CF 102697112 - Unit Conversions

The task is a small text-parsing and arithmetic problem. We are given one sentence describing a length, such as I am 1.5 meters, how many feet am I. The number and its original unit appear immediately before the comma.

codeforcescompetitive-programming

CF 102697112 - Unit Conversions

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Solve time: 1m 8s
Verified: yes

Solution

Problem Understanding

The task is a small text-parsing and arithmetic problem. We are given one sentence describing a length, such as I am 1.5 meters, how many feet am I. The number and its original unit appear immediately before the comma. After the comma, the phrase contains the unit we want to convert to. We must calculate the equivalent length and print that value followed by the requested unit. The statement defines all units relative to feet: an inch is 0.08333 feet, a meter is 3.28084 feet, a centimeter is 0.0328084 feet, a mile is 5280 feet, and a kilometer is 3280.84 feet.

The six supported units are feet, inches, meters, centimeters, miles, and kilometers. There is only one input line, and the statement does not give a numeric bound for the value. The time limit is 1 second and the memory limit is 256 MB, but the amount of work required is constant regardless of the numerical value because there are only six possible units and one conversion is requested.

The central representation is to store, for every unit, how many feet correspond to one unit. If the input is x units of A, then x * factor[A] gives the amount in feet. To reach the requested unit B, divide by factor[B]. Thus the entire calculation is

[ answer = x \times \frac{factor[A]}{factor[B]}. ]

The most common parsing mistake is taking the wrong word from the sentence. For example,

I am 2 miles, how many kilometers am I

must produce approximately

3.21869 kilometers

The source unit is miles, while the target is kilometers. A parser that simply takes the last word would read I as the unit and fail. The target unit is the word immediately after how many.

Another edge case is converting a unit to itself. For example,

I am 7 feet, how many feet am I

should produce

7.00000 feet

A solution that always assumes the two units differ can accidentally apply the wrong conversion formula. The factor representation handles this naturally because factor[feet] / factor[feet] = 1.

A third case is a conversion in the direction opposite to the stored factors. For example,

I am 5280 feet, how many miles am I

should produce

1.00000 miles

The given constants describe how many feet are in one unit, so converting feet to miles requires division by 5280, not multiplication. Using the same multiplication operation in both directions is a common source of incorrect answers.

Approaches

A direct brute-force implementation could write a separate formula for every ordered pair of units. With six units there are only (6 \times 5 = 30) distinct conversions between different units, or 36 if self-conversions are included. Such a program is still constant time and would easily fit the 1 second limit, so unlike many algorithmic problems, there is no realistic input size at which this brute-force method becomes too slow. The difficulty is correctness and maintainability rather than running time. Thirty separate formulas create many opportunities to reverse a conversion factor or mistype a constant.

The observation that every supported unit is already expressed relative to feet removes the need for all those individual formulas. We only need to convert the input to feet and then convert feet to the requested unit. If one unit contains a feet and the target unit contains b feet, then one input unit is a / b target units. This turns every possible pair into the same formula.

The input is also structured enough that no sophisticated parser is necessary. Splitting at the comma isolates the source description from the question. The final source number and source unit are the last two tokens before the comma, while the target unit is the token immediately after how many.

Approach Time Complexity Space Complexity Verdict
Brute Force O(1), at most 36 explicit cases O(1) Accepted, but unnecessarily repetitive
Optimal O(1) O(1) Accepted

Algorithm Walkthrough

  1. Read the complete input line instead of reading individual whitespace-separated tokens. The sentence contains meaningful punctuation and several words, so treating the entire line as a string makes the structure easy to parse.
  2. Split the line at the comma. The part before the comma contains the original measurement, while the part after it contains the requested unit.
  3. Split the left part into words. The last two tokens are the numeric value and the original unit because the statement guarantees that they occur immediately before the comma.
  4. Split the right part into words and take the token immediately after how many. That token is the target unit.
  5. Store the number of feet represented by one unit for every supported unit. For example, meters maps to 3.28084 and miles maps to 5280.
  6. Convert the original quantity to feet by multiplying the input value by its source factor.
  7. Convert feet to the target unit by dividing by the target factor. Combining the two operations gives value * source_factor / target_factor.
  8. Print the converted value followed by the requested unit, using five digits after the decimal point to match the precision shown by the sample.

Why it works

For every supported unit (U), let (F(U)) be the number of feet equivalent to one unit of (U). A measurement of (x) units of the source unit represents exactly (xF(source)) feet. Since one target unit represents (F(target)) feet, the same physical length contains (xF(source)/F(target)) target units. The algorithm computes exactly this quantity, so every supported source and target pair uses the correct conversion relationship.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    line = input().strip()

    left, right = line.split(",", 1)

    left_tokens = left.split()
    value = float(left_tokens[-2])
    source = left_tokens[-1]

    right_tokens = right.split()
    target = right_tokens[2]

    to_feet = {
        "feet": 1.0,
        "inches": 0.08333,
        "meters": 3.28084,
        "centimeters": 0.0328084,
        "miles": 5280.0,
        "kilometers": 3280.84,
    }

    answer = value * to_feet[source] / to_feet[target]

    print(f"{answer:.5f} {target}")

if __name__ == "__main__":
    solve()

The first split uses split(",", 1) so that the sentence is divided into exactly the two logical parts required by the problem. There is no reason to inspect the words before the measurement individually because the statement guarantees that the final two tokens before the comma are the value and source unit.

The source value is parsed with float, since the example contains 1.5 and the input can represent non-integer measurements. Python's floating-point arithmetic is more than sufficient for the decimal conversion factors supplied by the statement.

The dictionary stores every conversion relative to feet. This is the key simplification. Instead of maintaining separate meters_to_miles, miles_to_meters, feet_to_inches, and similar rules, there is only one factor per unit.

The expression

value * to_feet[source] / to_feet[target]

first moves the quantity into the common reference unit, feet, and then moves it into the requested unit. The order matters because the factors describe how many feet are in one source or target unit.

The target is obtained with right_tokens[2]. After the comma, the guaranteed structure is how many <unit> ..., so indexes 0, 1, and 2 correspond to how, many, and the requested unit.

Five decimal places are printed because the supplied sample uses five digits after the decimal point. The calculation itself is performed using full Python floating-point precision before formatting.

Worked Examples

Sample 1

Input:

I am 1.5 meters, how many feet am I

The important state changes are:

Step Value Source Target Source Factor Target Factor Answer
Parse source 1.5 meters feet 3.28084 1.0
Convert 1.5 meters feet 3.28084 1.0 4.92126
Format 1.5 meters feet 3.28084 1.0 4.92126

The calculation is (1.5 \times 3.28084 / 1 = 4.92126), giving the required output 4.92126 feet. This demonstrates the normal source-to-feet-to-target path when the target itself is the reference unit.

Example 2

Input:

I am 5280 feet, how many miles am I
Step Value Source Target Source Factor Target Factor Answer
Parse source 5280 feet miles 1.0 5280.0
Convert to feet 5280 feet miles 1.0 5280.0 5280
Convert to target 5280 feet miles 1.0 5280.0 1

The calculation is (5280 \times 1 / 5280 = 1), so the output is 1.00000 miles. This example exercises the division direction and catches implementations that multiply by the target factor instead of dividing by it.

Complexity Analysis

Measure Complexity Explanation
Time O(1) The input has one line and the algorithm performs a fixed number of string operations and arithmetic operations.
Space O(1) The dictionary contains exactly six conversion factors, and the parsed sentence has constant structure.

The problem has only one conversion request and six fixed units, so the algorithm is comfortably within the 1 second and 256 MB limits. The numerical value itself does not affect the amount of work performed.

Test Cases

# helper: run solution on input string, return output string
import sys
import io

def solve():
    line = input().strip()

    left, right = line.split(",", 1)

    left_tokens = left.split()
    value = float(left_tokens[-2])
    source = left_tokens[-1]

    right_tokens = right.split()
    target = right_tokens[2]

    to_feet = {
        "feet": 1.0,
        "inches": 0.08333,
        "meters": 3.28084,
        "centimeters": 0.0328084,
        "miles": 5280.0,
        "kilometers": 3280.84,
    }

    answer = value * to_feet[source] / to_feet[target]
    print(f"{answer:.5f} {target}")

def run(inp: str) -> str:
    global input
    old_stdin = sys.stdin
    old_input = input

    sys.stdin = io.StringIO(inp)
    input = sys.stdin.readline

    output = io.StringIO()
    old_stdout = sys.stdout
    sys.stdout = output

    try:
        solve()
        return output.getvalue()
    finally:
        sys.stdin = old_stdin
        sys.stdout = old_stdout
        input = old_input

# provided sample
assert run(
    "I am 1.5 meters, how many feet am I\n"
) == "4.92126 feet\n", "sample 1"

# minimum-size style conversion
assert run(
    "I am 1 feet, how many feet am I\n"
) == "1.00000 feet\n", "self conversion"

# all-equal units, using a non-reference unit
assert run(
    "I am 1 miles, how many miles am I\n"
) == "1.00000 miles\n", "same unit"

# reverse conversion through the reference unit
assert run(
    "I am 5280 feet, how many miles am I\n"
) == "1.00000 miles\n", "feet to miles"

# metric to imperial
assert run(
    "I am 100 centimeters, how many feet am I\n"
) == "3.28084 feet\n", "centimeters to feet"

# large value, exercising the kilometer factor
assert run(
    "I am 2 kilometers, how many feet am I\n"
) == "6561.68000 feet\n", "kilometers to feet"
Test input Expected output What it validates
I am 1 feet, how many feet am I 1.00000 feet Self-conversion and minimum-sized numerical value
I am 1 miles, how many miles am I 1.00000 miles Equal source and target units
I am 5280 feet, how many miles am I 1.00000 miles Correct division when converting from the reference unit
I am 100 centimeters, how many feet am I 3.28084 feet Metric-to-imperial conversion
I am 2 kilometers, how many feet am I 6561.68000 feet Larger magnitude and kilometer factor

Edge Cases

For a self-conversion such as

I am 7 feet, how many feet am I

the source and target factors are both 1.0. The algorithm computes 7 * 1.0 / 1.0, giving 7.00000 feet. No special branch is needed, because the general formula already contains the identity conversion.

For the reverse reference conversion,

I am 5280 feet, how many miles am I

the source factor is 1.0 and the target factor is 5280.0. The algorithm first interprets 5280 feet as 5280 feet, then divides by the number of feet in one mile, producing exactly 1.00000 miles. This is the case most likely to expose a multiplication-versus-division error.

For a conversion such as

I am 100 centimeters, how many feet am I

the source factor is 0.0328084 and the target factor is 1.0. The result is 100 * 0.0328084 / 1 = 3.28084, so the program prints 3.28084 feet. The decimal factor is used exactly as supplied by the problem rather than replacing it with a separately derived value.

For a conversion in the other direction,

I am 1 feet, how many centimeters am I

the source factor is 1.0 and the target factor is 0.0328084. The formula produces approximately 30.479999..., which is formatted as 30.48000 centimeters. The floating-point formatting handles the representation noise while preserving the required displayed precision.

Finally, the wording itself can vary while preserving the guaranteed structure. For example,

The road is 1.5 miles, how many kilometers is it

has extra words before the measurement, but the last two tokens before the comma are still 1.5 miles, and the token after how many is still kilometers. The parser deliberately relies on these guarantees rather than depending on the particular phrase I am, so it works for the general sentence form described by the problem.