Counting letters in a string using two for-loops

Question

I have to read the string "hello world" and output each letter's frequency using only for loops. The instructor hinted that I'd need to use two loops and gave us the following code to begin:

int ch, count;
for (ch ='a'; ch <='z'; ch++) {
  //count the number of occurrences in a line
  //Print the count>0
}

Edit: I figured I'd necro this question and post the solution I found a year ago due to the fact that this question has been getting a decent amount of hits.

int count;
int value;
for (int i=65; i<91; i++) {
    count=0;
    for (int j=0; j<S.length; j++) {
        value=(int)S[j];
        if (value == i) {
             count++;
        }
    }
    if (count>0) 
       System.out.println((char)i+" -- "+count);
}

For each character, iterate through the string and increment your counter if you see it. — sdasdadas, Mar 20 '13 at 19:27
@ajp15243 I'd say so. Also, I'd personally declare `ch` in the for clause, not above it. — Duncan Jones, Mar 20 '13 at 19:30

Jean Waghetti · Answer 1 · 2013-03-21T20:00:57.157

5

On the second for loop, just go through each character of string and compare it with the current character of first for loop.

(not the solution I would do, just following your instructors hint)

Another way is to store values within a map with the chars as key and a counter of occurrences as value.

HashMap<Character,Integer> map = new HashMap<>();

for (int ii=0; ii<string.length; ii++) {
   char c = string.charAt(ii);
   if (map.containsKey(c)) {
      map.put(c, get(c)++);
   } else {
      map.put(c, 1);
   }
}

UPDATE:

//iterating on the map to output values:
for (char key : map.keySet()) {
   System.out.println(key+": "+map.get(key));
}

edited Mar 21 '13 at 20:00

answered Mar 20 '13 at 19:27

Jean Waghetti

4,711
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Yup, good hint. Don't forget to reset the count after each character. – Duncan Jones Mar 20 '13 at 19:31
It will be a bit excessive approach. For example letter "l" will show up twice. You need one moe cycle before, to eliminate duplicates, the use the string with eliminated duplicates in outer for, where in inner for use full string. – Alex Kreutznaer Mar 20 '13 at 19:40
How would i output from this? – Tdorno Mar 21 '13 at 18:00
There was a error in my code: key and values on the hashmap need to be objects. Look at the edit for your answer. – Jean Waghetti Mar 21 '13 at 19:58

score 1 · Answer 2 · answered Mar 20 '13 at 19:37

1

I would use a simple array. Simply convert each letter to an index and increment the array at that index.

int letters[26];
int index = ch - 'a';
letters[index]++;

answered Mar 20 '13 at 19:37

andre

7,018
4
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yes but how do i interweave a counter that increases by an increment of 1 (starting from 0) when reading over that given index value? This is the part i'm confused on. – Tdorno Mar 20 '13 at 19:52
What about other characters, like '!', '+', '>'? – Jean Waghetti Mar 20 '13 at 19:53
@Tdorno `letters[index]++;` is the counter. – andre Mar 20 '13 at 20:21

score 1 · Answer 3 · edited May 23 '17 at 11:57

To build off of sdasdadas's comment and Jean's respective answer:

The outer for loop would rotate through each character in the alphabet, maintaining a count (which needs to be reset each time the outer loop executes.) The inner loop cycles through the "hello world" string, incrementing the counter if the character serving as the current argument of the outer for loop is found.

UPDATE I can't comment below Andre's answer, but I can provide some pseudocode to address what I think you meant in your comment regarding the counter.

int i;
for (ch characterOuter : alphabet){ //for each character in the alphabet
    i = 0 //i starts at zero, and returns to zero for each iteration  <-----THIS
    for (ch characterInner : "hello world"){
        if (characterOuter == characterInner){
            i++; //increase i by 1 <-----AND THIS
        }//end if
    }//end  innerfor
    if (i > 0) {
        print(characterOuter + " -- " + i);
    } //end if;   <---------------- this if statement was missing
}//end outer for

Also, see this question.

I like Ben's approach but it is repeating every char over displaying once — Tdorno, Mar 21 '13 at 17:58
Do you mean your execution gets caught in an infinite loop, or do you mean that it prints each instance of each character? — eenblam, Mar 22 '13 at 03:16
Oh, I see. You just need another if statement. I'll adjust the code. — eenblam, Mar 22 '13 at 03:20

score 0 · Accepted Answer · answered Oct 28 '13 at 03:14

int count;
int value;
   for (int i=65; i<91; i++) {
      count=0;
      for (int j=0; j<S.length; j++) {
      value=(int)S[j];
      if (value == i) {
         count++;
      }
   }
   if (count>0) 
      System.out.println((char)i+" -- "+count);
}

Counting letters in a string using two for-loops

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