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javier.hdvs.marino.bn

NumbersVsLetters

You will be given a String which will be composed of different groups of letters and numbers separated by a blank space.

You will have to calculate for each group whether the value of the sum of the letters is greater, equal or less than the multiplication of the numbers. When the values are equal the group will not count towards the final value.

The value of each letter corresponds to its position in the English alphabet so a=1, b=2, c=3 ... x=24, y=25 and z=26. If there are more groups that have a greater numeric value you will return 1. If the alphabetic value is greater for more groups, you will return -1. If the sum of groups that have greater numeric value is equal to the sum of groups that have greater alphabetic value you will return 0.

Example:

("a1b 42c") ->
a1b(lettersValue(1(a)+2(b)=3)>numbersValue(1))>-1(letters win)
42c(lettersValue(3(c))<numbersValue(4*2=8))->1(numbers win)
returns 0 since 1-1=0(draw)

("12345 9812") ->returns 1
("abc def") -> returns -1
("abcdef 12345") ->returns 0 since the first group has
a greater alphabetic value and the second group has a 
greater numeric value
("a1b2c3") ->returns 0 since the sum of the letters 
(1(a)+2(b)+3(c)) is equal to the multiplication of the
numbers (1*2*3)
("z27jk 99sd")->
//z27jk
26('z')+10('j')+11('k')>2*7-> 47>14 -> -1
//99sd
9*9>19('s')+4('d')->81>23->1
//returns
1-1->0
("a1 b3 44e")->returns 0+1+1=2->1

Notes:

  • There won't be any uppercase letters
  • Special characters such as # or $ do not have any value ("$@#") returns 0
  • There might be empty Strings
  • There can be more than one blank space splitting each group and there can be trailling spaces, for example ("aei 213 ")
  • Watch out for groups that only contain a zero!
javier.hdvs.marino.bn

NumbersVsLetters

You will be given a String which will be composed of different groups of letters and numbers separated by a blank space. You will have to calculate for each group whether the value of the sum of the letters is greater, equal or less than the multiplication of the numbers. When the values are equal the group will not count towards the final value. The value of each letter corresponds to its position in the English alphabet so a=1, b=2, c=3 ... x=24, y=25 and z=26. If there are more groups that have a greater numeric value you will return 1. If the alphabetic value is greater for more groups, you will return -1. If there are the same number of groups with greater numeric and alphabetic values, you will return 0.

Example:

("12345 9812") -> 1
("abc def") -> -1
("abcdef 12345") -> 0 since the first group has a greater letter value and the second group has a greater numeric value
("a1b2c3") -> 0 since the sum of the letters (1(a)+2(b)+3(c)) is equal to the multiplication of the numbers (1*2*3)

Notes:

  • There won't be any uppercase letters
  • Special characters such as # or $ do not have any value
  • There might be empty Strings
Code
Diff
  • import java.util.HashMap;
import java.util.Map;
public class NumbersVsLetters{
  
  public static int numbersVsLetters(String str) {
	    if (str.isEmpty() || str.isBlank()) return 0;
	    Map<Character, Integer> allChars = fillMap();
	    int score = 0;
	    int numberMultiplication = -1;
	    int characterSum = 0;
	    for (char c : str.toCharArray()) {
	        if (c == ' ') {
	            if (numberMultiplication > 0) score += Integer.compare(numberMultiplication, characterSum);
	            else if (characterSum > 0) score--;
	            numberMultiplication = -1;
	            characterSum = 0;
	        } else if (Character.isDigit(c)) numberMultiplication = Character.getNumericValue(c) * Math.abs(numberMultiplication);
            else if (Character.isAlphabetic(c)) characterSum += allChars.get(c);
	    }
	    if (numberMultiplication > 0) score += Integer.compare(numberMultiplication, characterSum);
	    else if (characterSum > 0) score--;
	    return Integer.compare(score, 0);
	}

	public static Map<Character, Integer> fillMap(){
		Map<Character, Integer> allChars = new HashMap<>();
		for (int i = 0; i < 26; i++) {
			char c = (char) ('a' + i);
		    allChars.put(c, i + 1);
		}
		return allChars;
	}
  
}
    • import java.util.HashMap;
    • import java.util.Map;
    • public class NumbersVsLetters{
    • public static int numbersVsLetters(String str) {
    • if (str.isEmpty() || str.isBlank()) return 0;
    • Map<Character, Integer> allChars = fillMap();
    • int score = 0;
    • int numberMultiplication = -1;
    • int characterSum = 0;
    • for (char c : str.toCharArray()) {
    • if (c == ' ') {
    • if (numberMultiplication > 0) score += Integer.compare(numberMultiplication, characterSum);
    • else if (characterSum > 0) score--;
    • numberMultiplication = -1;
    • characterSum = 0;
    • } else if (Character.isDigit(c)) {
    • numberMultiplication = Character.getNumericValue(c) * Math.abs(numberMultiplication);
    • } else if (Character.isAlphabetic(c)) {
    • characterSum += allChars.get(c);
    • }
    • } else if (Character.isDigit(c)) numberMultiplication = Character.getNumericValue(c) * Math.abs(numberMultiplication);
    • else if (Character.isAlphabetic(c)) characterSum += allChars.get(c);
    • }
    • if (numberMultiplication > 0) score += Integer.compare(numberMultiplication, characterSum);
    • else if (characterSum > 0) score--;
    • return Integer.compare(score, 0);
    • }
    • public static Map<Character, Integer> fillMap(){
    • Map<Character, Integer> allChars = new HashMap<>();
    • for (int i = 0; i < 26; i++) {
    • char c = (char) ('a' + i);
    • allChars.put(c, i + 1);
    • }
    • return allChars;
    • }
    • }
javier.hdvs.javier.hdFailed Tests

Discover the value

Code
Diff
  • public class pp{
  
  public static int hello(String s) {
		Map<String, Integer> map = new HashMap<>();
		int result = 0,num=1,letr=0;
		boolean correcto= false;
		for (int i = 0; i < 26; i++) {
			map.put(Character.toString('a' + i), i + 1);
		}

		// map.forEach((a,b)->System.out.println(a+"-"+b));

		String numeros = "", letras = "";
		for (int c = 0; c < s.length(); c++) {
			if (!String.valueOf(s.charAt(c)).equals(" ") && s.length()!=c+1) {
				if (String.valueOf(s.charAt(c)).matches("^\\d$")) {
					//numeros = numeros + s.charAt(c);
					correcto=true;
					num=num*s.charAt(c);
				} else if (String.valueOf(s.charAt(c)).matches("^[a-z]$")) {
					//letras = letras + s.charAt(c);
				}

			} else {
				
				
				correcto=false;
				numeros="";letras="";
			}

		}

		System.out.println(numeros);
		System.out.println(letras);

		return result;
	}
}
    • public class pp{
    • public static int hello(String s) {
    • Map<String, Integer> map = new HashMap<>();
    • int result = 0,num=1,letr=0;
    • boolean correcto= false;
    • for (int i = 0; i < 26; i++) {
    • map.put(Character.toString('a' + i), i + 1);
    • }
    • // map.forEach((a,b)->System.out.println(a+"-"+b));
    • String numeros = "", letras = "";
    • for (int c = 0; c < s.length(); c++) {
    • if (!String.valueOf(s.charAt(c)).equals(" ") && s.length()!=c+1) {
    • if (String.valueOf(s.charAt(c)).matches("^\\d$")) {
    • //numeros = numeros + s.charAt(c);
    • correcto=true;
    • num=num*s.charAt(c);
    • } else if (String.valueOf(s.charAt(c)).matches("^[a-z]$")) {
    • //letras = letras + s.charAt(c);
    • }
    • } else {
    • correcto=false;
    • numeros="";letras="";
    • }
    • }
    • System.out.println(numeros);
    • System.out.println(letras);
    • return result;
    • }
    • }
javier.hd

Discover the value

Se trata de un ejercicio en el que se le pasa a una función un String que tenga una estructura así -> "pe1321pe3 zw2d1", como se puede ver es una cadena formada por caracteres del alfabeto inglés y números del 0 al 9, el objetivo del ejercicio es determinar si la suma de los valores de las letras (serán del 1 al 26) es mayor a la multiplicación de los números, en caso de que haya más palabras cuyo valor de letras sumadas sea mayor al valor de números multiplicados se devolverá un -1, en caso de que haya más palabras con un valor mayor de multiplicación de números que suma de letras devolveremos un 1 y en caso de que haya el mismo número retornaremos 0. En caso de que se introduzca una cadena de texto vacía devolveremos 0, y adicionalmente podemos ver si añadimos signos especiales como !$# que no cuenten en las cuentas pero que estén por molestar.

public class pp{
  
}