Alternate CodingBat sumNumbers exercise solution

Question

I was trying to get around the below problem in codingBat for Java: http://codingbat.com/prob/p121193

Given a string, return the sum of the numbers appearing in the string, ignoring all other characters. A number is a series of 1 or more digit chars in a row. (Note: Character.isDigit(char) tests if a char is one of the chars '0', '1', .. '9'. Integer.parseInt(string) converts a string to an int.) sumNumbers("abc123xyz") → 123 sumNumbers("aa11b33") → 44 sumNumbers("7 11") → 18

Below is my solution

public int sumNumbers(String str) {
  final int len=str.length();

  int[] numbers=new int[len];
  int count=0;
  String temp="";
  int sum=0;

  for(int i=0;i<len;i++)
  {
     if(Character.isDigit(str.charAt(i)))
     {
        temp=temp+str.substring(i, i+1);

        if(i==len-1)
        {
          numbers[count]=Integer.parseInt(temp);
          break;
         }
        if(Character.isDigit(str.charAt(i+1)))
        {
           continue;
        }
        else
        {
          numbers[count]=Integer.parseInt(temp);
          count++;
          temp="";
        } 
     }   

   }
   for(int j=0;j<numbers.length;j++)
   {
      sum=sum+numbers[j];
    }
    return sum;  

}

It's a simple problem please provide any alternative efficient answers using regex or anyway other PLEASE DO NOT USE ANYTHING FROM COLLECTIONS FRAMEWORK.

Okay!! Thanks for focusing on code review part rather than reading the actual question....Dear sir what I meant was if there is any code enthusiast who wants to review the code..then they can provide the nitpicks!! — Anirudh, Nov 30 '13 at 13:24
Provide a good reason then why stack overflow has the [code-review] tag?? — Anirudh, Nov 30 '13 at 13:28
In the tag description for the [code-review] tag, which you can read by hovering over the tag with your mouse or clicking through to the tag wiki, it says "Code reviews are off-topic on Stack Overflow, please use codereview.stackexchange.com to request a code review of otherwise working code." I know, it is stupid that StackOverflow.com doesn't show you this message as an error message, we asked for this feature over 4 years ago but Jeff Atwood denied us this feature. I think it is time to revisit that decision. — Robin Green, Nov 30 '13 at 13:31
Okay I have edited the question and removed the tag, please provide and answer!! — Anirudh, Nov 30 '13 at 13:34
The problem is NOT that you added the code review tag, the problem is that you have posted the question on the wrong site. — Robin Green, Nov 30 '13 at 13:35

score 1 · Accepted Answer · answered Nov 30 '13 at 14:36

Here's my solution. It's similar to yours.

public int sumNumbers(String str) {
    int sPos = -1;
    int ePos = -1;
    int sum = 0;

    for (int i = 0; i < str.length(); i++) {
        char c = str.charAt(i);
        if (Character.isDigit(c)) {
            if (sPos < 0) {
                sPos = i;
                ePos = i;
            } else {
                ePos = i;
            }
        } else {
            sum = add(str, sum, sPos, ePos);
            sPos = -1;
        }
    }

    sum = add(str, sum, sPos, ePos);

    return sum;
}

private int add(String str, int sum, int sPos, int ePos) {
    if (sPos >= 0) {
        sum += Integer.parseInt(str.substring(sPos, ePos + 1));
    }
    return sum;
}

score 1 · Answer 2 · answered Nov 30 '13 at 14:39

1

public int sumNumbers(String str) {
    int sum = 0;
    StringTokenizer st = new StringTokenizer(str,"$!;Cabcdefghijklmnopqrstuvwxyz ");
    while (st.hasMoreTokens()) {
         sum += Integer.parseInt(st.nextToken());
    }
    return sum;
}

I wrote it a long time ago, and wonder now what I thought when writing this piece of code back in the days.

answered Nov 30 '13 at 14:39

jcklie

4,054
3
24
42

Can you use regex as an alternate solution? Wouldn't it be better than stringTokenizer performance wise? I am not asserting just guessing! – Anirudh Nov 30 '13 at 16:25
1

I cared about linecount. Regex or tokenizer will degrade the performance I assume, so a solution just based on array access + loops should normally be faster. – jcklie Nov 30 '13 at 17:39

score 1 · Answer 3 · answered Dec 18 '13 at 15:11

I wrote this one. I have no idea does it look better or not :)

    public int sumNumbers(String str) {
String justLetter = "";
int answer = 0; 
int number = 0; 
int factor = 1; 

for (int a = str.length()-1; a >= 0 ; a--) {

justLetter = justLetter + str.charAt(a);

if (Character.isDigit(justLetter.charAt(0))) {
number = Integer.parseInt(justLetter) * factor; 
factor = factor * 10; 
answer = answer + number;
}

else {
factor = 1;
}
justLetter = "";
}
return answer;
}

enducat · Answer 4 · 2014-04-24T21:20:37.597

Here's another solution:

public int sumNumbers(String str) {
      if (str.length()<1)
        return 0;

      int sum=0;
      int lengthOfDigits;

      for (int i=0;i<str.length();i++)
      {
        char currentChar = str.charAt(i);
        lengthOfDigits=1;
        if (Character.isDigit(currentChar))
        {
            for (int j=i;j<str.length()-1;j++)
            {
                char nextChar = str.charAt(j+1);
                if (Character.isDigit(nextChar))
                    lengthOfDigits++;
                else
                    break;
            }   
            String number = str.substring(i,i+lengthOfDigits);
            sum+=Integer.parseInt(number);          

            //dont double count, in the case of 123, skip to 3 instead of count 23 again
            i+=lengthOfDigits;
        }                       
      }
      return sum;
    }

score 1 · Answer 5 · edited Dec 18 '18 at 08:53

1

Simple and short solution in a single for-loop:

   public int sumNumbers(String str) {
    int sum = 0;
    String num = "0";
    for (int i = 0; i < str.length(); i++) {
        char ch = str.charAt(i);
        if (Character.isDigit(ch)) {
            num += ("" + ch);
        } else {
            sum += Integer.parseInt(num);
            num = "0";
        }
    }
    sum += Integer.parseInt(num);
    return sum;
}

Note: It is recommended to use StringBuilder.append() instead of concatenating new strings using +.

edited Dec 18 '18 at 08:53

mel3kings

8,857
3
60
68

answered Dec 07 '16 at 08:59

user3437460

17,253
15
58
106

1

This is a hidden gem here, ingenious way of using String to hold the ints using "0", deserves more upvote – mel3kings Dec 18 '18 at 08:51

Rahul Gheewala · Answer 6 · 2020-05-23T17:54:53.207

My solution

public int sumNumbers(String str) {

    int sum = 0;

    if (!str.matches(".*\\d.*")) return 0;

    for (int i = 0; i < str.length(); i++) {
        if (!Character.isDigit(str.charAt(i))) {
            str = str.replace(str.charAt(i), ' ');
        }

    }
    str = str.trim();

    String[] splitStr = str.split("\\s+");

    for (int z = 0; z < splitStr.length; z++) {
        sum += Integer.parseInt(splitStr[z]);
    }

    return sum;
}

score 0 · Answer 7 · answered Jan 03 '15 at 14:50

public int sumNumbers(String str) {
    int sum = 0;
    int tmp = 0;
    str = str + " ";
    for (int i = 0; i < str.length(); i++) {
            if(Character.isDigit(str.charAt(i))){
                tmp = 0;
                for(int j = i; j < str.length(); j++){
                    if(!Character.isDigit(str.charAt(j))){
                        tmp = Integer.parseInt(str.substring(i,j));
                        i=j;
                        break;
                    }
            }   
                sum += tmp; 
        }
    }
    return sum;
}

My solution for this exercise. – Leon Lai Jan 03 '15 at 14:55 — Leon Lai, Jan 03 '15 at 14:55

score 0 · Answer 8 · answered Apr 02 '15 at 12:24

Alternative solution with single loop:

public int sumNumbers(String str) {
    boolean b = false;
    String con = null;
    int sum = 0;
    int index = 0;

    for (int i = 0; i <= str.length(); i++) {
        if (i < str.length() && Character.isDigit(str.charAt(i))) {
            if (!b) {
                index = i;
                b = true;
            }
        } else if (b) {
            con = str.substring(index, i);
            b = false;
            sum += Integer.parseInt(con);
        }
    }
    return sum;
}

score 0 · Answer 9 · answered May 24 '15 at 02:36

public int sumNumbers(String str) {

  int sum = 0;
  for(int i = 0 ; i < str.length(); i++ ) {
    if(Character.isDigit(str.charAt(i))) {
       int count = 0;
       for (int j = i; j < str.length(); j ++ ) {
           if (Character.isDigit(str.charAt(j))) {
              count++;
           } else {
              break;
           }
       }
       sum += Integer.parseInt(str.substring(i, i + count));
       i += count;
    }
  }


  return sum;
}

score 0 · Answer 10 · edited Jul 13 '18 at 04:45

0

In a single loop:

public int sumNumbers(String str) {

    int sum=0;
    String temp="";
    str=str+" ";
    for (int i = 0; i < str.length(); i++) {
        if(Character.isDigit(str.charAt(i))){
            temp+=str.charAt(i);
        }else if(temp!=""){
                sum+=Integer.parseInt(temp);
                temp="";
        }
    }

    return sum; 
}

edited Jul 13 '18 at 04:45

Community

1
1

answered Oct 20 '15 at 12:19

Pradeep

1
1
4

score 0 · Answer 11 · edited May 23 '17 at 12:02

0

public int sumNumbers(String str) {
if(str.length()==0){return 0;}
 String [] s = str.split("(?<=\\D)(?=\\d)|(?<=\\d)(?=\\D)");
     int sum = 0;
     for (int i = 0; i < s.length; i++) {
        char c = s[i].charAt(0);
        if(Character.isDigit(c)){
            sum += Integer.valueOf(s[i]);
        }

    }

    return sum;
     }

based on this link

edited May 23 '17 at 12:02

Community

1
1

answered Feb 14 '16 at 17:55

ML13

895
8
15

Rk R Bairi · Answer 12 · 2016-11-15T16:53:32.117

    public int sumNumbers(String str) {
    StringBuilder sb = new StringBuilder(); 
    int count = 0;

    for (int i = 0; i < str.length(); i++) {
        if (Character.isDigit(str.charAt(i)) {                           
            sb.append(str.charAt(i));
        } else {                             //clear buffer to take next num
            if (sb.length() != 0) {
                count += (Integer.parseInt(sb.toString()));
                sb.setLength(0);
            }
        }
    }
    if (sb.length() != 0) {
        count += (Integer.parseInt(sb.toString()));
    }

    return count;
    }

score 0 · Answer 13 · answered Dec 23 '16 at 06:28

Here's how I did mine. (I have comments inside).

public int sumNumbers(String str) 
  {
    //Create a sum for the program.
    int sum = 0;
    //Create a blank string
    String temp = "";
    //Create a for loop with i iterations, the amount of str.length().
    for (int i = 0; i < str.length(); i++)
    {
      //If we are not on the last iteration,
      if (i >= 0 && i < str.length() - 1)
      {
        //If the current character is a digit,
        if (Character.isDigit(str.charAt(i)))
        {
          //We add the current character to the temporary String
          temp = temp + str.charAt(i);
          //If the next character is not a digit, we add the temporary String to the sum.
          //And wipe out the temporary string so that when we have another number, we can add it.
          if (!Character.isDigit(str.charAt(i + 1)))
          {
            sum = sum + Integer.parseInt(temp);
            temp = "";
          }
        }
      }
      //If we are on the last iteration, we do all three without conditions because
      //We know it ends with the last character in the String.
      if (i == str.length() - 1)
      {
        if (Character.isDigit(str.charAt(i)))
        {
          temp = temp + str.charAt(i);
          sum = sum + Integer.parseInt(temp);
          temp = "";
        }
      }
    }
    //We return the sum of the numbers.
    return sum;
  }

jegadeesh · Answer 14 · 2017-03-08T09:56:48.707

0

Using String builder the performance is slightly improved 


public static int sumNumbers(String str) {
int sum = 0;
StringBuilder sb = new StringBuilder();
for (int i = 0; i < str.length(); i++) {
char c = str.charAt(i);

if (Character.isDigit(c)) {
sb.append(c);
} else {
if (sb.length() > 0) {
sum = sum + Integer.parseInt(sb.toString());
sb.setLength(0);
}
}
}
return sum + Integer.parseInt(sb.toString());
}

edited Mar 08 '17 at 09:56

answered Mar 07 '17 at 03:04

jegadeesh

109
2
8

Hello, and welcome to StackOverflow. Please format your code, and add some explanation to the answer. – Chait Mar 07 '17 at 03:34

score 0 · Answer 15 · answered Mar 07 '17 at 03:18

Solution with single for loop. Advantage of Using string builder will improve performance by 20%.

public static int sumNumbers(String str) {
        int sum = 0;
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < str.length(); i++) {
            char c = str.charAt(i);

            if (Character.isDigit(c)) {
                sb.append(c);
            } else {
                if (sb.length() > 0) {
                    sum = sum + Integer.parseInt(sb.toString());
                    sb.setLength(0);
                }
            }
        }
        return sum + Integer.parseInt(sb.toString());
    }

Taladork · Answer 16 · 2018-04-25T01:52:31.423

This is a solution based off another post that I'm happy with. What's happening is appending each number to a StringBuilder and adding a space for non-numbers. Then, replace all double+ spaces with a single space and parse + sum it from a stream of the split..

public  int sumNumbers(String str){
  int len = str.length();
  StringBuilder nums = new StringBuilder();

  for(int i=0; i< len; i++){
      char c = str.charAt(i);
      if(Character.isDigit(c)) nums.append(c);
      else nums.append(" ");
  }
  String result = nums.toString().trim().replaceAll(" +", " ");
  return result.equals("") ? 0 : Arrays.stream(result.split(" ")).mapToInt(Integer::parseInt).sum();
}

score 0 · Answer 17 · answered Mar 20 '19 at 07:52

public int sumNumbers(String str) {
int count = 0;
int i =0;
while(i<str.length()){
boolean isDigit = Character.isDigit(str.charAt(i));
int j =i;
String tote = "";
while(isDigit==true&&j<str.length()){
 if(Character.isDigit(str.charAt(j))==true)
 {tote += str.charAt(j);
 j++;}
 else
 break;
}
if(tote.length() > 0)
{count += Integer.parseInt(tote);
i+=tote.length();}
else
{i++;}

}
return count;}

This was the easiest way for me!

Varad Paralikar · Answer 18 · 2019-10-30T18:51:33.170

0

Here is my answer

public int sumNumbers(String str) {
  String num = "";
  int total= 0;

  for(int i=0;i<str.length();i++){

    if(Character.isDigit(str.charAt(i))){ // detected a number
        int pos = i;
        while(pos < str.length() && Character.isDigit(str.charAt(pos))){
          pos++; // running the loop until we reach the end of the digits that comprise the number

        }
          //Now adding the number to the total
         total += Integer.parseInt(str.substring(i,pos));  
          i = pos; // skipping the position of iterator to past the obtained number 
    }


  }

  return total;//returning result


}

edited Oct 30 '19 at 18:51

answered Oct 30 '19 at 17:15

Varad Paralikar

81
1
11

Plese explain you answer with proper comment. – Kaushik Burkule Oct 30 '19 at 18:22
Well In the for loop I first detect if a char is digit then I run a while loop for the chars that are in front of our digit char to get the whole number then I just add the number obtained to the total and skip the loop iterator (here i) to past the position of obtained number ( i = pos) simple :) – Varad Paralikar Oct 30 '19 at 18:46

score 0 · Answer 19 · answered Nov 02 '19 at 19:35

public static int sumNumbers(String str) {
        int result = 0;
        StringBuilder sb = new StringBuilder();
        if (str.length() < 1) return 0;
        for (int i = 0; i < str.length(); i++) {
            if (Character.isDigit(str.charAt(i))) {
                sb.append(str.charAt(i));
            } else {
                if (sb.length() > 0) {
                    result += Integer.parseInt(sb.toString());
                    sb.delete(0, sb.length());  //we need to clear the stringBuilder to account for new Numbers
                }
            }
        }
        if (sb.length() > 0) {
            result += Integer.parseInt(sb.toString());
        }
        return result;
    }

justNate · Answer 20 · 2020-08-07T13:23:47.810

0

public int sumNumbers(String str) {
      String nums = "";
      int sum = 0;
      for (int i = 0; i < str.length(); i++){
        if (Character.isDigit(str.charAt(i))){
          nums += str.charAt(i);
          if (i == str.length()-1 && Character.isDigit(str.charAt(i)))
            sum += Integer.parseInt(nums);
        }
        else if (i > 0 && !Character.isDigit(str.charAt(i)) && Character.isDigit(str.charAt(i-1))){
          sum += Integer.parseInt(nums);
          nums = "";
        }
      }
      return sum;
    }

edited Aug 07 '20 at 13:23

answered Jul 20 '20 at 15:31

justNate

1
2

this uses a for loop that checks whether a certain index is a character and if it is, then parse that character to a integer. Since Integer.parseInt() takes in a string we need to turn the character into a string. – justNate Jul 20 '20 at 15:34
This does not solve the problem. Several tests fail with this answer. (https://codingbat.com/prob/p121193) – tim-montague Aug 06 '20 at 07:56
@tfmontague this should fix that – justNate Aug 07 '20 at 13:26

tim-montague · Answer 21 · 2020-08-10T04:38:15.707

CodingBat sumNumbers solution

Uses one for-loop, and avoids special-case logic
Uses integers, and avoids method calls

public int sumNumbers(String str) {
  int sum = 0;
  int num = 0;

  // If the string is closed,
  // we don't have to write a special case
  // to sum in numbers at the end of a string
  String closedString = str + ".";

  for (int i = 0; i < closedString.length(); i += 1) {
    // Characters are just integers
    int c = (int) closedString.charAt(i);

    if (c >= '0' && c <= '9') {
      // ParseInt alternative
      // Decimals are base 10
      num = num * 10 + c - '0';
    } else {
      sum += num;
      num = 0;
    }
  }

  return sum;
}

Or maybe you have other stuff to do?

public int sumNumbers(String str) {
  return Arrays
    .stream(str.split("\\D+"))
    .filter(s -> !s.equals(""))
    .mapToInt(Integer::parseInt)
    .sum();
}

score 0 · Answer 22 · answered May 31 '22 at 17:21

My solution:

public int sumDigits(String str) {
  int counter = 0;
  
  for (int i = 0; i < str.length(); i++) {
    if (Character.isDigit(str.charAt(i))) {
      counter += Integer.parseInt(str.valueOf(str.charAt(i)));
    }
  }
  
  return counter;
}

It passed all the tests on CodingBat.

score -1 · Answer 23 · answered Dec 29 '17 at 14:50

See My code.

    public static int sumNumbers(String str) {
        char[] characters= str.toCharArray();
        int start=-1,end,sum=0,prev=0;
        for(int i=0; i<str.length(); i++){
            if(i==0) {
                if (Character.isDigit(characters[0])) {
                    start = 0;
                }
            }else{
                if (Character.isDigit(characters[i]) && !Character.isDigit(characters[i-1])) {
                    start = i;
                }
                if (!Character.isDigit(characters[i]) && Character.isDigit(characters[i-1])) {
                    end = i;
                    if(start>-1){
                        sum+=Integer.parseInt(str.substring(start,end));
                    }
                }
            }
            if(i==str.length()-1){
                if(start>-1) {
                    sum += Integer.parseInt(str.substring(start));
                }
            }
        }
        return sum;
    }

This is not a solution. Tests fail (https://codingbat.com/prob/p121193) — tim-montague, Aug 06 '20 at 08:00

Alternate CodingBat sumNumbers exercise solution

23 Answers23

CodingBat sumNumbers solution