CODEWITHSUNDEEP
java
Rasika Joshi
Rasika Joshi

Reputation: 33

Trying to count occurence of each letter in String in Java

I am not able to get the correct occurrence of each letter present in the string. What is wrong with my code?

Code:

public class consecutiveCharacters {
    public static String solve(String A, int B) {
        HashMap<String, Integer> map = new HashMap<>();
        for (int i=0;i<A.length();i++) {
            Integer value = map.get(A.charAt(i));
            if(value == null){
                map.put(String.valueOf(A.charAt(i)),1);
            }
            else {
                map.put(String.valueOf(A.charAt(i)),i+1);
            }
        }
        System.out.println(map); 
        String text = "i feel good";
        return text;
    }

    public static void main(String[] args) {
        solve("aabbccd",2);
    }
}

The output of the map should be:

{a=2, b=2, c=2, d=1}

Yet the output that the above code is giving is:

{a=1, b=1, c=1, d=1}

Upvotes: 0

Views: 800

Answers (7)

Shtefan
Shtefan

Reputation: 808

Simple as this:

String s = "aabbcd";

Map<Character, Integer> map = new HashMap<>();
    for (var c: s.toCharArray()){
        map.put(c, (int)Arrays.stream(s.chars().toArray()).filter((i)->i == c).count());
    }

Output: {a=2, b=2, c=1, d=1}

Upvotes: 0

Behnam Nikbakht
Behnam Nikbakht

Reputation: 61

Try this using getOrDefault instead of checking if the map contains the key:

    Map<Character, Integer> map = new HashMap<>();
    for(char c: A.toCharArray()){
        map.put(c, map.getOrDefault(c, 0) + 1);
    }

Upvotes: 0

ashish malgawa
ashish malgawa

Reputation: 187

The reason why it is not working is that you are putting a key of type String and you are trying to get using a character value. The fastest way to solve the bug would be to change the get to

map.get(String.valueOf(A.charAt(i)))

Another way to do a check if a key does not exist is to use the contains method.

map.containsKey(String.valueOf(A.charAt(i)))

Upvotes: 0

Tanbir Ahmed
Tanbir Ahmed

Reputation: 320

The problem seems to be in this line map.put(String.valueOf(A.charAt(i)),i+1) where it is resetting the count every time it found a recurrence of a character. You need to get the previous count of a character if it recurred. Try changing the line to map.put(String.valueOf(A.charAt(i)),map.get(String.valueOf(A.charAt(i)))+1).

Also you can use Character directly as the key in your HashMap object, which will make the code much simpler. Hope the following code will be helpful:

public static String solve(String A, int B) {
    HashMap<Character, Integer> count = new HashMap<>();
    for(Character ch: A.toCharArray()){
        if(count.containsKey(ch) == false){
            count.put(ch, 0);
        }
        count.put(ch, count.get(ch)+1);
    }
    
    System.out.println(count);
    String text = "i feel good";
    return text;
}

// A = "aabbccd"
// count = {a=2, b=2, c=2, d=1}

Upvotes: 0

Oleg Cherednik
Oleg Cherednik

Reputation: 18245

A more generic solution is to use Map<Character, Integer>:

public static void solve(String str) {
    Map<Character, Integer> map = new HashMap<>();
    
    for(int i = 0; i < str.length(); i++) {
        char ch = Character.toLowerCase(str.charAt(i));
        map.merge(ch, 1, Integer::sum);
    }
    
    System.out.println(map);
}

In case you have only English letter, then solution could be more memory effective:

public static void solve(String str) {
    str = str.toLowerCase();
    int[] letters = new int[26];

    for (int i = 0; i < str.length(); i++)
        letters[str.charAt(i) - 'a']++;

    for (int i = 0; i < letters.length; i++)
        if (letters[i] > 0)
            System.out.format("%c=%d\n", 'a' + i, letters[i]);
}

Upvotes: 0

Thomas
Thomas

Reputation: 88707

Using Map.merge() you can rid of checking this yourself:

Map<Character, Integer> map = new HashMap<>();
for( char c : A.toCharArray()) {
    map.merge(c, 1, Integer::sum);                      
}

merge() takes 3 parameters:

  • A key which is the character in your case
  • A value to put for the key, which is 1 in your case
  • A merge bi-function to combine the new and existing values if the entry already existed. Integer::sum refers to Integer's method static int sum(int a, int b) which basically is the same as your value + 1.

Upvotes: 3

Tim Biegeleisen
Tim Biegeleisen

Reputation: 520958

I would use the following logic:

public static void solve(String A) {
    Map<Character, Integer> map = new HashMap<>();
    for (int i=0; i < A.length(); i++) {
        char letter = A.charAt(i);
        int count = map.get(letter) != null ? map.get(letter) : 0;
        map.put(letter, ++count);
    }

    System.out.println(map);  // prints {a=2, b=2, c=2, d=1}
}

public static void main(String[] args) {
    solve("aabbccd");
}

The main changes I made were to switch to using a Map<Character, Integer>, since we want to keep track of character key counts, rather than strings. In addition, I tightened the lookup logic to set the initial count at either zero, in case a new letter comes along, or the previous value in the map, in case the letter be already known.

Upvotes: 1

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