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Check for Contradictions in Equations

Hard

Amazon

1 view

Topics:

GraphsArrays

You are given a list of equations represented as strings. Each equation consists of two variables (represented by lowercase letters) and an operator ('==' or '!=') indicating the relationship between them. The task is to determine whether there is any contradiction within the set of equations.

For example:

equations = [["a","==","b"],["b","!=","a"]] should return true because it states that 'a' is equal to 'b' and 'b' is not equal to 'a', which is a contradiction.
equations = [["a","==","b"],["b","==","c"],["a","==","c"]] should return false because there is no contradiction. 'a' is equal to 'b', 'b' is equal to 'c', and 'a' is equal to 'c', which is consistent.
equations = [["a","==","b"],["b","==","c"],["a","!=","c"]] should return true because 'a' is equal to 'b', 'b' is equal to 'c', implying 'a' should be equal to 'c', but the equation states 'a' is not equal to 'c', which is a contradiction.
What is the most efficient way to determine if there are any contradictions in the system of equations? Explain the algorithm, including time and space complexity, and provide example code.

Check for Contradictions in Equations

Hard

Amazon

1 view

Topics:

GraphsArrays

For example:

equations = [["a","==","b"],["b","!=","a"]] should return true because it states that 'a' is equal to 'b' and 'b' is not equal to 'a', which is a contradiction.
equations = [["a","==","b"],["b","==","c"],["a","==","c"]] should return false because there is no contradiction. 'a' is equal to 'b', 'b' is equal to 'c', and 'a' is equal to 'c', which is consistent.
equations = [["a","==","b"],["b","==","c"],["a","!=","c"]] should return true because 'a' is equal to 'b', 'b' is equal to 'c', implying 'a' should be equal to 'c', but the equation states 'a' is not equal to 'c', which is a contradiction.
What is the most efficient way to determine if there are any contradictions in the system of equations? Explain the algorithm, including time and space complexity, and provide example code.

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 range of values for the constants in the equations? Are they integers, floats, or another data type?
Can the coefficients in the equations be zero? What happens if a coefficient is zero in an equation like '0 * x + y = 5'?
If a contradiction is found, what should the function return? Should it be a boolean (`true` if a contradiction exists, `false` otherwise`), an error message string, or something else?
How are the equations represented in the input? Is it a list of strings, a custom object, or something else?
Are there any limits on the number of variables or equations in the input? For example, can I assume it's always just two variables, x and y?

Brute Force Solution

Approach

The brute force way to find contradictions is to try all possible value assignments to variables. We check each assignment against all the equations to see if any equation is violated. If even one assignment violates an equation, there's a contradiction.

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

First, make a list of all the different variables involved in your equations.
Then, for each variable, consider all possible numbers it could be equal to.
Start trying combinations of values for the variables. Imagine assigning a number to the first variable, then a number to the second variable, and so on.
For each complete set of values assigned to the variables, check if those values satisfy every single equation.
If even one equation is not true using those assigned values, then that combination of variable assignments doesn't work, and there's a potential contradiction.
Keep doing this process, trying every possible combination of numbers for the variables.
If you try all combinations, and in every case at least one equation is never satisfied, then you have found a definite contradiction. If there is even one combination that satisfies every equation, then there is no contradiction.

Code Implementation

def check_for_contradictions(equations):    all_variables = set()
    for equation in equations:
        for variable in equation:
            all_variables.add(variable)

    all_variables_list = list(all_variables)
    number_of_variables = len(all_variables_list)

    # Explore up to this number to assign to variables, exhaustive search
    for value1 in range(-5, 6): 
        for value2 in range(-5, 6):
            for value3 in range(-5, 6):
                variable_values = {}

                # Assign the tested values to each variable present
                if number_of_variables >= 1:
                    variable_values[all_variables_list[0]] = value1
                if number_of_variables >= 2:
                    variable_values[all_variables_list[1]] = value2
                if number_of_variables >= 3:
                    variable_values[all_variables_list[2]] = value3
                
                satisfies_all_equations = True

                # Verify if the assigned values satisfy the equations
                for equation in equations:
                    if len(equation) == 3:
                        left_side = variable_values.get(equation[0], 0)
                        operator = equation[1]
                        right_side = variable_values.get(equation[2], 0)

                        if operator == '=':
                            if left_side != right_side:
                                satisfies_all_equations = False
                                break
                        elif operator == '>':
                            if left_side <= right_side:
                                satisfies_all_equations = False
                                break
                        elif operator == '<':
                            if left_side >= right_side:
                                satisfies_all_equations = False
                                break
                    elif len(equation) == 1:
                        if not variable_values.get(equation[0], True):
                            satisfies_all_equations = False
                            break

                # If an assignment is found, it means no contradiction
                if satisfies_all_equations:
                    return False

    # If no assignments satisfies equations, there is contradiction
    return True

Big(O) Analysis

Time Complexity

O(V^N) – The described brute force approach involves assigning values to N distinct variables. Let's say each variable can take on V possible values. We are essentially exploring all possible combinations of values for the variables. This means we have V choices for the first variable, V choices for the second, and so on, up to V choices for the Nth variable, leading to V * V * ... * V (N times) possible combinations. After choosing the assignment, we check if these values satisfy the equations. Therefore, the time complexity is driven by the need to evaluate V to the power of N different assignments making the overall runtime O(V^N).

Space Complexity

O(V * R^V) – The described brute force approach requires storing a list of unique variables which takes O(V) space, where V is the number of unique variables in the equations. The algorithm then iterates through all possible assignments of values to these variables. If each variable can take R possible values, the total number of combinations is R^V. Checking each combination does not require significant extra memory, but the algorithm conceptually explores all possible combinations of values which contributes to the potential combinations that must be considered. Because it recursively assigns the R values to each of the V variables, it could have a recursion stack of size V at any given time. Therefore, the space complexity can be thought of as influenced by the number of variables and the range of their possible values, leading to O(V * R^V).

Optimal Solution

Approach

The most efficient way to solve this problem is to treat each equation as a relationship between variables and use a technique to find if relationships contradict each other. We will use a data structure that efficiently represents these relationships, and an algorithm that quickly finds inconsistencies.

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

Imagine each variable in the equations as a separate node in a network.
Whenever you see an equation saying two variables are equal, link their nodes together in the network. This means they belong to the same group, or represent the same value.
When you come across an equation saying two variables are NOT equal, check if those two variables are already in the same group in the network.
If the two variables that are not equal are already in the same group, it means we've found a contradiction!
If you process all the equations without finding any contradictions, it means the equations are consistent.

Code Implementation

def check_for_contradictions(equations):

    parent = {}

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

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

    for equation_type, variable_one, variable_two in equations:
        if equation_type == "equal":
            union(variable_one, variable_two)

    # Need a second pass to check for contradictions after union
    for equation_type, variable_one, variable_two in equations:

        if equation_type == "not_equal":
            # If variables are in same set, contradiction found
            if find(variable_one) == find(variable_two):

                return True

    return False

Big(O) Analysis

Time Complexity

O(n α(n)) – The input consists of n equations. We iterate through each equation once. For equations asserting equality, we perform a union operation, which, using union-by-rank and path compression, has a time complexity of approximately α(n) where α(n) is the inverse Ackermann function, which grows extremely slowly and is effectively constant for all practical input sizes. For equations asserting inequality, we perform a find operation twice to check if the variables are in the same connected component, each find operation also taking α(n) time. Thus, the total time complexity is driven by iterating through the n equations and performing a near-constant α(n) operation for each, resulting in O(n α(n)). Because α(n) grows so slowly, it's often treated as constant.

Space Complexity

O(N) – The auxiliary space is dominated by the data structure used to represent the network of variable relationships. In the worst case, this network can store a separate node for each variable encountered in the input equations, where N is the number of unique variables across all equations. The union-find data structure (used to track groups of connected variables) typically requires an array of size N to store the parent pointers or rank/size information for each node. Therefore, the space complexity is O(N).

Edge Cases

Case	How to Handle
Empty input equations list	Return true (no contradictions) as there are no equations to contradict each other.
Equations list contains only a single equation	Process the single equation normally; if it leads to a contradiction (e.g., x = y and x != y), return false, otherwise true.
Equations list contains redundant equations (e.g., x = y, y = x)	Union-find or other equivalence checking should handle this without error, and the result will be consistent.
Equation variables contain cycles (e.g., x = y, y = z, z = x but x != x)	Union-find collapses the connected components; the final check of equality/inequality will catch contradictions.
Large number of variables and equations, potentially leading to memory exhaustion	Use efficient data structures (e.g., hash maps) and consider garbage collection to minimize memory footprint, potentially failing if memory limits are truly exceeded.
Conflicting equations (e.g., x = y and x != y) where x and y are the same variable.	The algorithm should detect that the assigned groups don't align during the equality/inequality check and immediately return false.
Equations refer to undefined variables that have no other constraints.	Introduce these variables into the Union-Find data structure as standalone sets; they will not influence conflict detection until other equations involve them.
Integer overflow when representing variable IDs	Use sufficiently large integer types (e.g., long) or use hashing to map variable names to indices to avoid overflow.

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 range of values for the constants in the equations? Are they integers, floats, or another data type?
Can the coefficients in the equations be zero? What happens if a coefficient is zero in an equation like '0 * x + y = 5'?
If a contradiction is found, what should the function return? Should it be a boolean (`true` if a contradiction exists, `false` otherwise`), an error message string, or something else?
How are the equations represented in the input? Is it a list of strings, a custom object, or something else?
Are there any limits on the number of variables or equations in the input? For example, can I assume it's always just two variables, x and y?

Brute Force Solution

Approach

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

First, make a list of all the different variables involved in your equations.
Then, for each variable, consider all possible numbers it could be equal to.
Start trying combinations of values for the variables. Imagine assigning a number to the first variable, then a number to the second variable, and so on.
For each complete set of values assigned to the variables, check if those values satisfy every single equation.
If even one equation is not true using those assigned values, then that combination of variable assignments doesn't work, and there's a potential contradiction.
Keep doing this process, trying every possible combination of numbers for the variables.
If you try all combinations, and in every case at least one equation is never satisfied, then you have found a definite contradiction. If there is even one combination that satisfies every equation, then there is no contradiction.

Code Implementation

def check_for_contradictions(equations):    all_variables = set()
    for equation in equations:
        for variable in equation:
            all_variables.add(variable)

    all_variables_list = list(all_variables)
    number_of_variables = len(all_variables_list)

    # Explore up to this number to assign to variables, exhaustive search
    for value1 in range(-5, 6): 
        for value2 in range(-5, 6):
            for value3 in range(-5, 6):
                variable_values = {}

                # Assign the tested values to each variable present
                if number_of_variables >= 1:
                    variable_values[all_variables_list[0]] = value1
                if number_of_variables >= 2:
                    variable_values[all_variables_list[1]] = value2
                if number_of_variables >= 3:
                    variable_values[all_variables_list[2]] = value3
                
                satisfies_all_equations = True

                # Verify if the assigned values satisfy the equations
                for equation in equations:
                    if len(equation) == 3:
                        left_side = variable_values.get(equation[0], 0)
                        operator = equation[1]
                        right_side = variable_values.get(equation[2], 0)

                        if operator == '=':
                            if left_side != right_side:
                                satisfies_all_equations = False
                                break
                        elif operator == '>':
                            if left_side <= right_side:
                                satisfies_all_equations = False
                                break
                        elif operator == '<':
                            if left_side >= right_side:
                                satisfies_all_equations = False
                                break
                    elif len(equation) == 1:
                        if not variable_values.get(equation[0], True):
                            satisfies_all_equations = False
                            break

                # If an assignment is found, it means no contradiction
                if satisfies_all_equations:
                    return False

    # If no assignments satisfies equations, there is contradiction
    return True

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Imagine each variable in the equations as a separate node in a network.
Whenever you see an equation saying two variables are equal, link their nodes together in the network. This means they belong to the same group, or represent the same value.
When you come across an equation saying two variables are NOT equal, check if those two variables are already in the same group in the network.
If the two variables that are not equal are already in the same group, it means we've found a contradiction!
If you process all the equations without finding any contradictions, it means the equations are consistent.

Code Implementation

def check_for_contradictions(equations):

    parent = {}

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

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

    for equation_type, variable_one, variable_two in equations:
        if equation_type == "equal":
            union(variable_one, variable_two)

    # Need a second pass to check for contradictions after union
    for equation_type, variable_one, variable_two in equations:

        if equation_type == "not_equal":
            # If variables are in same set, contradiction found
            if find(variable_one) == find(variable_two):

                return True

    return False

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty input equations list	Return true (no contradictions) as there are no equations to contradict each other.
Equations list contains only a single equation	Process the single equation normally; if it leads to a contradiction (e.g., x = y and x != y), return false, otherwise true.
Equations list contains redundant equations (e.g., x = y, y = x)	Union-find or other equivalence checking should handle this without error, and the result will be consistent.
Equation variables contain cycles (e.g., x = y, y = z, z = x but x != x)	Union-find collapses the connected components; the final check of equality/inequality will catch contradictions.
Large number of variables and equations, potentially leading to memory exhaustion	Use efficient data structures (e.g., hash maps) and consider garbage collection to minimize memory footprint, potentially failing if memory limits are truly exceeded.
Conflicting equations (e.g., x = y and x != y) where x and y are the same variable.	The algorithm should detect that the assigned groups don't align during the equality/inequality check and immediately return false.
Equations refer to undefined variables that have no other constraints.	Introduce these variables into the Union-Find data structure as standalone sets; they will not influence conflict detection until other equations involve them.
Integer overflow when representing variable IDs	Use sufficiently large integer types (e.g., long) or use hashing to map variable names to indices to avoid overflow.