A Simple Interview Question

August 14, 2018

This is easy:

(define (sum x y)
  (while (< x 0) (set! x (add1 x)) (set! y (sub1 y)))
  (while (< 0 x) (set! x (sub1 x)) (set! y (add1 y)))
  y)

Here are some examples:

> (add 3 5)
8
> (add 3 -5)
-2
> (add -3 5)
2
&gr; (add -3 -5)
-8

You can run the program at https://ideone.com/PXMVRM.

Pages: 1 2

Posted by programmingpraxis
Filed in Exercises
8 Comments »

8 Responses to “A Simple Interview Question”

  1. James Curtis-Smith said
    August 14, 2018 at 1:08 PM 
    my($x,$y) = @ARGV;
$x>0 ? ($y++,$x--) : ($y--,$x++) while $x;
print "$y\n"
  2. Zack said
    August 14, 2018 at 2:53 PM

    Here is my take with Julia (v. 1.0). Optimizing for number of iterations:

    function add(x::Int64, y::Int64)
    if x == 0
    return y
    elseif y == 0
    return x
    else
    x_ = abs(x)
    y_ = abs(y)
    ind = argmin([x_, y_])

        if ind == 1
        if x > 0
            for i = 1:x_
                y += 1
            end
        else
            for i = 1:x_
                y -= 1
            end
        end

        return y
    else
        if y > 0
            for i = 1:y_
                x += 1
            end
        else
            for i = 1:y_
                x -= 1
            end
        end

        return x
    end
end

    end

  3. John Cowan said
    August 14, 2018 at 10:02 PM

    The expression (- x (- y)), equivalently x – -y for infix programmers, seems to meet the conditions of the problem.

  4. Globules said
    August 15, 2018 at 1:29 AM

    Haskell.

    decr :: Integral a => a -> a
decr = subtract 1

incr :: Integral a => a -> a
incr = (+1)

add :: Integral a => a -> a -> a
add x y | y  < 0 = add (decr x) (incr y)
        | y == 0 = x
        | y  > 0 = add (incr x) (decr y)

main = do
  print $ add (-4) (-5)
  print $ add (-4)   5
  print $ add   4  (-5)
  print $ add   4    5

    $ ./add2
-9
1
-1
9
  5. Daniel said
    August 15, 2018 at 5:00 AM

    Here’s a solution in C.

    #include <stdio.h>
#include <stdlib.h>

int add(int a, int b) {
    while (a > 0) {
        --a;
        ++b;
    }
    while (a < 0) {
        ++a;
        --b;
    }
    return b;
}

int main(int argc, char* argv[]) {
    if (argc != 3) {
        fprintf(stderr, "Usage: %s INT INT\n", argv[0]);
        return EXIT_FAILURE;
    }
    int a = atoi(argv[1]);
    int b = atoi(argv[2]);
    int result = add(a, b);
    printf("%d\n", result);
    return EXIT_SUCCESS;
}

    Examples:

    $ ./add -2 5
3

$ ./add -2 -7
-9

$ ./add 0 8
8

$ ./add 1 11
12
  6. Jonathan Fuentes said
    August 15, 2018 at 12:38 PM

    function sum(a, b) {
    return a – (-b);
    }

  7. kernelbob said
    August 15, 2018 at 1:39 PM

    Instead of the add instruction, let’s use <<, >>, &, ==, ++, and — to implement a classical binary adder. Because Python automatically promotes ints to bignums, we special case -1 to prevent infinite regress.

    def add(a, b, carry=0):
    if a == -1:
        carry -= 1
        a = 0
    if b == -1:
        carry -= 1
        b = 0
    if a == b == 0:
        return carry
    if a & 1:
        carry += 1
    if b & 1:
        carry += 1
    r = add(a >> 1, b >> 1, carry >> 1) << 1
    if carry & 1:
        r += 1
    return r
  8. Paul said
    August 17, 2018 at 6:19 PM

    Another approach in Python.

    def add(a, b):
    return 2*a if a == b else (a*a - b*b) // (a - b)

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