'''
CMSI 3300 - Classwork 1
Author: SOLUTION

Complete each exercise as described in the Classwork spec, ensuring that its subsequent
unit test is satisfied by running the associated review_tests.py as indicated in the spec.
'''

from typing import *
import itertools


def get_conflict(dict1: dict[str, bool], dict2: dict[str, bool]) -> Optional[str]:
    '''
    Given 2 dictionaries of string keys mapped to boolean values, returns the first
    key that is mapped to True in one dictionary but False in the other. In the event
    of no such "conflicts," returns None, and if multiple conflicts exists, can return
    any one of them.
    
    Parameters:
        dict1, dict2: dict[str, bool]:
            Dictionaries of string keys mapped to boolean values.
    
    Returns:
          Optional[str]:
              The first key with a conflicting boolean value between dictionaries, or
              None if no such key exists. 
        
    Examples:
        get_conflict({"A": True, "B": False, "C": True}, {"A": False, "B": False, "C": True}) => "A"
        get_conflict({"A": True}, {"B": False}) => None
    '''
    for symbol, truth_val in dict1.items():
        if symbol in dict2 and not dict2[symbol] == truth_val:
            return symbol
    return None


def count_set_member_pairs(sets: list[set[str]]) -> int:
    '''
    Given a list of sets of strings, counts the number of pairs that can be made between
    these sets where a pair can be formed whenever at least one string is common between
    both sets. Should count each possible pair exactly once.
    
    [!] Hint: use a particular method in the itertools package to simplify your life!
    
    Parameters:
        sets: list[set[str]]:
            A list of sets of strings.
    
    Returns:
          int:
              The number of pairs that could be formed between sets in the given list.
        
    Examples:
        count_set_member_pairs([
            {"A"},
            {"B", "A"},
            {"C", "B"}
        ]) => 2     # (the first two and last two sets can be paired)
        
        count_set_member_pairs([
            {"A", "C"},
            {"B", "A"},
            {"C", "B"},
            {"A", "B", "C"}
        ]) => 6    # (can you count all of the pairs?)
        
        self.assertEqual(0, count_set_member_pairs([
            {"A", "D"},
            {"B", "E"},
            {"C", "G"},
            {"X", "Y", "Z"}
        ])) => 0    # (no common strings in these sets)
    '''
    combos = itertools.combinations(sets, 2)
    count = 0
    
    # Note: beware playing programming golf below and using something like set intersection
    # or the "any" method, each of which will not stop comparing set elements after the first
    # match is found, which is computationally wasteful
    for set1, set2 in combos:
        for str1 in set1:
            if str1 in set2:
                count += 1
                break
    return count