriendly = 0, maxPrice = 30, maxDistance = 3
Output: [4,5]

Constraints:

  • 1 <= restaurants.length <= 10^4
  • restaurants[i].length == 5
  • 1 <= idi, ratingi, pricei, distance<= 10^5
  • 1 <= maxPrice, maxDistance <= 10^5
  • veganFriendlyi and veganFriendly are 0 or 1.
  • All idi are distinct.

Solution

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

C++

花花酱 LeetCode 1329. Sort the Matrix Diagonally

Given a m * n matrix mat of integers, sort it diagonally in ascending order from the top-left to the bottom-right then return the sorted array.

Example 1:

Input: mat = [[3,3,1,1],[2,2,1,2],[1,1,1,2]]
Output: [[1,1,1,1],[1,2,2,2],[1,2,3,3]]

Constraints:

  • m == mat.length
  • n == mat[i].length
  • 1 <= m, n <= 100
  • 1 <= mat[i][j] <= 100

Solution: HashTable

Collect each diagonal’s (keyed by i – j) elements into an array and sort it separately.
If we offset the key by n, e.g. i – j + n, we can use an array instead of a hashtable.

Time complexity: O(m*n + (m+n) * (m+n) * log(m + n))) = (n^2*logn)
Space complexity: O(m*n)

C++

花花酱 LeetCode 1328. Break a Palindrome

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++

花花酱 LeetCode 1325. Delete Leaves With a Given Value

Given a binary tree root and an integer target, delete all the leaf nodes with value target.

Note that once you delete a leaf node with value targetif it’s parent node becomes a leaf node and has the value target, it should also be deleted (you need to continue doing that until you can’t).

Example 1:

Input: root = [1,2,3,2,null,2,4], target = 2
Output: [1,null,3,null,4]
Explanation: Leaf nodes in green with value (target = 2) are removed (Picture in left). 
After removing, new nodes become leaf nodes with value (target = 2) (Picture in center).

Example 2:

Input: root = [1,3,3,3,2], target = 3
Output: [1,3,null,null,2]

Example 3:

Input: root = [1,2,null,2,null,2], target = 2
Output: [1]
Explanation: Leaf nodes in green with value (target = 2) are removed at each step.

Example 4:

Input: root = [1,1,1], target = 1
Output: []

Example 5:

Input: root = [1,2,3], target = 1
Output: [1,2,3]

Constraints:

  • 1 <= target <= 1000
  • Each tree has at most 3000 nodes.
  • Each node’s value is between [1, 1000].

Solution: Recursion

Post-order traversal

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

C++