Counting occurences of unique letters in a string

Question

Let's say I have the string "UNDERSTAND". Now let's say I want to count the occurrences of each letter in the string. So from my example I would want something like this,

U - 1
N - 2
D - 2
E - 1
R - 1
S - 1
T - 1
A - 1

How can I do this?

Answer 1

You could convert the string to a char array, and add those to a map where the key is the character and the value its count.

Map<Character, Integer> count = new TreeMap<Character, Integer>();
for (char c : yourString.toCharArray()) {
    if (count.containsKey(c)) {
        count.put(c, (int) count.get(c) + 1);
    } else {
        count.put(c, 1);
    }
}

Printing count when yourString = "UNDERSTAND" gives:

{A=1, D=2, E=1, N=2, R=1, S=1, T=1, U=1}

As seen on ideone.

---

If you don't get the idea, here's a pseudo-code version of it (aka Python):

string = 'UNDERSTAND'
count = {} # new map (dict)

for char in string:
    if char in count:
        count[char] = count[char] + 1
    else:
        count[char] = 1

Also available on ideone

Answer 2

You can use a HashMultiset from the guava libraries.

Multiset<Character> set = new HashMultiset<Character>();
for (char c : string.toCharArray()) {
   set.add(c);
}

The Multiset stores each item (character) together with the number of times it was added.

So you can iterate:

for (Character c : set) {
    System.out.println(c + " - " + set.getcount(c));
}

But in case it is a homework, I assume this approach would be frowned upon by the teachers. Figure out how to implement it without the help of additional libraries.

Answer 3

Java answer

Note: please study this code, don't just copy.

public static void main(String[] args) {
    String text = "UNDERSTAND";
    Map<String,Integer> countMap = new HashMap<String,Integer>();
    for(int i=0; i < text.length(); i++) {
        String letter = String.valueOf(text.charAt(i));
        if (countMap.get(letter) == null) {
            countMap.put(letter, 1);
        } else {
            Integer temp = countMap.get(letter);
            temp++;
            countMap.put(letter, temp++);
        }
    }
    System.out.println(countMap);
}

Javascript answer (just for quick test)

var text = "UNDERSTAND";
var count = new Array();
for(i=0; i < text.length; i++) {
    var letter = text[i];
    if (count[letter] == null) {
        count[letter] = 1;
    } else {
        count[letter]++;
    }
}
console.log(count);

Answer 4

Why use complex data structures that require hashing/pointers...etc, when you can solve it with the most simple (and fast) algorithm.

public static void main(String[] args) {

    String s = "UNDERSTOOD";
    int res[]  = new int[256];

    for(int i=0 ; i<s.length() ; i++)
    {
        res[s.charAt(i)]++;
    }

    for(int i=0 ; i<res.length ; i++)
    {
        if(res[i] !=0)
        {
            System.out.println(""+(char)i+" : "+res[i]);
        }
    }
}

The print:
D : 2
E : 1
N : 1
O : 2
R : 1
S : 1
T : 1
U : 1

Roni