November 28, 2021 – My Programming Notes

Daily Archives: November 28, 2021

C# || How To Find The Largest Component Size By Common Factor Using C#

November 28, 2021 admin No comments

The following is a module with functions which demonstrates how to find the largest component size by common factor using C#.

1. Largest Component Size – Problem Statement

You are given an integer array of unique positive integers nums. Consider the following graph:

There are nums.length nodes, labeled nums[0] to nums[nums.length – 1],
There is an undirected edge between nums[i] and nums[j] if nums[i] and nums[j] share a common factor greater than 1.

Return the size of the largest connected component in the graph.

Example 1:

Input: nums = [4,6,15,35] Output: 4

Example 2:

Input: nums = [20,50,9,63] Output: 2

Example 3:

Input: nums = [2,3,6,7,4,12,21,39] Output: 8

2. Largest Component Size – Solution

The following is a solution which demonstrates how to find the largest component size by common factor.

The following solution uses a union find set to group connections together.





		
		
			2. Largest Component Size - Solution
			
C#
			
			// ============================================================================
//    Author: Kenneth Perkins
//    Date:   Nov 28, 2021
//    Taken From: http://programmingnotes.org/
//    File:  Solution.cs
//    Description: Demonstrates how to find largest common factor
// ============================================================================
public class Solution {
    public int LargestComponentSize(int[] nums) {
        var set = new UnionFindSet(nums.Max() + 1);

        // Calculate primes set for all elements
        foreach (var num in nums) {
            for (var k = 2; k * k <= num; ++k) {
                if (num % k == 0) {
                    set.Union(num, k);
                    set.Union(num, num / k);
                }
            }
        }

        // Count the apperance of parents, return the maxium one.
        // All connected nodes will point to same parent
        var map = new Dictionary<int, int>();
        var result = 1;
        foreach (var num in nums) {
            var count = set.Find(num);
            map[count] = (map.ContainsKey(count) ? map[count] : 0) + 1;
            result = Math.Max(result, map[count]);
        }
        return result;
    }

    public class UnionFindSet {
        private int[] parent;

        public UnionFindSet(int size) {
            parent = new int[size];
            for (int i = 0; i < parent.Length; ++i) {
                parent[i] = i;
            }
        }
        public void Union(int x, int y) {
            parent[Find(x)] = parent[Find(y)];
        }
        public int Find(int x) {
            if (parent[x] != x) {
                parent[x] = Find(parent[x]);
            }
            return parent[x];
        }
    }
}// http://programmingnotes.org/
			
				
					
				
					1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
				
						// ============================================================================
//    Author: Kenneth Perkins
//    Date:   Nov 28, 2021
//    Taken From: http://programmingnotes.org/
//    File:  Solution.cs
//    Description: Demonstrates how to find largest common factor
// ============================================================================
public class Solution {
    public int LargestComponentSize(int[] nums) {
        var set = new UnionFindSet(nums.Max() + 1);
 
        // Calculate primes set for all elements
        foreach (var num in nums) {
            for (var k = 2; k * k <= num; ++k) {
                if (num % k == 0) {
                    set.Union(num, k);
                    set.Union(num, num / k);
                }
            }
        }
 
        // Count the apperance of parents, return the maxium one.
        // All connected nodes will point to same parent
        var map = new Dictionary<int, int>();
        var result = 1;
        foreach (var num in nums) {
            var count = set.Find(num);
            map[count] = (map.ContainsKey(count) ? map[count] : 0) + 1;
            result = Math.Max(result, map[count]);
        }
        return result;
    }
 
    public class UnionFindSet {
        private int[] parent;
 
        public UnionFindSet(int size) {
            parent = new int[size];
            for (int i = 0; i < parent.Length; ++i) {
                parent[i] = i;
            }
        }
        public void Union(int x, int y) {
            parent[Find(x)] = parent[Find(y)];
        }
        public int Find(int x) {
            if (parent[x] != x) {
                parent[x] = Find(parent[x]);
            }
            return parent[x];
        }
    }
}// http://programmingnotes.org/

QUICK NOTES:
The highlighted lines are sections of interest to look out for.

The code is heavily commented, so no further insight is necessary. If you have any questions, feel free to leave a comment below.

Once compiled, you should get this as your output for the example cases:

4 2 8

C# || Accounts Merge – How To Merge A List Of Emails Using C#

November 28, 2021 admin No comments

The following is a module with functions which demonstrates how to merge a list of emails using C#.

1. Accounts Merge – Problem Statement

Given a list of accounts where each element accounts[i] is a list of strings, where the first element accounts[i][0] is a name, and the rest of the elements are emails representing emails of the account.

Now, we would like to merge these accounts. Two accounts definitely belong to the same person if there is some common email to both accounts. Note that even if two accounts have the same name, they may belong to different people as people could have the same name. A person can have any number of accounts initially, but all of their accounts definitely have the same name.

After merging the accounts, return the accounts in the following format: the first element of each account is the name, and the rest of the elements are emails in sorted order. The accounts themselves can be returned in any order.

Example 1:

Input: accounts = [["John","[email protected]","[email protected]"],["John","[email protected]","[email protected]"],["Mary","[email protected]"],["John","[email protected]"]] Output: [["John","[email protected]","[email protected]","[email protected]"],["Mary","[email protected]"],["John","[email protected]"]] Explanation: The first and second John's are the same person as they have the common email "[email protected]". The third John and Mary are different people as none of their email addresses are used by other accounts. We could return these lists in any order, for example the answer [['Mary', '[email protected]'], ['John', '[email protected]'], ['John', '[email protected]', '[email protected]', '[email protected]']] would still be accepted.

Example 2:

Input: accounts = [["Gabe","[email protected]","[email protected]","[email protected]"],["Kevin","[email protected]","[email protected]","[email protected]"],["Ethan","[email protected]","[email protected]","[email protected]"],["Hanzo","[email protected]","[email protected]","[email protected]"],["Fern","[email protected]","[email protected]","[email protected]"]] Output: [["Ethan","[email protected]","[email protected]","[email protected]"],["Gabe","[email protected]","[email protected]","[email protected]"],["Hanzo","[email protected]","[email protected]","[email protected]"],["Kevin","[email protected]","[email protected]","[email protected]"],["Fern","[email protected]","[email protected]","[email protected]"]]

2. Accounts Merge – Solution

The following is a solution which demonstrates how to merge a list of emails.

The following solution uses a union find set to group accounts with matching emails together.





		
		
			2. Accounts Merge - Solution
			
C#
			
			// ============================================================================
//    Author: Kenneth Perkins
//    Date:   Nov 28, 2021
//    Taken From: http://programmingnotes.org/
//    File:  Solution.cs
//    Description: Demonstrates how to merge a list of emails
// ============================================================================
public class Solution {
    public IList<IList<string>> AccountsMerge(IList<IList<string>> accounts) {
        var set = new UnionFindSet(accounts.Count);

        // Map email to their component index
        var emailComponents = new Dictionary<string, int>();
        for (int i = 0; i < accounts.Count; ++i) {
            var account = accounts[i];

            for (int j = 1; j < account.Count; ++j) {
                var email = account[j];

                // Assign component group as the account index
                var group = i;
                if (!emailComponents.ContainsKey(email)) {
                    emailComponents[email] = group;
                } else {
                    // Union this group with the previous group of the email
                    set.Union(group, emailComponents[email]);
                }
            }
        }

        // Store emails corresponding to the components parent
        var emailGroups = new Dictionary<int, List<string>>();
        foreach (var email in emailComponents.Keys) {
            var group = emailComponents[email];
            var groupParent = set.Find(group);

            if (!emailGroups.ContainsKey(groupParent)) {
                emailGroups[groupParent] = new List<string>();
            }
            emailGroups[groupParent].Add(email);
        }

        // Sort the emails and add the account name
        var result = new List<IList<string>>();
        foreach (var group in emailGroups.Keys) {
            var emails = emailGroups[group];
            emails.Sort(StringComparer.Ordinal);
            emails.Insert(0, accounts[group][0]);
            result.Add(emails);
        }
        return result;
    }

    public class UnionFindSet {
        int[] parent;
        int[] size;

        public UnionFindSet(int count) {
            parent = new int[count];
            size = new int[count];
            for (int index = 0; index < count; ++index) {
                parent[index] = index;
                size[index] = 1;
            }
        }
        public int Find(int x) {
            if (parent[x] != x) {
                parent[x] = Find(parent[x]);
            }
            return parent[x];
        }
        public void Union(int x, int y) {
            var parentX = Find(x);
            var parentY = Find(y);
            if (parentX == parentY) {
                return;
            }
            if (size[parentX] >= size[parentY]) {
                size[parentX] += size[parentY];
                parent[parentY] = parentX;
            } else {
                size[parentY] += size[parentX];
                parent[parentX] = parentY;
            }
        }
    }
}// http://programmingnotes.org/
			
				
					
				
					1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
				
						// ============================================================================
//    Author: Kenneth Perkins
//    Date:   Nov 28, 2021
//    Taken From: http://programmingnotes.org/
//    File:  Solution.cs
//    Description: Demonstrates how to merge a list of emails
// ============================================================================
public class Solution {
    public IList<IList<string>> AccountsMerge(IList<IList<string>> accounts) {
        var set = new UnionFindSet(accounts.Count);
 
        // Map email to their component index
        var emailComponents = new Dictionary<string, int>();
        for (int i = 0; i < accounts.Count; ++i) {
            var account = accounts[i];
 
            for (int j = 1; j < account.Count; ++j) {
                var email = account[j];
 
                // Assign component group as the account index
                var group = i;
                if (!emailComponents.ContainsKey(email)) {
                    emailComponents[email] = group;
                } else {
                    // Union this group with the previous group of the email
                    set.Union(group, emailComponents[email]);
                }
            }
        }
 
        // Store emails corresponding to the components parent
        var emailGroups = new Dictionary<int, List<string>>();
        foreach (var email in emailComponents.Keys) {
            var group = emailComponents[email];
            var groupParent = set.Find(group);
 
            if (!emailGroups.ContainsKey(groupParent)) {
                emailGroups[groupParent] = new List<string>();
            }
            emailGroups[groupParent].Add(email);
        }
 
        // Sort the emails and add the account name
        var result = new List<IList<string>>();
        foreach (var group in emailGroups.Keys) {
            var emails = emailGroups[group];
            emails.Sort(StringComparer.Ordinal);
            emails.Insert(0, accounts[group][0]);
            result.Add(emails);
        }
        return result;
    }
 
    public class UnionFindSet {
        int[] parent;
        int[] size;
 
        public UnionFindSet(int count) {
            parent = new int[count];
            size = new int[count];
            for (int index = 0; index < count; ++index) {
                parent[index] = index;
                size[index] = 1;
            }
        }
        public int Find(int x) {
            if (parent[x] != x) {
                parent[x] = Find(parent[x]);
            }
            return parent[x];
        }
        public void Union(int x, int y) {
            var parentX = Find(x);
            var parentY = Find(y);
            if (parentX == parentY) {
                return;
            }
            if (size[parentX] >= size[parentY]) {
                size[parentX] += size[parentY];
                parent[parentY] = parentX;
            } else {
                size[parentY] += size[parentX];
                parent[parentX] = parentY;
            }
        }
    }
}// http://programmingnotes.org/

QUICK NOTES:
The highlighted lines are sections of interest to look out for.

The code is heavily commented, so no further insight is necessary. If you have any questions, feel free to leave a comment below.

Once compiled, you should get this as your output for the example cases:

[["John","[email protected]","[email protected]","[email protected]"],["Mary","[email protected]"],["John","[email protected]"]] [["Gabe","[email protected]","[email protected]","[email protected]"],["Kevin","[email protected]","[email protected]","[email protected]"],["Ethan","[email protected]","[email protected]","[email protected]"],["Hanzo","[email protected]","[email protected]","[email protected]"],["Fern","[email protected]","[email protected]","[email protected]"]]

C# || How To Find All Paths From Source To Target In Graph Using C#

November 28, 2021 admin No comments

The following is a module with functions which demonstrates how to find all paths from source to target in a graph using C#.

1. All Paths Source Target – Problem Statement

Given a directed acyclic graph (DAG) of n nodes labeled from 0 to n – 1, find all possible paths from node 0 to node n – 1 and return them in any order.

The graph is given as follows: graph[i] is a list of all nodes you can visit from node i (i.e., there is a directed edge from node i to node graph[i][j]).

Example 1:

Input: graph = [[1,2],[3],[3],[]] Output: [[0,1,3],[0,2,3]] Explanation: There are two paths: 0 -> 1 -> 3 and 0 -> 2 -> 3.

Example 2:

Input: graph = [[4,3,1],[3,2,4],[3],[4],[]] Output: [[0,4],[0,3,4],[0,1,3,4],[0,1,2,3,4],[0,1,4]]

Example 3:

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

Example 4:

Input: graph = [[1,2,3],[2],[3],[]] Output: [[0,1,2,3],[0,2,3],[0,3]]

Example 5:

Input: graph = [[1,3],[2],[3],[]] Output: [[0,1,2,3],[0,3]]

2. All Paths Source Target – Solution

The following is a solution which demonstrates how to find all paths from source to target in a graph.

This solution uses Breadth First Search and backtracking when looking for paths.





		
		
			2. All Paths Source Target - Solution
			
C#
			
			// ============================================================================
//    Author: Kenneth Perkins
//    Date:   Nov 28, 2021
//    Taken From: http://programmingnotes.org/
//    File:  Solution.cs
//    Description: Demonstrates how to find all paths in a graph
// ============================================================================
public class Solution {
    public IList<IList<int>> AllPathsSourceTarget(int[][] graph) {
        var result = new List<IList<int>>();
        var queue = new Queue<List<int>>();

        queue.Enqueue(new List<int>{0});

        while (queue.Count > 0) {
            var currentPath = queue.Dequeue();
            var currentNode = currentPath[currentPath.Count - 1];

            if (currentNode == graph.Length - 1) {
                result.Add(currentPath);
            } else {
                foreach (var child in graph[currentNode]) {
                    currentPath.Add(child);
                    queue.Enqueue(new List<int>(currentPath));
                    currentPath.RemoveAt(currentPath.Count - 1);
                }
            }
        }
        return result;
    }
}// http://programmingnotes.org/
			
				
					
				
					1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
				
						// ============================================================================
//    Author: Kenneth Perkins
//    Date:   Nov 28, 2021
//    Taken From: http://programmingnotes.org/
//    File:  Solution.cs
//    Description: Demonstrates how to find all paths in a graph
// ============================================================================
public class Solution {
    public IList<IList<int>> AllPathsSourceTarget(int[][] graph) {
        var result = new List<IList<int>>();
        var queue = new Queue<List<int>>();
 
        queue.Enqueue(new List<int>{0});
 
        while (queue.Count > 0) {
            var currentPath = queue.Dequeue();
            var currentNode = currentPath[currentPath.Count - 1];
 
            if (currentNode == graph.Length - 1) {
                result.Add(currentPath);
            } else {
                foreach (var child in graph[currentNode]) {
                    currentPath.Add(child);
                    queue.Enqueue(new List<int>(currentPath));
                    currentPath.RemoveAt(currentPath.Count - 1);
                }
            }
        }
        return result;
    }
}// http://programmingnotes.org/

QUICK NOTES:
The highlighted lines are sections of interest to look out for.

The code is heavily commented, so no further insight is necessary. If you have any questions, feel free to leave a comment below.

Once compiled, you should get this as your output for the example cases:

[[0,1,3],[0,2,3]] [[0,4],[0,3,4],[0,1,4],[0,1,3,4],[0,1,2,3,4]] [[0,1]] [[0,3],[0,2,3],[0,1,2,3]] [[0,3],[0,1,2,3]]

Daily Archives: November 28, 2021

C# || How To Find The Largest Component Size By Common Factor Using C#

C# || Accounts Merge – How To Merge A List Of Emails Using C#

C# || How To Find All Paths From Source To Target In Graph Using C#

Search

Compile Your Code Now

Translate Page

Popular Pages

Recent

Archives

Categories