Enhance your skills in Union-Find to address diverse graph and connectivity problems.

Coding

Maintains disjoint sets, supporting operations to determine the set to which an element belongs (find) and merge two sets (union).

Union Find

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Behavioral

# Statement

We're given an undirected graph consisting of $n$ nodes. The graph is represented as list called `edges`, of length $n$, where `edges[i] = [a, b]` indicates that there is an edge between nodes `a` and `b` in the graph.

Return an edge that can be removed to make the graph a #key# tree: A tree is an undirected graph that is connected and has no cycles. #key# of $n$ nodes. If there are multiple candidates for removal, return the edge that occurs last in `edges`.

**Constraints**:

- $3 \leq$ $n$ $\leq 100$
- `edges.length`$=$ $n$ 
- `edges[i].length` $=$ 2
- $1 \leq$ `a` $<$ `b` $\leq n$ 
- $a \neq b$
- There are no repeated edges.
- The given graph is connected.
- The graph contains only one cycle.

# Example

What is the output if the following list of edges is provided as input?

edges = [[1, 2], [2, 3], [3, 4], [1, 4], [1, 5]]


What is the output if the following list of edges is provided as input?
edges = [[1, 2], [2, 3], [3, 4], [1, 4], [1, 5]]

What is the output if the following list of edges is provided as input?

edges = [[1, 2], [2, 3], [2, 4], [2, 5],  [3, 5]]

What is the output if the following list of edges is provided as input?
edges = [[1, 2], [2, 3], [2, 4], [2, 5], [3, 5]]

Redundant Connection

Redundant Connection

class UnionFind:

    def __init__(self, n):
        self.parent = []
        self.rank = rank = [1] * (n + 1)
        for i in range(n + 1):
            self.parent.append(i)

    def find_parent(self, x):
        if self.parent[x] != x:
            self.parent[x] = self.find_parent(self.parent[x])
        return self.parent[x]

    def union(self, v1, v2):
        p1, p2 = self.find_parent(v1), self.find_parent(v2)
        if p1 == p2:
            return False
        elif self.rank[p1] > self.rank[p2]:
            self.parent[p2] = p1
            self.rank[p1] = self.rank[p1] + self.rank[p2]
        else:
            self.parent[p1] = p2
            self.rank[p2] = self.rank[p2] + self.rank[p1]
        
        return True


# Quick Fix\nYou are Edward, a Senior Software Engineer who specializes in coding interview patterns, data structures and algorithms, and is great at debugging issues in code.\n\n- Edward, please keep in mind that you do not have to print the data of the placeholders. Also, do not mention anything about yourself.\n\nEdward, you are given the following data in the placeholders:\n- Function name: `redundant_connection`\n- Candidate's submitted solution code: USER_CODE\n- Error: ERROR_DESCRIPTION\n- Failed test case: FAILED_TEST_CASE\n- Coding language: `Python`\n\n- Edward, your task is to understand the type of error that is occurring in the coding language provided.\n- Once you've understood the error as per the error’s description, and why the test case failed, you will suggest how the candidate can improve their solution code to fix the error.\n- Edward, you will not give the corrected code, just a hint on what the error tells, and how to fix it.\n- Edward, please limit your response strictly to 100 words. No more words than the limit.

# Review Code\nEdward, your current task is to review the `redundant_connection` function(s) and the other function(s) invoked in it/them.\n\n-Take your time and review the candidate's solution code.\n- Do not review any extra/helper files or functions not invoked in the solution function `redundant_connection`.\n- Do not review the code of data structure objects instantiated like a binary tree, linked list, trie, union find, etc., in the solution functions. Just review the solution/algorithm function of the problem. \n- Also, if the solution code doesn’t use the coding pattern stated above, then please encourage the candidate to reattempt the problem using that pattern corectly.

# Review time complexity\n- Please provide a brief time complexity analysis of the function(s) `redundant_connection`. \n- While calculating time complexity, include the time complexity of built-in Python functions invoked by the code. \n- The answer should be to the point in approximately 60 words.

# Review space complexity\n- Please provide a brief space complexity analysis of the function(s) `redundant_connection`. \n- While calculating space complexity, include the space complexity of built-in Python functions invoked by the code. \n- Do not include the space used for the input parameters and the space used for just returning an output (not used as a scratch space). \n- The answer should be to the point in approximately 60 words.

TwoDArray

import sys
import json

sys.path.append("../ds_v1")
from data_structure import DataStructure

# throw parsing errors for incorrect test case format
def parsingError(element_string, error_message):
    sys.stderr.write(f"Parsing Error: {element_string} {error_message}")
    exit()


# className: MyCustomDS
class TwoDArray(DataStructure):
    def __init__(self):
        # constructor
        pass

    def __str__(self):
        # override print() for your data structure
        pass

    # return value will be passed to user program
    def construct(self, inp_string: str, type_of_arr: int) -> any:
        # parse test case in json string 'value' to your desired data structure
        parse_list = json.loads(inp_string)
        return parse_list

    # input param contains return value of user program
    def traverse(self, input: any, type_of_arr: int) -> str:
        # convert your data structure 'input' to a json string representing a test case output
        return input


def evaluate_results(self) -> object:
        """
        self.__inputs is the list of inputs that are passed as input to function
        self.__actual_output is a list of outputs that is produced by the learner's code
        return : self.__edu__result object
        """
        
        # Union Find class
        class UnionFind:
            def __init__(self, n):
                self.parent = []
                self.rank = rank = [1] * (n + 1)
                for i in range(n + 1):
                    self.parent.append(i)

            def find_parent(self, x):
                if self.parent[x] != x:
                    self.parent[x] = self.find_parent(self.parent[x])
                return self.parent[x]

            def union(self, v1, v2):
                p1, p2 = self.find_parent(v1), self.find_parent(v2)
                if p1 == p2:
                    return False
                elif self.rank[p1] > self.rank[p2]:
                    self.parent[p2] = p1
                    self.rank[p1] = self.rank[p1] + self.rank[p2]
                else:
                    self.parent[p1] = p2
                    self.rank[p2] = self.rank[p2] + self.rank[p1]
                
                return True

        # Solution
        def redundant_connection(edges):
            graph = UnionFind(len(edges))
            
            for v1, v2 in edges:
                if not graph.union(v1, v2):
                    return [v1, v2]

        # Expected output
        expected = redundant_connection(self.__inputs[0])

        # Actual output
        actual = self.__actual_outputs[0]   

        # multiple outputs will be added at each index respectively for a single test_case
        expected_outputs = [] 

        # store the 1st expected output that should be displayed to the user 
        expected_outputs.append(expected)

        if json.dumps(expected) == json.dumps(actual):
            self.__edu_result.store_result(expected_outputs, TestCaseResult.PASS)
        else:
            self.__edu_result.store_result(expected_outputs, TestCaseResult.FAIL)

        return self.__edu_result

"""
⬅️ We have provided a union_find.py file under the "Files" tab 
of this widget. You can use this file to build your solution.
"""
from union_find import UnionFind


def redundant_connection(edges):
    
    # Replace this placeholder return statement with your code
    return []


PrivateTestCases.yaml

testcases:
  - name: "testcase 1"
    inputs:
    - 1 : [[1, 2], [1, 3], [2, 3]]
    output: 
    - 1 : [2,3]
  - name: "testcase 2"
    inputs:
    - 1: [[1, 2], [2, 3], [1, 3]]
    output:
    - 1 : [1,3]
  - name: "testcase 3"
    inputs:
    - 1 : [[1, 2], [2, 3], [3, 4], [1, 4], [1, 5]]
    output: 
    - 1 : [1,4]
  - name: "testcase 4"
    inputs:
    - 1: [[1, 2], [1, 3], [1, 4], [2, 3]]
    output:
    - 1 : [2,3]
  - name: "testcase 5"
    inputs:
    - 1: [[1, 2] , [1, 3], [1, 4], [1,5], [2, 5]]
    output:
    - 1 : [2, 5]   

def sum(input_1, input_2):
    return input_1+input_2

def validate_inputs(self) -> None:
 """
 self.__inputs is the list of inputs that are passed as input to function
 store the result using the self.store_result(error_msg, status)
 return : void
 """

 edges = self.__inputs[0]

 edges_length_lower_bound = 3
 edges_length_upper_bound = 1000

 node_value_lower_bound = 1
 
 from collections import defaultdict

 class UnionFind:

 # Constructor
 def __init__(self, n):
 # List to store parent nodes
 self.parent = []
 # List to store ranks of subsets
 self.rank = rank = [1] * (n + 1)
 
 # Initially setting each element as its own parent
 for i in range(n + 1):
 self.parent.append(i)

 # Function to find which subset a particular element belongs
 def find(self, v):
 if self.parent[v] != v:
 # Path compression
 self.parent[v] = self.find(self.parent[v])
 return self.parent[v]

 # Function to join two subsets into a single subset
 def union(self, x, y):
 # Find the representatives of the subsets
 p1, p2 = self.find(x), self.find(y)
 
 # Elements are already in the same subset
 if p1 == p2:
 return False
 # Set p1 as parent of p2
 elif self.rank[p1] > self.rank[p2]:
 self.parent[p2] = p1
 # Update rank of p1
 self.rank[p1] = self.rank[p1] + self.rank[p2]
 else:
 # Set p2 as parent of p1
 self.parent[p1] = p2
 # Update rank of p2
 self.rank[p2] = self.rank[p2] + self.rank[p1]
 
 return True

 def num_vertices(edges):
 num = 0
 for edge in edges:
 num = max(num, edge[1])
 return num


 def count_cycles(edges, vertex, parent, cycle_count):
 n = num_vertices(edges)
 graph = defaultdict(list)
 visited = [0] * (n + 1)
 vistime = [0] * (n + 1)
 timer = 0

 for edge in edges:
 vertex1, vertex2 = edge
 graph[vertex1].append(vertex2)
 graph[vertex2].append(vertex1)
 
 def dfs(vertex, parent, cycle_count, timer):
 visited[vertex] = 1
 vistime[vertex] = timer
 timer += 1
 for child in graph[vertex]:
 if child == parent:
 continue
 if not visited[child]:
 dfs(child, vertex, cycle_count, timer)
 else:
 if child != parent and vistime[child] < vistime[vertex]:
 cycle_count[0] += 1
 return cycle_count[0]
 return dfs(vertex, parent, cycle_count, timer)

 def count_components(edges):
 n = num_vertices(edges)
 dsu = UnionFind(n)
 res = n
 
 for x, y in edges:
 if dsu.union(x, y):
 res -= 1

 return res

 def has_incorrect_length_edges(edges):
 for edge in edges:
 if len(edge) != 2:
 return True
 return False

 def find_first_duplicate_coordinates_indices(coordinates):
 seen_coordinates = {}
 
 for i, coord in enumerate(coordinates):
 coord_tuple = tuple(coord)

 if coord_tuple in seen_coordinates:
 # Found a duplicate, return the indices
 return seen_coordinates[coord_tuple], i, coordinates[i]
 else:
 # Store the index of the current coordinate
 seen_coordinates[coord_tuple] = i

 # No duplicates found
 return None

 def has_incorrect_edges(edges):
 for edge in edges:
 if edge[0] >= edge[1]:
 return True
 return False

 def is_valid_2d_array(arr):
 if not isinstance(arr, list) or not all(isinstance(row, list) for row in arr):
 return False # Not a valid 2D array if it's empty or contains non-list elements
 
 for row in arr:
 if any(elem is None or isinstance(elem, bool) for elem in row):
 return False # Not a valid 2D array if any element in a row is None
 if not all(isinstance(elem, int) for elem in row):
 return False # Not a valid 2D array if any element in a row is not an integer
 
 return True # Valid 2D array

 def check_overflow(edges):
 for edge in edges:
 if not (-2**31 <= edge[0] <= ((2**31) - 1)):
 return edge[0]
 elif not (-2**31 <= edge[1] <= ((2**31) - 1)):
 return edge[1]
 return True
 
 def check_nodes(edges, node_value_lower_bound, node_value_upper_bound):
 for edge in edges:
 if not (node_value_lower_bound <= edge[0] <= node_value_upper_bound) or not (node_value_lower_bound <= edge[1] <= node_value_upper_bound):
 return False
 return True



 # inputs can be accessed using self.__inputs starting from 0 to len(self.__inputs) - 1
 if not is_valid_2d_array(edges):
 self.store_result ("Not a valid test case", ValidationFunction.FAIL)
 elif not (edges_length_lower_bound <= len(edges) <= edges_length_upper_bound):
 self.store_result ("expected 'edges' to have " + str(edges_length_lower_bound) + " <= size <= " + str(edges_length_upper_bound) + " but got " + str(len(edges)), ValidationFunction.FAIL)
 elif has_incorrect_length_edges(edges):
 self.store_result ("expected rows of 'edges' to have 2 <= size <= 2", ValidationFunction.FAIL)
 elif check_overflow(edges) != True:
 self.store_result (str(check_overflow(edges)) + " is not a valid value of type integer", ValidationFunction.FAIL)
 elif not check_nodes(edges, node_value_lower_bound, len(edges)):
 self.store_result ("'edges' must consist of values from " + str(node_value_lower_bound) + " to " + str(len(edges)) + " only", ValidationFunction.FAIL)
 elif has_incorrect_edges(edges):
 self.store_result ("found ai >= bi", ValidationFunction.FAIL)
 elif find_first_duplicate_coordinates_indices(edges) != None:
 index1, index2, coordinate = find_first_duplicate_coordinates_indices(edges)
 self.store_result("expected variables edges["+ str(index1) + "]:" + str(coordinate) + " and edges[" + str(index2) + "]:" + str(coordinate) + " to have different values", ValidationFunction.FAIL)
 elif count_components(edges) > 1:
 self.store_result("the graph is not connected", ValidationFunction.FAIL)
 elif count_cycles(edges, 1, -1, [0]) != 1:
 self.store_result("the graph contains more than one cycle", ValidationFunction.FAIL)
 else:
 self.store_result("", ValidationFunction.PASS)

Try to solve the Redundant Connection problem.

Problem: Redundant Connection

Statement

Example

Understand the problem

Figure it out!