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Check if the chain has the same amount

José Oscátegui on June 15, 2018

Check if a string has the same amount of 'x' and 'o'. The method must return a Boolean value and must not be case-sensitive. The string can contain any char.

Tata Ganesh • Jun 16 '18 • Edited

Python solution -

def equal_x_and_o(input_string):
    x_count = 0
    o_count = 0
    for c in input_string:
         if c == "x" or c == "X":
            x_count += 1
         elif c == "o" or c == "O":         
            o_count +=1
    return x_count == o_count

Thank you @Pierre for pointing out the error.

Pierre Bouillon • Jun 18 '18

Hey @ganeshtata !
I think you missed an uppercase in your elif :)

elif c == "o" or c == "o":

instead of

elif c == "o" or c == "O":

To dig a little deeper, I suggest you to see some very useful string methods in Python !

Tata Ganesh • Jun 18 '18 • Edited

Thank you for that! I had not noticed it. Yes, there are some awesome string methods in Python, but I didn't use any of them here because I wanted to finish the count(s) in one pass of the string. Doing a .lower() and then counting might involve two passes of the same string.

Pierre Bouillon • Jun 18 '18

I suggested it to reduce your conditions, lowering the string before checking would have allowed you to only check for o and x for example. This way you would have one pass on the string and a more concise condition :)

Tata Ganesh • Jun 18 '18 • Edited

Do you mean lowering the string before the loop or lowering the character in the loop before checking whether it is an 'x' or 'o'?

Lowering the string before the loop would involve one complete pass of the string. And then you would have to iterate through the string, again, to count the frequencies.
Lowering just the character inside the loop, is just equivalent to the if condition, right? I am probably saving on a function call by not calling .lower() on the character.

Pierre Bouillon • Jun 18 '18

Oh right, nevermind !

Oskar Larsson • Jun 15 '18

In Haskell, with what I thought was kind of a neat method:

import Data.Char

val 'x' = 1
val 'o' = -1
val other = 0

check str
 | (sum $ map (val . toLower) str) == 0 = True
 | otherwise = False

As a javascript one-liner:

string.toLowerCase().split('').map(a => a == 'x' ? 1 : a == 'o' ? -1 : 0).reduce((acc, a) => acc+a) == 0 ? true : false

kip • Jun 17 '18

A simple solution with Rust.

fn main() {
    println!(
        "{} the same amount",
        if has_same_amount("$XxkkLLOoox-", "x", "o") {
            "Has"
        } else {
            "Not has"
        }
    )
}

fn has_same_amount(str: &str, ch1: &str, ch2: &str) -> bool {
    let str = str.to_lowercase();

    str.matches(ch1).count() == str.matches(ch2).count()
}

Has the same amount

Pierre Bouillon • Jun 16 '18

Python to the rescue !

def dev(inp: str) -> bool:
    return inp.lower().count('x') == inp.lower().count('o')

output

>>> dev('xoxo')
True
>>> dev('xaabboccddxeeffo')
True
>>> dev('x')
False
>>> dev('ooooox')
False

Adrian B.G. • Jun 17 '18

Time for a Go kata you say?


func IsBalanced(dic map[rune]int, s string) bool {
        b := 0
        for _, r := range strings.ToLower(s) {
            i, ok := dic[r]
            if ok == false {
                continue
            }
            b += i
        }
        return b == 0
    }   

func main() {
    //because the world is bigger than Latin
    var step = map[rune]int{'x': 1, '🗴': 1, 'o': -1, 'ອ': -1}
    table := map[string]bool{
        "":true, //0x == 0o
        "x0x0": true,
        "ອXອXອ":  false,
        "H🗴H🗴H(╯° °）╯︵ ┻━┻)HoHo":true,
    }

    for input, exp := range table {
        fmt.Printf("'%s', exp: '%v', got: '%v'\n", input, exp, IsBalanced(step, input))
    }

}

play.golang.org/p/n8q9BYrsW4a

Jay • Jun 17 '18

Rust Solution.

fn check_x_o(s: &str) -> bool {
  s.to_lowercase()
    .chars()
    .map(|c| match c {
      'x' => 1,
      'o' => -1,
      _ => 0,
    })
    .sum::<i32>() == 0
}

// Usage
fn main() {
  let st = "xoxABCxo";
  println!("{}", check_x_o(&st));    // false
}

Corey Alexander • Jun 16 '18

Ruby Solution!

def character_counts(str)
  str.chars.each_with_object(Hash.new(0)) { |char, hash| hash[char] += 1 }
end

def check(str)
  str.downcase!
  char_count = character_counts(str)
  char_count['o'] == char_count['x']
end

irb(main):003:0> check('xoxo')
=> true
irb(main):004:0> check('xaabboccddxeeffo')
=> true
irb(main):005:0> check('x')
=> false
irb(main):006:0> check('ooooox')
=> false