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Posts tagged as “graph”

花花酱 LeetCode 1584. Min Cost to Connect All Points

You are given an array points representing integer coordinates of some points on a 2D-plane, where points[i] = [xi, yi].

The cost of connecting two points [xi, yi] and [xj, yj] is the manhattan distance between them: |xi - xj| + |yi - yj|, where |val| denotes the absolute value of val.

Return the minimum cost to make all points connected. All points are connected if there is exactly one simple path between any two points.

Example 1:

Input: points = [[0,0],[2,2],[3,10],[5,2],[7,0]]
Output: 20
Explanation:

We can connect the points as shown above to get the minimum cost of 20.
Notice that there is a unique path between every pair of points.

Example 2:

Input: points = [[3,12],[-2,5],[-4,1]]
Output: 18

Example 3:

Input: points = [[0,0],[1,1],[1,0],[-1,1]]
Output: 4

Example 4:

Input: points = [[-1000000,-1000000],[1000000,1000000]]
Output: 4000000

Example 5:

Input: points = [[0,0]]
Output: 0

Constraints:

  • 1 <= points.length <= 1000
  • -106 <= xi, yi <= 106
  • All pairs (xi, yi) are distinct.

Solution: Minimum Spanning Tree

Kruskal’s algorithm
Time complexity: O(n^2logn)
Space complexity: O(n^2)
using vector of vector, array, pair of pair, or tuple might lead to TLE…

C++

Prim’s Algorithm
ds[i] := min distance from i to ANY nodes in the tree.

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

C++

花花酱 LeetCode 1579. Remove Max Number of Edges to Keep Graph Fully Traversable

Alice and Bob have an undirected graph of n nodes and 3 types of edges:

  • Type 1: Can be traversed by Alice only.
  • Type 2: Can be traversed by Bob only.
  • Type 3: Can by traversed by both Alice and Bob.

Given an array edges where edges[i] = [typei, ui, vi] represents a bidirectional edge of type typei between nodes ui and vi, find the maximum number of edges you can remove so that after removing the edges, the graph can still be fully traversed by both Alice and Bob. The graph is fully traversed by Alice and Bob if starting from any node, they can reach all other nodes.

Return the maximum number of edges you can remove, or return -1 if it’s impossible for the graph to be fully traversed by Alice and Bob.

Example 1:

Input: n = 4, edges = [[3,1,2],[3,2,3],[1,1,3],[1,2,4],[1,1,2],[2,3,4]]
Output: 2
Explanation: If we remove the 2 edges [1,1,2] and [1,1,3]. The graph will still be fully traversable by Alice and Bob. Removing any additional edge will not make it so. So the maximum number of edges we can remove is 2.

Example 2:

Input: n = 4, edges = [[3,1,2],[3,2,3],[1,1,4],[2,1,4]]
Output: 0
Explanation: Notice that removing any edge will not make the graph fully traversable by Alice and Bob.

Example 3:

Input: n = 4, edges = [[3,2,3],[1,1,2],[2,3,4]]
Output: -1
Explanation: In the current graph, Alice cannot reach node 4 from the other nodes. Likewise, Bob cannot reach 1. Therefore it's impossible to make the graph fully traversable.

Constraints:

  • 1 <= n <= 10^5
  • 1 <= edges.length <= min(10^5, 3 * n * (n-1) / 2)
  • edges[i].length == 3
  • 1 <= edges[i][0] <= 3
  • 1 <= edges[i][1] < edges[i][2] <= n
  • All tuples (typei, ui, vi) are distinct.

Solution: Greedy + Spanning Tree / Union Find

Use type 3 (both) edges first.

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

C++

python3

花花酱 LeetCode 1568. Minimum Number of Days to Disconnect Island

Given a 2D grid consisting of 1s (land) and 0s (water).  An island is a maximal 4-directionally (horizontal or vertical) connected group of 1s.

The grid is said to be connected if we have exactly one island, otherwise is said disconnected.

In one day, we are allowed to change any single land cell (1) into a water cell (0).

Return the minimum number of days to disconnect the grid.

Example 1:

Input: grid = [[0,1,1,0],[0,1,1,0],[0,0,0,0]]
Output: 2
Explanation: We need at least 2 days to get a disconnected grid.
Change land grid[1][1] and grid[0][2] to water and get 2 disconnected island.

Example 2:

Input: grid = [[1,1]]
Output: 2
Explanation: Grid of full water is also disconnected ([[1,1]] -> [[0,0]]), 0 islands.

Example 3:

Input: grid = [[1,0,1,0]]
Output: 0

Example 4:

Input: grid = [[1,1,0,1,1],
               [1,1,1,1,1],
               [1,1,0,1,1],
               [1,1,0,1,1]]
Output: 1

Example 5:

Input: grid = [[1,1,0,1,1],
               [1,1,1,1,1],
               [1,1,0,1,1],
               [1,1,1,1,1]]
Output: 2

Constraints:

  • 1 <= grid.length, grid[i].length <= 30
  • grid[i][j] is 0 or 1.

Solution: Brute Force

We need at most two days to disconnect an island.
1. check if we have more than one islands. (0 days)
2. For each 1 cell, change it to 0 and check how many islands do we have. (1 days)
3. Otherwise, 2 days

Time complexity: O(m^2*n^2)
Space complexity: O(m*n)

C++

花花酱 LeetCode 1559. Detect Cycles in 2D Grid

Given a 2D array of characters grid of size m x n, you need to find if there exists any cycle consisting of the same value in grid.

A cycle is a path of length 4 or more in the grid that starts and ends at the same cell. From a given cell, you can move to one of the cells adjacent to it – in one of the four directions (up, down, left, or right), if it has the same value of the current cell.

Also, you cannot move to the cell that you visited in your last move. For example, the cycle (1, 1) -> (1, 2) -> (1, 1) is invalid because from (1, 2) we visited (1, 1) which was the last visited cell.

Return true if any cycle of the same value exists in grid, otherwise, return false.

Example 1:

Input: grid = [["a","a","a","a"],["a","b","b","a"],["a","b","b","a"],["a","a","a","a"]]
Output: true
Explanation: There are two valid cycles shown in different colors in the image below:

Example 2:

Input: grid = [["c","c","c","a"],["c","d","c","c"],["c","c","e","c"],["f","c","c","c"]]
Output: true
Explanation: There is only one valid cycle highlighted in the image below:

Example 3:

Input: grid = [["a","b","b"],["b","z","b"],["b","b","a"]]
Output: false

Constraints:

  • m == grid.length
  • n == grid[i].length
  • 1 <= m <= 500
  • 1 <= n <= 500
  • grid consists only of lowercase English letters.

Solution: DFS

Finding a cycle in an undirected graph => visiting a node that has already been visited and it’s not the parent node of the current node.
b b
b b
null -> (0, 0) -> (0, 1) -> (1, 1) -> (1, 0) -> (0, 0)
The second time we visit (0, 0) which has already been visited before and it’s not the parent of the current node (1, 0) ( (1, 0)’s parent is (1, 1) ) which means we found a cycle.

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

C++

花花酱 LeetCode 1557. Minimum Number of Vertices to Reach All Nodes

Given a directed acyclic graph, with n vertices numbered from 0 to n-1, and an array edges where edges[i] = [fromi, toi] represents a directed edge from node fromi to node toi.

Find the smallest set of vertices from which all nodes in the graph are reachable. It’s guaranteed that a unique solution exists.

Notice that you can return the vertices in any order.

Example 1:

Input: n = 6, edges = [[0,1],[0,2],[2,5],[3,4],[4,2]]
Output: [0,3]
Explanation: It's not possible to reach all the nodes from a single vertex. From 0 we can reach [0,1,2,5]. From 3 we can reach [3,4,2,5]. So we output [0,3].

Example 2:

Input: n = 5, edges = [[0,1],[2,1],[3,1],[1,4],[2,4]]
Output: [0,2,3]
Explanation: Notice that vertices 0, 3 and 2 are not reachable from any other node, so we must include them. Also any of these vertices can reach nodes 1 and 4.

Constraints:

  • 2 <= n <= 10^5
  • 1 <= edges.length <= min(10^5, n * (n - 1) / 2)
  • edges[i].length == 2
  • 0 <= fromi, toi < n
  • All pairs (fromi, toi) are distinct.

Solution: In degree

Nodes with 0 in degree will be the answer.
Time complexity: O(E+V)
Space complexity: O(V)

C++