Evaluate Boolean Expression - Practice Coding Problems

Evaluate Boolean Expression - Problem

Database Medium

Evaluate Boolean Expression

You are given two tables: Variables and Expressions. Your task is to evaluate boolean expressions by looking up variable values and applying comparison operators.

The Variables table stores variable names and their integer values:

name	value
x	66
y	77

The Expressions table contains boolean expressions to evaluate:

left_operand	operator	right_operand
x	<	y

Goal: For each expression, substitute the variable names with their values from the Variables table, then evaluate the comparison. Return 'true' or 'false' for each expression.

In the example above: x < y becomes 66 < 77 which evaluates to 'true'.

Input & Output

example_1.sql — Basic Comparison

$ Input: Variables: [(x,66), (y,77)] Expressions: [(x,<,y)]

› Output: [(x,<,y,'true')]

💡 Note: x = 66 and y = 77, so x < y evaluates to 66 < 77 which is true

example_2.sql — Multiple Operations

$ Input: Variables: [(x,66), (y,77), (z,66)] Expressions: [(x,>,y), (x,=,z), (y,>,z)]

› Output: [(x,>,y,'false'), (x,=,z,'true'), (y,>,z,'true')]

💡 Note: x > y: 66 > 77 = false, x = z: 66 = 66 = true, y > z: 77 > 66 = true

example_3.sql — Edge Case Equal Values

$ Input: Variables: [(a,50), (b,50)] Expressions: [(a,<,b), (a,=,b), (a,>,b)]

› Output: [(a,<,b,'false'), (a,=,b,'true'), (a,>,b,'false')]

💡 Note: When both variables have equal values: < and > are false, = is true

Constraints

1 ≤ Variables.length ≤ 100
1 ≤ Expressions.length ≤ 100
-10⁹ ≤ Variables.value ≤ 10⁹
All variable names in Expressions are guaranteed to exist in Variables
Variable names contain only lowercase English letters

Visualization

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Understanding the Visualization

Load Variables

Store all variable names and their values in memory

Process Expression

Take an expression like 'x < y' and look up both values

Substitute Values

Replace variable names with their stored values: 'x < y' becomes '66 < 77'

Evaluate Comparison

Apply the operator logic and return boolean result as string

Key Takeaway

🎯 Key Insight: Double JOIN eliminates subqueries by efficiently matching variable names to values in a single query execution

Asked in

a Amazon 35 ⊞ Microsoft 28 G Google 22 f Meta 18

The optimal solution uses double JOIN approach: JOIN the Expressions table with Variables table twice (once for left operand, once for right operand) to get both values, then use CASE statement to evaluate the comparison operator and return 'true' or 'false'.

Common Approaches

Approach	Time	Space	Notes
✓ Double JOIN Approach (Optimal)	O(n + m)	O(1)	JOIN Variables table twice to get both operand values in one query
Nested Subqueries (Brute Force)	O(n×m)	O(1)	Use separate subqueries to look up each operand value

Double JOIN Approach (Optimal) — Algorithm Steps

JOIN Expressions with Variables to get left operand value
JOIN again with Variables to get right operand value
Use CASE statement to evaluate the operator
Return the boolean result as 'true' or 'false'

Visualization

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Step-by-Step Walkthrough

Start with Expressions

Begin with the Expressions table as base

First JOIN

JOIN with Variables to get left operand value

Second JOIN

JOIN again with Variables to get right operand value

Evaluate

Apply CASE logic to compare values and return result

Code -

solution.c — C

// Optimal SQL solution (C string format)
const char* sqlQuery = "\
SELECT \
    e.left_operand, \
    e.operator, \
    e.right_operand, \
    CASE \
        WHEN e.operator = '<' AND v1.value < v2.value THEN 'true' \
        WHEN e.operator = '>' AND v1.value > v2.value THEN 'true' \
        WHEN e.operator = '=' AND v1.value = v2.value THEN 'true' \
        ELSE 'false' \
    END AS value \
FROM Expressions e \
JOIN Variables v1 ON e.left_operand = v1.name \
JOIN Variables v2 ON e.right_operand = v2.name;";

Time & Space Complexity

Time Complexity

⏱️

O(n + m)

Linear scan through expressions (n) and variables (m) with efficient hash joins

✓ Linear Growth

Space Complexity

O(1)

No additional storage beyond result set

✓ Linear Space

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Double JOIN Approach (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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