Leetcode in Ada (#9)

Last Updated 2023 March 02

• Why I picked this one
• Palindrome Number
• The First Solution

Why I picked this one

Happy 3/2/23 (m/d/yyyy, palindrome) or 2/3/23 (d/m/yyyy, repeating digits).

Because I don’t like to talk about myself much, I had trouble coming up with what to write for blogs. That is until I noticed people write blogs about solving Leetcode problems.

Weeks ago, I was outside and away from my desktop. There was an intermission of not doing anything. So I thought I could add to my collection of leetcode solutions even if the problem was easy.

The page for problems on leetcode wouldn’t load correctly on my phone, so I searched for a scraper. I didn’t run the scraper myself because the requirements seemed high. But the scraper did have a sample page of problems in simple HTML. This challenge was on there.

Palindrome Number

Given an integer x, return true if x is a palindrome, and false otherwise.

Follow up: Could you solve it without converting the integer to a string?

I can see how solutions like return (str(x) == (str(x))[::-1]) (python) defeat the purpose of the chalenge. For the original solution, I put the digits of the number into a container.¹ I technically did not convert it to a string as a part of the solution, But I later decided to revisit the problem and write a solution with no arrays or containers.

There is a small story behind the check for x < 100 instead of sigs = 2. When I didn’t use float'floor() before converting to an integer, numbers greater than 44 were treated as though they had three digits. Turns out, integer() will round up or round down decimal values (integer(0.5)= 1) which was a slight curveball coming from languages that round down in all cases for float-to-integer conversions.

    function is_palindrome_no_array 
        (x : in integer)
    return boolean is
        result : boolean := false;
        sigs: integer; -- significant digits
        msd : integer; -- more significant digit
        lsd : integer; -- less significant digit
        a   : integer;
    begin
        if x < 0 then
            return false;
        end if;
        if x < 10 then
            return true;
        end if;
        if x < 100 then
            return (x mod 11 = 0);
        end if;
        sigs := integer(float'floor(math.log(float(x), 10.0))) + 1;
        for i in 0..(sigs/2) loop
            lsd := (x/(10 ** i)) rem 10;
            a   := (x/(10 ** (sigs-(i+1))));
            msd :=  a rem 10;
            if msd /= lsd then
                return false;
            end if;
        end loop;
        return true;
    end;

The First Solution

    function is_palindrome 
        (x : in integer) return boolean is
        v : vector;
        operand : integer := x;
        i : integer := 1;
    begin
        if x < 0 then
            return false;
        end if;
        if x < 10 then
            return true;
        end if;
        while operand > 0 loop
            v.append(operand rem 10);
            operand := (operand/10);
        end loop;
        print(v);
        --put_line(integer'image(v.element(0)));
        --put_line(integer'image(v.last_index));
        if v.first_element /= v.last_element then
            return false;
        end if;
        while i < ((v.last_index)/2) loop
            if v.element(v.first_index + i) 
            /= v.element(v.last_index - i) then
                return false;
            end if;
            i := i + 1;
        end loop;
        return true;
    end;