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Satisfiability of Equality Equations

Medium

Microsoft

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Topics:

ArraysStringsGraphsGreedy Algorithms

You are given an array of strings equations that represent relationships between variables where each string equations[i] is of length 4 and takes one of two different forms: "x_i==y_i" or "x_i!=y_i".Here, x_i and y_i are lowercase letters (not necessarily different) that represent one-letter variable names.

Return true if it is possible to assign integers to variable names so as to satisfy all the given equations, or false otherwise.

Example 1:

Input: equations = ["a==b","b!=a"]
Output: false
Explanation: If we assign say, a = 1 and b = 1, then the first equation is satisfied, but not the second.
There is no way to assign the variables to satisfy both equations.

Example 2:

Input: equations = ["b==a","a==b"]
Output: true
Explanation: We could assign a = 1 and b = 1 to satisfy both equations.

Constraints:

1 <= equations.length <= 500
equations[i].length == 4
equations[i][0] is a lowercase letter.
equations[i][1] is either '=' or '!'.
equations[i][2] is '='.
equations[i][3] is a lowercase letter.

Satisfiability of Equality Equations

Medium

Microsoft

0 views

Topics:

ArraysStringsGraphsGreedy Algorithms

Return true if it is possible to assign integers to variable names so as to satisfy all the given equations, or false otherwise.

Example 1:

Input: equations = ["a==b","b!=a"]
Output: false
Explanation: If we assign say, a = 1 and b = 1, then the first equation is satisfied, but not the second.
There is no way to assign the variables to satisfy both equations.

Example 2:

Input: equations = ["b==a","a==b"]
Output: true
Explanation: We could assign a = 1 and b = 1 to satisfy both equations.

Constraints:

1 <= equations.length <= 500
equations[i].length == 4
equations[i][0] is a lowercase letter.
equations[i][1] is either '=' or '!'.
equations[i][2] is '='.
equations[i][3] is a lowercase letter.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the maximum number of equations I should expect in the input array?
Are the variable names in the equations limited to single lowercase letters?
If the equations are contradictory (e.g., 'a==b' and 'a!=b'), should I return false immediately, or process all equations before determining satisfiability?
What should I return if the input array is empty or null?
Is it possible for an equation to compare the same variable to itself with '!=' (e.g., 'a!=a')?

Brute Force Solution

Approach

We want to check if a bunch of equations (like a==b, b!=c) can all be true at the same time. A brute force approach tries out all possible scenarios for whether variables are equal or not to find a satisfying combination. It methodically checks if each potential arrangement makes all given equations true.

Here's how the algorithm would work step-by-step:

First, consider every possible way the variables could be equal or not equal to each other.
For each of these possibilities, go through the list of equations.
Check if the current possibility makes each equation true. For example, if a==b is in the list, and we assumed a and b are equal in this possibility, then this equation is satisfied.
If you find an equation that's NOT true under the current possibility, then this whole possibility is bad. Discard it and move on to the next one.
If all the equations are true for a specific possibility, then you've found a solution, meaning the original set of equations is satisfiable.
If you've gone through ALL the possibilities and none of them work, that means no combination of variable equalities can satisfy the equations, so the original set of equations is not satisfiable.

Code Implementation

def solve_equations_brute_force(equations):
    # Extract all unique variables from the equations
    variables = set()
    for equation in equations:
        variables.add(equation[0])
        variables.add(equation[3])

    variables = list(variables)
    num_variables = len(variables)

    # Iterate through all possible equality scenarios
    for i in range(2**(num_variables*(num_variables-1)//2)):
        variable_equality = {}
        temp_value = i

        # Determine the equality relationships based on the binary representation of i.
        index = 0
        for first_variable_index in range(num_variables):
            for second_variable_index in range(first_variable_index + 1, num_variables):
                variable_equality[(variables[first_variable_index], variables[second_variable_index])] = temp_value % 2
                temp_value //= 2

        # Check if the current equality scenario satisfies all equations
        satisfies_all = True
        for equation in equations:
            first_variable = equation[0]
            operator = equation[1:3]
            second_variable = equation[3]

            if first_variable == second_variable:
                if operator == '==':
                    continue
                else:
                    satisfies_all = False
                    break
            else:
                # Retrieve equality for a pair of variables
                equal = are_equal(first_variable, second_variable, variable_equality, variables)
                if operator == '==':
                    if not equal:
                        satisfies_all = False
                        break
                else:
                    if equal:
                        satisfies_all = False
                        break

        # If the current scenario satisfies all equations, return True
        if satisfies_all:
            return True

    # If no scenario satisfies all equations, return False
    return False

def are_equal(first_variable, second_variable, variable_equality, variables):
    if (first_variable, second_variable) in variable_equality:
        return variable_equality[(first_variable, second_variable)] == 1
    elif (second_variable, first_variable) in variable_equality:
        return variable_equality[(second_variable, first_variable)] == 1
    else:
        # Need to recursively check for transitive equality.
        for other_variable in variables:
            if other_variable != first_variable and other_variable != second_variable:
                if are_equal(first_variable, other_variable, variable_equality, variables) and \
                   are_equal(other_variable, second_variable, variable_equality, variables):
                    return True
        return False

Big(O) Analysis

Time Complexity

O(2^(n*m)) – Let n be the number of unique variables and m be the number of equations. The algorithm explores all possible equality assignments among the n variables. There are potentially n*(n-1)/2 or approximately n*n/2 possible pairs of variables to consider for equality. For each pair, a decision must be made whether the variables are equal or not, leading to 2^(n*n/2) or 2^(n*m) possible configurations of equalities. For each of these configurations, the algorithm checks all m equations. Therefore, the total time complexity is O(m * 2^(n*m)). Since 2^(n*m) grows faster than m, we simplify to O(2^(n*m)).

Space Complexity

O(1) – The brute force approach, as described, explores different possibilities for variable equalities. The explanation doesn't explicitly mention using any auxiliary data structures like lists, hash maps, or sets to store these possibilities. It implicitly suggests iterating through all possible scenarios which could be tracked using a constant number of variables to represent the current 'possibility' being evaluated. Therefore, the auxiliary space used is constant, independent of the number of equations or variables.

Optimal Solution

Approach

This problem involves determining if a set of equality and inequality equations can be true at the same time. The clever trick is to treat the equality equations as instructions for grouping things together, and then check if any of the inequality equations create a contradiction within those groups.

Here's how the algorithm would work step-by-step:

Imagine each variable is a separate object. The goal is to figure out if all the equations make sense together.
First, go through all the equality equations (the ones with '=='). For each of these, merge the objects that are said to be equal into a single group. Think of it like combining LEGO bricks that are supposed to be the same.
After grouping all the equal objects, go through all the inequality equations (the ones with '!=').
For each inequality equation, check if the two objects being compared are in the same group. If they are, it means the equation is saying that something is NOT equal to itself, which is a contradiction. So, immediately determine that the set of equations cannot be true.
If you go through all the inequality equations and never find a contradiction (an object not being equal to itself), then all the equations can be true together.

Code Implementation

def solve_equality_equations(equations):
    parent = {}

    def find(variable):
        if variable not in parent:
            parent[variable] = variable
        if parent[variable] != variable:
            parent[variable] = find(parent[variable])
        return parent[variable]

    def union(variable_one, variable_two):
        root_one = find(variable_one)
        root_two = find(variable_two)
        parent[root_one] = root_two

    # Process equality equations to group variables.
    for equation in equations:
        if '==' in equation:
            variable_one, variable_two = equation.split('==')
            union(variable_one, variable_two)

    # Check for contradictions in inequality equations.
    for equation in equations:
        if '!=' in equation:
            variable_one, variable_two = equation.split('!=')
            # If two unequal variables have the same root, there's a contradiction.
            if find(variable_one) == find(variable_two):

                return False

    return True

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input equations twice. First, it processes equality equations for grouping, and then it processes inequality equations to check for contradictions. Assuming the union-find operations within the grouping are nearly constant time due to path compression and union by rank (or treated as amortized constant time), each iteration through the n equations takes O(n) time. Therefore, the overall time complexity is O(n) + O(n), which simplifies to O(n).

Space Complexity

O(26) – The primary space complexity comes from the need to group variables based on equality. In the worst case, each variable is distinct. Since the variables are represented by lowercase English letters, there are at most 26 unique variables. We need a data structure (implicitly a list or dictionary) to store the group/parent each variable belongs to, and this data structure will have a maximum size of 26 (or a small constant), one entry for each possible variable. Therefore, the auxiliary space used is O(26) which simplifies to O(1) since it is constant regardless of the number of equations.

Edge Cases

Case	How to Handle
Empty input equation list	Return true immediately as there are no constraints to violate.
Input list contains only one equation	Process the single equation and return its truthiness.
Input contains equations like 'a==a' and 'b!=b'	These equations should be handled correctly; 'a==a' is always true, and 'b!=b' is always false, potentially leading to early termination.
Input contains contradictory equations such as 'a==b' and 'a!=b'	The algorithm should detect this contradiction and return false.
Input has a large number of variables and equations	Ensure the chosen algorithm (e.g., Union-Find) scales efficiently, avoiding quadratic time complexity.
Input contains equations with only inequality constraints	Process equality constraints first before applying inequality constraints.
Equations use a large number of distinct variables	Ensure sufficient memory allocation for the data structures representing variables and their relationships.
Integer overflow when representing variable indices	Use appropriate data types (e.g., long) to represent variable indices or consider using a hash map to map variables to indices.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the maximum number of equations I should expect in the input array?
Are the variable names in the equations limited to single lowercase letters?
If the equations are contradictory (e.g., 'a==b' and 'a!=b'), should I return false immediately, or process all equations before determining satisfiability?
What should I return if the input array is empty or null?
Is it possible for an equation to compare the same variable to itself with '!=' (e.g., 'a!=a')?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

First, consider every possible way the variables could be equal or not equal to each other.
For each of these possibilities, go through the list of equations.
Check if the current possibility makes each equation true. For example, if a==b is in the list, and we assumed a and b are equal in this possibility, then this equation is satisfied.
If you find an equation that's NOT true under the current possibility, then this whole possibility is bad. Discard it and move on to the next one.
If all the equations are true for a specific possibility, then you've found a solution, meaning the original set of equations is satisfiable.
If you've gone through ALL the possibilities and none of them work, that means no combination of variable equalities can satisfy the equations, so the original set of equations is not satisfiable.

Code Implementation

def solve_equations_brute_force(equations):
    # Extract all unique variables from the equations
    variables = set()
    for equation in equations:
        variables.add(equation[0])
        variables.add(equation[3])

    variables = list(variables)
    num_variables = len(variables)

    # Iterate through all possible equality scenarios
    for i in range(2**(num_variables*(num_variables-1)//2)):
        variable_equality = {}
        temp_value = i

        # Determine the equality relationships based on the binary representation of i.
        index = 0
        for first_variable_index in range(num_variables):
            for second_variable_index in range(first_variable_index + 1, num_variables):
                variable_equality[(variables[first_variable_index], variables[second_variable_index])] = temp_value % 2
                temp_value //= 2

        # Check if the current equality scenario satisfies all equations
        satisfies_all = True
        for equation in equations:
            first_variable = equation[0]
            operator = equation[1:3]
            second_variable = equation[3]

            if first_variable == second_variable:
                if operator == '==':
                    continue
                else:
                    satisfies_all = False
                    break
            else:
                # Retrieve equality for a pair of variables
                equal = are_equal(first_variable, second_variable, variable_equality, variables)
                if operator == '==':
                    if not equal:
                        satisfies_all = False
                        break
                else:
                    if equal:
                        satisfies_all = False
                        break

        # If the current scenario satisfies all equations, return True
        if satisfies_all:
            return True

    # If no scenario satisfies all equations, return False
    return False

def are_equal(first_variable, second_variable, variable_equality, variables):
    if (first_variable, second_variable) in variable_equality:
        return variable_equality[(first_variable, second_variable)] == 1
    elif (second_variable, first_variable) in variable_equality:
        return variable_equality[(second_variable, first_variable)] == 1
    else:
        # Need to recursively check for transitive equality.
        for other_variable in variables:
            if other_variable != first_variable and other_variable != second_variable:
                if are_equal(first_variable, other_variable, variable_equality, variables) and \
                   are_equal(other_variable, second_variable, variable_equality, variables):
                    return True
        return False

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

Imagine each variable is a separate object. The goal is to figure out if all the equations make sense together.
First, go through all the equality equations (the ones with '=='). For each of these, merge the objects that are said to be equal into a single group. Think of it like combining LEGO bricks that are supposed to be the same.
After grouping all the equal objects, go through all the inequality equations (the ones with '!=').
For each inequality equation, check if the two objects being compared are in the same group. If they are, it means the equation is saying that something is NOT equal to itself, which is a contradiction. So, immediately determine that the set of equations cannot be true.
If you go through all the inequality equations and never find a contradiction (an object not being equal to itself), then all the equations can be true together.

Code Implementation

def solve_equality_equations(equations):
    parent = {}

    def find(variable):
        if variable not in parent:
            parent[variable] = variable
        if parent[variable] != variable:
            parent[variable] = find(parent[variable])
        return parent[variable]

    def union(variable_one, variable_two):
        root_one = find(variable_one)
        root_two = find(variable_two)
        parent[root_one] = root_two

    # Process equality equations to group variables.
    for equation in equations:
        if '==' in equation:
            variable_one, variable_two = equation.split('==')
            union(variable_one, variable_two)

    # Check for contradictions in inequality equations.
    for equation in equations:
        if '!=' in equation:
            variable_one, variable_two = equation.split('!=')
            # If two unequal variables have the same root, there's a contradiction.
            if find(variable_one) == find(variable_two):

                return False

    return True

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty input equation list	Return true immediately as there are no constraints to violate.
Input list contains only one equation	Process the single equation and return its truthiness.
Input contains equations like 'a==a' and 'b!=b'	These equations should be handled correctly; 'a==a' is always true, and 'b!=b' is always false, potentially leading to early termination.
Input contains contradictory equations such as 'a==b' and 'a!=b'	The algorithm should detect this contradiction and return false.
Input has a large number of variables and equations	Ensure the chosen algorithm (e.g., Union-Find) scales efficiently, avoiding quadratic time complexity.
Input contains equations with only inequality constraints	Process equality constraints first before applying inequality constraints.
Equations use a large number of distinct variables	Ensure sufficient memory allocation for the data structures representing variables and their relationships.
Integer overflow when representing variable indices	Use appropriate data types (e.g., long) to represent variable indices or consider using a hash map to map variables to indices.