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Render A Config Template

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Trains the technique from

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A deployment tool renders a config template. text is the template and replacements is a list of rules; rule replacements[i] is a pair [name, body] saying that the placeholder named name stands for the string body.

A placeholder is written as a rule name wrapped in braces, for example {PORT}. Braces are used for nothing else: in text and in every body, each { opens a placeholder that the very next } closes, and the name between them is the name of exactly one rule. Rule names are distinct and are made of uppercase letters, digits and underscores.

A body may itself contain placeholders, and those must be rendered too, as must any placeholders that appear in their bodies, and so on until no braces are left. The rules come in no particular order, so a body may name a rule that is listed later or earlier, and some rules may go unused. No rule ever depends on itself, directly or through a chain of other rules, so the rendering always finishes.

Return the rendered template.

Examples

Example 1

Input: replacements = [["HOST", "db.{ZONE}.internal"], ["ZONE", "eu-west"]], text = "url=db_at_{HOST}:5432"
Output: "url=db_at_db.eu-west.internal:5432"

`{HOST}` stands for `db.{ZONE}.internal`, whose own `{ZONE}` stands for `eu-west`, so the placeholder renders as `db.eu-west.internal` and the rest of the template is copied through unchanged.

Example 2

Input: replacements = [["ROOT", "/srv"], ["LOGS", "{ROOT}/logs"]], text = "{LOGS} then {LOGS}"
Output: "/srv/logs then /srv/logs"

Here the body of the second rule names the first rule, so `{LOGS}` renders as `/srv/logs`. Both appearances of the placeholder are rendered.

Example 3

Input: replacements = [["A1", "keep"]], text = "no braces at all"
Output: "no braces at all"

The template holds no braces, so nothing is replaced and the single rule goes unused.

Constraints

1 <= replacements.length <= 100
replacements[i].length == 2
1 <= replacements[i][j].length <= 100
1 <= text.length <= 1000
Rule names are distinct, and every placeholder names one of the rules.
No rule depends on itself directly or through a chain of other rules.
The rendered template is at most 5000 characters long.

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def apply_substitutions(replacements: list[list[str]], text: str) -> str:

Java

public String applySubstitutions(String[][] replacements, String text)

The intended approach

Time O(B + T + S), where B is the total length of the rule bodies, T the length of `text` and S the total length of the resolved bodies: each body is scanned a bounded number of times and each character of the result is written once.Space O(S) for the resolved bodies, plus O(m) for the stack over m rules.

Put the rules in a dictionary, then resolve each body exactly once in dependency order. An explicit stack carries pairs of (rule, phase): on the first visit the rule is pushed back with phase 1 and all rules its body names are pushed with phase 0, so they finish first; on the phase-1 visit every placeholder in the body already has a finished replacement, and the body is spliced together in one pass. Splicing `text` with the same helper then gives the answer.

Pythonentry point: apply_substitutions

def apply_substitutions(replacements, text):
    # Resolve every rule body once, deepest rule first, using an explicit stack
    # so the nesting depth never touches the interpreter's recursion limit.
    # Time O(T + S), space O(S), where T is len(text) and S is the total length
    # of the fully resolved bodies.
    body = {}
    for name, value in replacements:
        body[name] = value

    def slots(s):
        found = []
        i = 0
        while i < len(s):
            if s[i] == "{":
                j = s.index("}", i)
                found.append((i, j, s[i + 1:j]))
                i = j + 1
            else:
                i += 1
        return found

    def splice(s, table):
        out = []
        at = 0
        for i, j, name in slots(s):
            out.append(s[at:i])
            out.append(table[name])
            at = j + 1
        out.append(s[at:])
        return "".join(out)

    done = {}
    for first in body:
        if first in done:
            continue
        # phase 0 = queue this rule's dependencies, phase 1 = build its text
        stack = [(first, 0)]
        while stack:
            name, phase = stack.pop()
            if name in done:
                continue
            if phase == 0:
                stack.append((name, 1))
                for _, _, dep in slots(body[name]):
                    if dep not in done:
                        stack.append((dep, 0))
            else:
                done[name] = splice(body[name], done)
    return splice(text, done)

The obvious version

Rewrite the string itself until it settles: find the leftmost `{`, cut out the placeholder it opens and paste that rule's raw body in its place, then look again. Because no rule depends on itself the loop always runs out of braces.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_apply_substitutions(replacements, text):
    body = {}
    for name, value in replacements:
        body[name] = value
    cur = text
    while "{" in cur:
        i = cur.index("{")
        j = cur.index("}", i)
        cur = cur[:i] + body[cur[i + 1:j]] + cur[j + 1:]
    return cur

How this problem was checked

Reference solution executed against 10 test cases — 10 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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