How should I practice Resolve Variable Equations with Dependency Errors?

Start by reading the full statement, solve the visible examples, then compare the pattern against related Applied Intuition phone screen problems in FastPrep.

Is Resolve Variable Equations with Dependency Errors associated with Applied Intuition?

FastPrep tags this as a Applied Intuition phone screen coding practice problem when the problem catalog has that company and stage context.

Resolve Variable Equations with Dependency Errors

Applied Intuition● MediumFULLTIMEPHONE SCREEN

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Problem statement

You are given an array of variable definitions. Each definition has the form name = term + term - term + ..., with addition or subtraction between consecutive terms. A term is either a signed integer or the name of another variable. Variable names contain only lowercase English letters. Whitespace may appear around names, integers, =, and operators. Expressions contain no parentheses.

Resolve every variable and return one string per definition, in the same order as the input. Each successful result must have the form name=value. Addition and subtraction are evaluated from left to right. A minus sign immediately before an integer belongs to that signed term, so foo - -3 subtracts negative three.

If the dependency graph among defined variables contains any cycle, return the one-element array ["Cyclic Dependency"]. Otherwise, if any expression references a variable that is not defined, return ["Unresolvable equations"]. A cycle takes precedence when both conditions occur.

Operator extension discussion

The exercise begins with addition, then adds subtraction. As a design follow-up, explain how you would add multiplication and division while making precedence, division-by-zero behavior and numeric precision explicit. These additional operators are a discussion follow-up; judged expressions use only addition and subtraction.

Function

resolveEquations(equations: String[]) → String[]

Examples

Example 1

equations = ["foo = bar + 5", "bar = 2", "abc = 3"]return = ["foo=7", "bar=2", "abc=3"]

bar resolves to 2, so foo resolves to 2 + 5 = 7. Results retain definition order.

Example 2

equations = ["g = abc + foo", "foo = bar + 5", "bar = 2", "abc = 3"]return = ["g=10", "foo=7", "bar=2", "abc=3"]

Depth-first dependency resolution gives foo = 7, then g = 3 + 7 = 10.

Example 3

equations = ["foo = bar + 3", "bar = abc + pqr + 2"]return = ["Unresolvable equations"]

Neither abc nor pqr is defined, so the equations cannot be resolved.

Example 4

equations = ["foo = bar + 3", "bar = foo + 2"]return = ["Cyclic Dependency"]

foo depends on bar, and bar depends on foo, forming a cycle.

Example 5

equations = ["foo = bar - 5","bar = 2","abc = foo - -3"]return = ["foo=-3","bar=2","abc=0"]

bar = 2 gives foo = 2 - 5 = -3. Then abc = -3 - (-3) = 0. Results keep the original definition order.

Constraints

1 <= equations.length <= 1000.
Every variable name contains between 1 and 30 lowercase English letters.
Each variable is defined exactly once.
Each expression contains between 1 and 100 signed-integer or variable terms joined by + or -; every expression follows the documented grammar.
Integer terms are in the range [-10^9, 10^9].
For this exercise, assume every intermediate arithmetic result and successfully resolved value fits in a signed 64-bit integer.