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Daily Challenge #130 - GCD Sum

thepracticaldev profile image dev.to staff ・1 min read

Given the sum and gcd of two numbers, return those two numbers in ascending order. If the numbers do not exist, return -1, (or return NULL in C).

For example:
Given sum = 12 and gcd = 4...

solve(12,4) = [4,8].
The two numbers 4 and 8 sum to 12 and have a gcd of 4.

solve(12,5) = -1.
No two numbers exist that sum to 12 and have gcd of 5.

solve(10,2) = [2,8].
Note that [4,6] is also a possibility but we pick the one with the lower first element: 2 < 4, so we take [2,8].

Good luck.


This challenge comes from KenKamau on CodeWars. Thank you to CodeWars, who has licensed redistribution of this challenge under the 2-Clause BSD License!

Want to propose a challenge idea for a future post? Email yo+challenge@dev.to with your suggestions!

Discussion

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avalander profile image
Avalander

Haskell.

A bit ineffective because it generates all possible solutions instead of stopping when it finds the first solution.

solve :: Int -> Int -> Maybe (Int, Int)
solve sum target
  | solutions == [] = Nothing
  | otherwise       = Just (head solutions)
  where
    solutions = [(a, b) | a <- [1..(sum `div` 2)],
                          let b = sum - a,
                          gcd a b == target ]
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craigmc08 profile image
Craig McIlwrath

Correct me if I'm wrong, but it doesn't generate all possible solutions. Haskell is lazy, and because the function only returns the head of the list, only the head can be forced into a value. So the evaluation of the list will stop as soon as a value for the head is found, meaning it does stop when the first solution is found.

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avalander profile image
Avalander

You might be right, I had forgotten about that tiny detail.

Update: I tried with an infinite sequence to generate the solutions and the program didn't get stuck in an infinite loop, so you are right, only the first element of the sequence is generated.

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rulikkk profile image
Rustem Mustafin

Assume 0 < N <= M
Sum = N + M
GCD = G, so that N = x * G, M = y * G and x, y are relatively prime integers > 0
=>
Sum = x * G + y * G = G * (x + y)
Sum / G should also be integer, otherwise no answer
x + y = Sum / G
We need to take smallest x, so that x, y are relatively prime, so x = 1
=> x = 1 => N = G => M = Sum - G
Answer: GCD, Sum - GCD
Test:
1) Solve(12, 4) = 4, 8; GCD(4, 8) = 4; 4 + 8 = 12; OK
2) Solve(12, 5): 12 % 5 > 0 => No Answer; OK
3) Solve(10, 2) = 2, 8; GCD(2, 8) = 2; 2 + 8 = 10; OK

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midhetfatema94 profile image
Midhet Sulemani

Reiterating to this logic in Swift:

import UIKit

func solve(sum: Int, hcf: Int) -> (Int, Int)? {
    let x = 1
    var y: Int!

    y = sum - hcf

    let verificationInt: Float = Float((hcf + y)/hcf)
    let verificationSum = Int(verificationInt + 1)
    if verificationSum % hcf == 0 {
        return (hcf*x, y)
    }
    return nil
}

if let solution = solve(sum: 12, hcf: 5) {
    print(solution)
} else {
    print(-1)
}
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_bkeren profile image
''

JavaScript ES6

const solve = (sum,gcd) => {
if(sum%gcd) return -1;
const constantSum = sum / gcd;
return [gcd,(constantSum-1) * gcd]
}

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juanrodriguezarc profile image
Juan Rodríguez
const solve = (s,g) => (s%g) ? -1 : [g,(s / g-1)*g]
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meave9786 profile image
meave9786

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