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Daily Challenge #219 - Compare Strings

dev.to staff on April 06, 2020

Create a function called that accepts 2 string arguments and returns an integer of the count of occurrences the 2nd argument is found in the first ...

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Empereol • Apr 6 '20

TypeScript

/**
 * Return the number of occurances of the search character in the target string.
 *
 * @param target String to search.
 * @param search Character to search string for.
 *
 * @throws {Error} If search params is not exactly one character.
 *
 * @example
 * strCount('Hello', 'o') // => 1
 * strCount('Hello', 'l') // => 2
 * strCount('', 'z')      // => 0
 */
function strCount(target: string, search: string): number {
  if (!search || search.length !== 1) {
    throw new Error('Search character be exactly one character');
  }

  return (target.match(new RegExp(search, 'g')) || []).length;
}

Robert • Apr 6 '20 • Edited

JavaScript

function strCount(text, letter) {
  if (!letter.length) {
    return 'a letter is required.';
  }

  return text.split('').filter(i => i === letter).length;
}

console.log(strCount('Hello world', '')) // a letter is required.
console.log(strCount('Hello world', 'l')) // 3

Gabi • Apr 6 '20

Java:
public static int strCount(String str, char letter) {
int count = 0;
for (char element : str.toCharArray()) {
if (element == letter) {
count++;
}
}
return count;
}

Oleg Aleksandrov • Apr 7 '20 • Edited

public static int strCount(String str, char letter) {
    return str.chars().filter(it -> it == letter).count();
}

Gabi • Apr 7 '20

Love it!

Sabin Pandelovitch • Apr 7 '20 • Edited

JS using a different approach, recursive function

The readable version

function strCompare(str, n, i = 0, c = 0) {
  const cond = str[i] === n ? c+1 : c;
  return i < str.length - 1 ? strCompare(str, n, i+1, cond) : cond;
}

One line version

const strCompare = (str, n, i = 0, c = 0) => (i < str.length - 1 ? strCompare(str, n, i+1, str[i] === n ? c+1 : c) : str[i] === n ? c+1 : c);

Amin • Apr 7 '20

Actually, string[index out of bound] will return undefined so you don't need to check that your index is lower than the length of the string.

Also, you don't need to keep track of the characters' count if you just return it right in your recursive call.

And you can increment your index right when you access your character for comparison.

const strCount =
    (string, character, index = 0) =>
        string[index] === undefined
            ? 0
            : string[index++] === character
                ? 1 + strCount(string, character, index)
                : strCount(string, character, index);

If you are about the bytes, this makes it for a shorter version of your solution.

const count = (s, c, i = 0) => !s[i] ? 0 : s[i++] === c ? 1 + count(s, c, i) : count(s, c, i);

Sabin Pandelovitch • Apr 7 '20

You are right, thanks!

Vidit Sarkar • Apr 6 '20

Python one liner,

strCount = lambda str1, str2: len(str1) and len(str2) and str1.count(str2)

Output,

print(strCount('Hello', 'o')) # output -> 1
print(strCount('Hello', 'l')) # output -> 2
print(strCount('', 'z')) # output -> 0
print(strCount('oh goodness gracious', 'o')) # output -> 4
print(strCount('Hello', '')) # output -> 0
print(strCount('howdy, pardner', 'd')) # output -> 2
print(strCount('Incumbent President', 'e')) # output -> 3
print(strCount('', '')) # output -> 0

Oleg Aleksandrov • Apr 6 '20 • Edited

Kotlin.

fun strCount(str1: String, str2: String): Int {
    var count = 0
    var matchIndex = 0
    var i = 0
    var indexOfFirstMatch = 0

    while (i < str1.length) {
        if (str1[i] == str2[matchIndex]) {
            if (matchIndex == 0) {
                indexOfFirstMatch = i
            }

            if ((++matchIndex) == str2.length) {
                count++
                matchIndex = 0
                i = indexOfFirstMatch + 1
            }
            else {
                i++
            }
        }
        else {
            i++
            matchIndex = 0
        }
    }

    return count
}

fun main() {
    println(strCount("Hello", "o"))
    println(strCount("Hello", "l"))
    println(strCount("", "z"))
    println(strCount("oh goodness gracious", "o"))
    println(strCount("howdy, pardner", "d"))
    println(strCount("Incumbent President", "e"))
 // also it's fine too
    println(strCount("Hello helllo", "ll"))
    println(strCount("eeeeeee", "ee"))
}

kesprit • Apr 7 '20

Swift solution :

func strCount(str: String, search: Character) -> Int {
    str.filter { $0 == search }.count
}

strCount(str:"Hello", search: "o") // => 1
strCount(str:"Hello", search: "l") // => 2
strCount(str:"", search: "z") // => 0

Lucas Frigo de Souza • Apr 7 '20 • Edited

public static int strCount(string text, char letter)
{
    var matches = Regex.Matches(text, $"{letter}");
    return matches.Count;
}

//Result
Console.WriteLine(strCount("Hello", 'o')); //1
Console.WriteLine(strCount("Hello", 'l')); //2
Console.WriteLine(strCount("", 'z')); //0
Console.WriteLine(strCount("oh goodness gracious", 'o')); //4
Console.WriteLine(strCount("howdy, pardner", 'd')); //2
Console.WriteLine(strCount("Incumbent President", 'e')); //3

Ahmad Ra'fat • Apr 6 '20 • Edited

Python 3

def strCount(str1, str2):
    if not str1 or str2 not in str1:
        return 0

    return str1.count(str2)


def strCount_2(str1, str2):
    return str1.count(str2)

if __name__ == "__main__":
    print(strCount('Hello', 'o'))
    print(strCount('Hello', 'l'))
    print(strCount('', 'z'))

    print(strCount('oh goodness gracious', 'o'))
    print(strCount('howdy, pardner', 'd'))
    print(strCount('Incumbent President', 'e'))

Davyd McColl • Apr 6 '20 • Edited

JavaScript:

function strCount(haystack, needle) {
  return (
    (haystack || "")
      .match(new RegExp(`${needle}`, "g")) || []
  ).length;
}