LeetCode 2469 - Convert the Temperature

Difficulty: 🟢 Easy
Topics: Math

Solution

Problem Understanding

The problem gives us a single floating point value named celsius, representing a temperature measured in degrees Celsius. Our task is to convert this temperature into two other temperature scales, Kelvin and Fahrenheit, and return both converted values in an array.

The conversion formulas are provided directly in the statement:

Kelvin = Celsius + 273.15
Fahrenheit = Celsius × 1.80 + 32.00

The expected output is an array of two floating point numbers:

[kelvin, fahrenheit]

The first element must contain the Kelvin conversion, and the second element must contain the Fahrenheit conversion.

The constraints are very small:

0 <= celsius <= 1000

This tells us several important things. First, the input is always non-negative, so we never need to handle temperatures below zero Celsius. Second, the value range is small enough that overflow is impossible in both Python and Go. Third, since there is only a single input value, performance is not a concern at all. Any correct implementation will run instantly.

The problem also specifies that answers within 10^-5 of the correct value are accepted. This means standard floating point arithmetic is completely sufficient. We do not need special precision handling or decimal libraries.

An important edge case is when the input is exactly 0. In that case:

Kelvin becomes 273.15
Fahrenheit becomes 32.0

Another edge case is the upper bound 1000, which verifies that the formulas still work correctly for large valid values. Since the input is already guaranteed to be rounded to two decimal places, we also do not need to perform any additional rounding ourselves.

Approaches

Brute Force Approach

A brute force approach would directly apply the given formulas one step at a time using temporary variables. We would calculate the Kelvin temperature first, then calculate the Fahrenheit temperature, and finally store both results in an array.

This approach is correct because the problem explicitly provides the mathematical relationships between Celsius and the other temperature scales. By substituting the input into those formulas, we obtain the exact required values.

Although this is called a brute force approach for completeness, there is no meaningful inefficiency here because the problem only requires a constant number of arithmetic operations.

Optimal Approach

The optimal approach is effectively the same as the brute force approach because the task itself is already minimal. The key observation is that there is no searching, iteration, recursion, or optimization problem involved. We simply evaluate two mathematical formulas.

Since each conversion takes constant time and constant memory, this is already the best possible solution.

Approach	Time Complexity	Space Complexity	Notes
Brute Force	O(1)	O(1)	Directly computes both formulas using temporary variables
Optimal	O(1)	O(1)	Same direct mathematical computation, minimal possible work

Algorithm Walkthrough

Receive the input floating point number celsius.

This value represents the temperature in the Celsius scale and serves as the base value for both conversions. 2. Compute the Kelvin temperature.

Use the formula:

$K = C + 273.15$

We add 273.15 because Kelvin and Celsius scales differ only by a constant offset. 3. Compute the Fahrenheit temperature.

Use the formula:

$F = C \times 1.80 + 32.00$

The multiplication by 1.80 adjusts the scale size, and the addition of 32.00 shifts the zero point. 4. Store both computed values in an array.

The required order is:

[kelvin, fahrenheit]

Return the resulting array.

Why it works

The algorithm works because it directly applies the exact mathematical conversion formulas provided in the problem statement. Since Kelvin and Fahrenheit are deterministic linear transformations of Celsius, substituting the input value into the formulas always produces the correct result.

Python Solution

from typing import List

class Solution:
    def convertTemperature(self, celsius: float) -> List[float]:
        kelvin = celsius + 273.15
        fahrenheit = celsius * 1.80 + 32.00
        
        return [kelvin, fahrenheit]

The implementation begins by calculating the Kelvin temperature using the provided formula. The result is stored in the variable kelvin.

Next, the code calculates the Fahrenheit temperature using the second formula and stores it in fahrenheit.

Finally, both values are returned in a list in the required order. Python floating point arithmetic naturally handles the required precision, so no additional formatting or rounding logic is necessary.

Go Solution

func convertTemperature(celsius float64) []float64 {
    kelvin := celsius + 273.15
    fahrenheit := celsius*1.80 + 32.00

    return []float64{kelvin, fahrenheit}
}

The Go implementation is nearly identical to the Python version. The input type is float64, which provides sufficient precision for the required calculations.

The result is returned as a slice of float64 values. Since the problem guarantees valid input sizes and ranges, no additional validation is required.

Worked Examples

Example 1

Input: celsius = 36.50

Step-by-step computation

Step	Calculation	Result
Input	celsius	36.50
Kelvin conversion	36.50 + 273.15	309.65
Fahrenheit conversion	36.50 × 1.80 + 32.00	97.70
Final output	[309.65, 97.70]	returned

Final result:

[309.65000,97.70000]

Example 2

Input: celsius = 122.11

Step-by-step computation

Step	Calculation	Result
Input	celsius	122.11
Kelvin conversion	122.11 + 273.15	395.26
Fahrenheit conversion	122.11 × 1.80 + 32.00	251.798
Final output	[395.26, 251.798]	returned

Final result:

[395.26000,251.79800]

Complexity Analysis

Measure	Complexity	Explanation
Time	O(1)	Only a fixed number of arithmetic operations are performed
Space	O(1)	Only two extra floating point variables are used

The algorithm performs two direct formula evaluations regardless of the input value. No loops, recursion, or additional data structures are involved. Because the amount of work never changes, both time and space complexity remain constant.

Test Cases

def almost_equal_list(a, b, eps=1e-5):
    return all(abs(x - y) < eps for x, y in zip(a, b))

solution = Solution()

# Provided example 1
assert almost_equal_list(
    solution.convertTemperature(36.50),
    [309.65, 97.70]
), "Example 1 failed"

# Provided example 2
assert almost_equal_list(
    solution.convertTemperature(122.11),
    [395.26, 251.798]
), "Example 2 failed"

# Minimum boundary value
assert almost_equal_list(
    solution.convertTemperature(0),
    [273.15, 32.0]
), "Zero Celsius conversion failed"

# Maximum boundary value
assert almost_equal_list(
    solution.convertTemperature(1000),
    [1273.15, 1832.0]
), "Maximum constraint value failed"

# Decimal input with small fractional part
assert almost_equal_list(
    solution.convertTemperature(0.01),
    [273.16, 32.018]
), "Small decimal input failed"

# Integer Celsius value
assert almost_equal_list(
    solution.convertTemperature(25),
    [298.15, 77.0]
), "Integer Celsius input failed"

# Mid-range floating point value
assert almost_equal_list(
    solution.convertTemperature(50.55),
    [323.70, 122.99]
), "Mid-range floating point input failed"

Test	Why
`36.50`	Validates the first provided example
`122.11`	Validates the second provided example
`0`	Tests the minimum allowed Celsius value
`1000`	Tests the maximum allowed Celsius value
`0.01`	Verifies precision handling for small decimals
`25`	Confirms correct handling of integer-like input
`50.55`	Tests general floating point arithmetic correctness

Edge Cases

One important edge case is when the input Celsius value is exactly 0. This case is significant because it represents a well-known physical reference point. A buggy implementation might accidentally return integer values or incorrectly apply offsets. The implementation handles this correctly by directly applying the formulas, producing 273.15 Kelvin and 32.0 Fahrenheit.

Another important edge case is the maximum allowed input value, 1000. Large values sometimes expose overflow issues or incorrect arithmetic ordering in weaker implementations. Here, the formulas remain simple linear calculations, and both Python and Go floating point types comfortably support these values.

A third important edge case involves fractional inputs such as 0.01 or 50.55. Floating point arithmetic can introduce precision concerns in some problems. However, the problem explicitly allows a tolerance of 10^-5, and standard floating point operations easily satisfy that requirement. The implementation performs no unnecessary rounding, which helps preserve accuracy.