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

Given `m = 3`, `n = 3`, and `positions = [[0,0], [0,1], [1,2], [2,1]]`, what is the output?

Given <code>m = 3</code>, <code>n = 3</code>, and <code>positions = [[0,0], [0,1], [1,2], [2,1]]</code>, what is the output?

Given `m = 3`, `n = 3`, and `positions = [[0,0], [0,1], [1,2], [2,1], [1,1]]`, what is the output?

Given <code>m = 3</code>, <code>n = 3</code>, and <code>positions = [[0,0], [0,1], [1,2], [2,1], [1,1]]</code>, what is the output?

Given `m = 2`, `n = 2` and `positions = [[0,0], [1,1]]`, what is the output?

Given <code>m = 2</code>, <code>n = 2</code> and <code>positions = [[0,0], [1,1]]</code>, what is the output?

Number of Islands II

Number of Islands II

class UnionFind:
 def __init__(self):
 self.parent = {}
 self.size = {}
 self.count = 0 # number of connected components (islands)

 def addLand(self, x):
 """Add a new node (land cell) if not already present."""
 if x not in self.parent:
 self.parent[x] = x
 self.size[x] = 1
 self.count += 1

 def find(self, x):
 """Find with path compression."""
 if self.parent[x] != x:
 self.parent[x] = self.find(self.parent[x])
 return self.parent[x]

 def union(self, a, b):
 """Union two nodes. If merged, island count decreases by 1."""
 ra, rb = self.find(a), self.find(b)
 if ra == rb:
 return False # already in same component
 # Union by size
 if self.size[ra] < self.size[rb]:
 ra, rb = rb, ra
 self.parent[rb] = ra
 self.size[ra] += self.size[rb]
 self.count -= 1
 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- Edward, please keep in mind that you do not have to print the data of the placeholders. Also, do not mention anything about yourself. \nEdward, you are given the following data in the placeholders: \n- Function name: `numIslands2` \n- User’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 understand the error using the error’s description, and why the test case failed. You will suggest how the learner 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 `numIslands2` function(s) and the other function(s) invoked in it/them.  \n-Take your time and review the user’s solution code. \n- Do not review any extra/helper files or functions not invoked in the solution function `numIslands2`. \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 user to reattempt the problem, to correctly use the coding pattern."

"# Review time complexity \n- Please provide a brief time complexity analysis of the function(s) `numIslands2`.  \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 and in a few lines."

"# Review space complexity \n- Please provide a brief space complexity analysis of the function(s) `numIslands2`.  \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 and in a few lines."

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: TwoDArray
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, value: str, type_of_arr: int) -> any:
        # parse test case in json string 'value' to your desired data structure
        return json.loads(value)

    # 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 json.dumps(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
 """
 
 class UnionFind:
 def __init__(self):
 self.parent = {}
 self.size = {}
 self.count = 0 # number of connected components (islands)
 
 def add(self, x):
 """Add a new node (land cell) if not already present."""
 if x not in self.parent:
 self.parent[x] = x
 self.size[x] = 1
 self.count += 1
 
 def find(self, x):
 """Find with path compression."""
 if self.parent[x] != x:
 self.parent[x] = self.find(self.parent[x])
 return self.parent[x]
 
 def union(self, a, b):
 """Union two nodes. If merged, island count decreases by 1."""
 ra, rb = self.find(a), self.find(b)
 if ra == rb:
 return False # already in same component
 # Union by size
 if self.size[ra] < self.size[rb]:
 ra, rb = rb, ra
 self.parent[rb] = ra
 self.size[ra] += self.size[rb]
 self.count -= 1
 return True 
 
 
 def numIslands2(m, n, positions):
 uf = UnionFind()
 land = set() 
 answer = []
 
 def nodeId(r, c):
 return r * n + c
 
 DIRS = [(1, 0), (-1, 0), (0, 1), (0, -1)]
 
 for r, c in positions:
 if (r, c) in land:
 answer.append(uf.count)
 continue
 
 land.add((r, c))
 node = nodeId(r, c)
 uf.add(node)
 
 for dr, dc in DIRS:
 nr, nc = r + dr, c + dc
 if 0 <= nr < m and 0 <= nc < n and (nr, nc) in land:
 uf.union(node, nodeId(nr, nc))
 
 answer.append(uf.count)
 
 return answer
 
 
 # multiple outputs will be added at each index respectively for a single test_case
 expected_outputs = [] 
 m = self.__inputs[0]
 n = self.__inputs[1]
 positions = self.__inputs[2]
 
 # store the 1st expected output that should be displayed to the user 
 expected = numIslands2(m, n, positions)
 expected_outputs.append(expected)
 
 actual = self.__actual_outputs[0]

 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

from UnionFind import UnionFind

def numIslands2(m, n, positions):

    # Replace this placeholder return statement with your code
    return []

Number_of_Islands_II.yaml

testcases:
  - name: "testcase 1"
    inputs:
    - 1 : 3
    - 2 : 3
    - 3 : [[0,0], [2,2], [1,2], [1,1], [1,0]]
    output: 
    - 1 : [1, 2, 2, 2, 1]

  - name: "testcase 2"
    inputs:
    - 1: 1
    - 2: 5
    - 3: [[0,0], [0,2], [0,1], [0,4], [0,3]]
    output:
    - 1: [1, 2, 1, 2, 1]
  
  - name: "testcase 3"
    inputs:
    - 1: 4
    - 2: 4
    - 3: [[1,1], [1,2], [2,2], [0,0], [3,3], [2,3]]
    output:
    - 1: [1, 1, 1, 2, 3, 2]
  
  - name: "testcase 4"
    inputs:
    - 1: 2
    - 2: 3
    - 3: [[0, 0], [1, 2], [0, 1], [1, 1]]
    output:
    - 1: [1, 2, 2, 1]
  
  - name: "testcase 5"
    inputs:
    - 1: 5
    - 2: 5
    - 3: [[4,4], [4,3], [3,3], [2,3], [2,2], [2,1], [0,0]]
    output:
    - 1: [1, 1, 1, 1, 1, 1, 2]


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
 """
 # inputs can be accessed using self.__inputs starting from 0 to len(self.__inputs) - 1
 if not isinstance(self.__inputs[0], int) or not isinstance(self.__inputs[1], int):
 self.store_result("Inputs 1 (m) and 2 (n) must be integers", ValidationFunction.FAIL)
 
 elif self.__inputs[0] < 1 or self.__inputs[1] < 1 or self.__inputs[0] > 10000 or self.__inputs[1] > 10000:
 self.store_result("Inputs 1 (m) and 2 (n) must each be between 1 and 10^4", ValidationFunction.FAIL)
 
 elif self.__inputs[0] * self.__inputs[1] > 10000:
 self.store_result("Product of m * n must be ≤ 10^4", ValidationFunction.FAIL)
 
 elif not isinstance(self.__inputs[2], list):
 self.store_result("Input 3 (positions) must be a list", ValidationFunction.FAIL)
 
 elif len(self.__inputs[2]) < 1 or len(self.__inputs[2]) > 10000:
 self.store_result("Input 3 (positions) length must be between 1 and 10^4", ValidationFunction.FAIL)
 
 elif any(not isinstance(p, list) or len(p) != 2 for p in self.__inputs[2]):
 self.store_result("Each position must be a 2-element list [ri, ci]", ValidationFunction.FAIL)
 
 elif any(not isinstance(p[0], int) or not isinstance(p[1], int) or 
 p[0] < 0 or p[0] >= self.__inputs[0] or 
 p[1] < 0 or p[1] >= self.__inputs[1] 
 for p in self.__inputs[2]):
 self.store_result("Each position must satisfy 0 <= ri < m and 0 <= ci < n", ValidationFunction.FAIL)
 
 else:
 self.store_result("", ValidationFunction.PASS)

Try to solve the Number of Islands II problem.

Problem: Number of Islands II

Statements

Examples

Understand the problem

Figure it out!