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Count the number of Duplicates

In any Programing Language, Write a function that will return the count of distinct case-insensitive alphabetic characters and numeric digits that occur more than once in the input string. The input string can be assumed to contain only alphabets (both uppercase and lowercase) and numeric digits.

Example

"abcde" -> 0 # no characters repeats more than once
"aabbcde" -> 2 # 'a' and 'b'
"aabBcde" -> 2 # 'a' occurs twice and 'b' twice (bandB)
"aA11" -> 2 # 'a' and '1'

Python

input = "aA11"
input = input.lower()
list = [0] * 36
for i in input:
    if i.isdigit():
        list[int(i)+26]+=1
    else:
        list[ord(i)-97]+=1
i=0
counter = 0
while i < len(input):
    if input[i].isdigit():
        counter += list[int(input[i])+26]-1
        i += list[int(input[i])+26]
    else:
        counter += list[ord(input[i])-97]-1
        i += list[ord(input[i])-97]
print counter

https://ideone.com/UstREc

Python 3

def solve(z):
p,count = z.upper(),0
for b in set(p):
if p.count(b) > 1:
count+= 1
return count

s = 'aabBcde'
print(solve(s))

https://ideone.com/DhTkdh
https://ideone.com/DhTkdh

[Modified]

Java...

public int counter(String word){
int count = 0; //this with count if letter or number occur twice
int wordCounter = 0; //this will let us know how many times a letter occur
for (int i = 1; i<word.length(); i++) {
String letter = String.valueOf(word.charAt(i)); //pick each letter of the word
for (int y = 0; y < word.length(); y++) {
//then check how many times that picked letter occur
if (String.valueOf(word.charAt(i)).equals(letter)) {
wordCounter++; //count the letter occurence
if (wordCounter == 2) {
count++; // if the letter occurs twice, we count
}
}
}

}
//then what we count earlier is returned
return count;
}

Tested OK with android studio, I just quickly tested that since I'm currently on a project and it worked..... 100%

^^^ With the above code, you can easily use as follow
.
.
.
int countWord = counter("java" ) ;
// answer will give you 1

SilverG33k:
^^^ With the above code, you can easily use as follow
.
.
.
int countWord = counter("java";
// answer will give you 1

check if d code is correct with ideon na

Javanian:
Python

input = "aA11"
input = input.lower()
list = [0] * 36
for i in input:
    if i.isdigit():
        list[int(i)+26]+=1
    else:
        list[ord(i)-97]+=1
i=0
counter = 0
while i < len(input):
    if input[i].isdigit():
        counter += list[int(input[i])+26]-1
        i += list[int(input[i])+26]
    else:
        counter += list[ord(input[i])-97]-1
        i += list[ord(input[i])-97]
print counter

https://ideone.com/UstREc

wHEN I TESTED YOUR CODE IT GAVE ME THIS....

>>> >>> >>> ... ... ... ... ...   File "<stdin>", line 6
    i=0
    ^
SyntaxError: invalid syntax
>>> >>> ... ... ... ... ... ... ...   File "<stdin>", line 8
    print counter
        ^
SyntaxError: Missing parentheses in call to 'print'
>>> 

SilverG33k:

//pardon me if the code is wrong but trust me, the logic is correct

You know its not actuall to find the length of the word but to find duplicate letters in the word

bash!
## yay bash is a juke box!
number_caught=0  # yap ..jukebox ..global variable
give_me_num () {
  word= $1
  lera= $2
  number_caught= $(echo $word | awk '{print tolower($0)}' | grep -o "lera" | wc -l)
   # ok pls  number_caught is global plss plss plss
}

give_me_num "yup life in programming is " "r"
echo $caught_num

antieverything:

bash!
## yay bash is a juke box!
number_caught=0  # yap ..jukebox ..global variable
give_me_num () {
  word= $1
  lera= $2
  number_caught= $(echo $word | awk '{print tolower($0)}' | grep -o "lera" | wc -l)
   # ok pls  number_caught is global plss plss plss
}

give_me_num "yup life in programming is " "r"
echo $caught_num

Is this for linux Terminal??

edicied:

wHEN I TESTED YOUR CODE IT GAVE ME THIS....

>>> >>> >>> ... ... ... ... ...   File "<stdin>", line 6
    i=0
    ^
SyntaxError: invalid syntax
>>> >>> ... ... ... ... ... ... ...   File "<stdin>", line 8
    print counter
        ^
SyntaxError: Missing parentheses in call to 'print'
>>> 

What interpreter are you using? Hope you are not screwing up the indentation? It runs on ideone, the link I pasted. Tested on 3 different interpreters online and couldn't replicate this error.

edicied:

You know its not actuall to find the length of the word but to find duplicate letters in the word

Finding the lenght is easy as word.lenght() but my code actually finds if the letters appear twice
e.g
counter("java"; // will return 1
according to what you asked, a is the only letter appearing twice

Javanian:


What interpreter are you using? Hope you are not screwing up the indentation? It runs on ideone, the link I pasted. Tested on 3 different interpreters online and couldn't replicate this error.

No, i didn't screw up the indentation and yes i also run it on ideone the link you posted and xame error

SilverG33k:

Finding the lenght is easier as word.lenght() but my code actually finds if the letters appear twice
e.g
counter("java"; // will return 1
according to what you asked, a as the only letter appears twice

It suppose to return 2

edicied:

No, i didn't screw up the indentation and yes i also run it on ideone the link you posted and xame error

What exactly are you talking about?

Javanian:


What exactly are you talking about?

ok

edicied:
It suppose to return 2

Why are you confusing everything, you said it yourself "aabbcde" -> 2 "aabBcde" -> 2 "aA11" -> 2 Then how the Bleep is "Java" supposed to return 2 ??

SilverG33k:

Why are you confusing everything, you said it yourself
"aabbcde" -> 2
"aabBcde" -> 2
"aA11" -> 2
Then how the Bleep is "Java" supposed to return 2 ??

Number of occurrence a appears twice in J(a)v(a) this is a codewars question. You read the question properly before answering online interviews are strict like that

Javanian:
Python

input = "aA11"
input = input.lower()
list = [0] * 36
for i in input:
    if i.isdigit():
        list[int(i)+26]+=1
    else:
        list[ord(i)-97]+=1
i=0
counter = 0
while i < len(input):
    if input[i].isdigit():
        counter += list[int(input[i])+26]-1
        i += list[int(input[i])+26]
    else:
        counter += list[ord(input[i])-97]-1
        i += list[ord(input[i])-97]
print counter

https://ideone.com/UstREc

This works accurately,

SilverG33k:

Why are you confusing everything, you said it yourself
"aabbcde" -> 2
"aabBcde" -> 2
"aA11" -> 2
Then how the Bleep is "Java" supposed to return 2 ??

I ment something like a = 2 and not the return statement

pcguru1:


This works accurately,

Yeah it does

edicied:

I ment something like a = 2 and not the return statement

pcguru1 and op, I thought he was talking about the return variable,,,, sorry mybad

edicied:

Yeah it does

The original comment was it wasn't working i thought i copied the other code but i copied your own code by mistake, so had no choice than to write "it works acurately"

pcguru1:


Number of occurrence a appears twice in J(a)v(a) this is a codewars question. You read the question properly before answering online interviews are strict like that

please when you guys post question, always state the instructions clearly and give good examples

according to the instruction

edicied:
Count the number of Duplicates

In any Programing Language, Write a function that will return the count of distinct case-insensitive alphabetic characters and numeric digits that occur more than once in the input string. The input string can be assumed to contain only alphabets (both uppercase and lowercase) and numeric digits.

Example

"abcde" -> 0 # no characters repeats more than once
"aabbcde" -> 2 # 'a' and 'b'
"aabBcde" -> 2 # 'a' occurs twice and 'b' twice (bandB)
"aA11" -> 2 # 'a' and '1'

the count of distinct character occuring more than once

in all the examples given, the returned value was the number of character occuring more than once (not the number of times they occur!)

java

should return 1 because 'a' is the only character occuring more than once

if we go by your reasoning that

a

occurs twice in

java

, what is going to be the returned value for

aabbbccccddddd

? (check the examples given in the op before answering)

[edit]
ittedly, the question did say the COUNT of the characters occuring more than once but the examples given did not reflect the instruction.
you didnt provide a correct returned value for any (but the first) of the examples

O(N) time

JS

let input = "aA11";
let [o, n, c] = [{}, {}, 0];
input = input.toLowerCase();
for (let i = 0, ch; i < input.length; i += 1) {
    ch = input[i];
    if (o[ch] && !n[ch]) {
        c += 1;
        n[ch] = 1;
    } else o[ch] = 1;
}

console.log(c); // => count

In Ruby:

def duplicate_count(word)
characters = word.downcase.split('')
characters.select{|c| characters.count(c) > 1}.uniq.count
end

edicied:

Is this for linux Terminal??

yeah!

edicied:
Count the number of Duplicates

In any Programing Language, Write a function that will return the count of distinct case-insensitive alphabetic characters and numeric digits that occur more than once in the input string. The input string can be assumed to contain only alphabets (both uppercase and lowercase) and numeric digits.

Example

"abcde" -> 0 # no characters repeats more than once
"aabbcde" -> 2 # 'a' and 'b'
"aabBcde" -> 2 # 'a' occurs twice and 'b' twice (bandB)
"aA11" -> 2 # 'a' and '1'

# PYTHON program

# this function handles the counting of the number of
# occurrence of each letter.
def countOccurrence(word):
WordSet = set([])
myDict = {}

for c in word:
WordSet.add(c)

for letter in WordSet:
myDict.update({letter : word.count(letter)})

return myDict


# Example words.
word1 = 'hippopotamus'
word2 = 'hippopotomonstrosesquipedaliophobia'
word3 = 'pneumonoumtramicroscopicsilicovocanokoniosis'

# create different dictionary type variables that takes the
# dictionary returned from the method call.
dict1 = countOccurrence(word1)
dict2 = countOccurrence(word2)
dict3 = countOccurrence(word3)

print (dict1, '\n' )
print (dict2, '\n' )
print (dict3, '\n' )


# the following allows the to enter the word they want the
# number of occurrence of each letter computed.

# this will prompt the to enter a word.
new_word = input ("Enter a word: " )

new_dict = countOccurrence(new_word)
print (new_dict, '\n')

Special respect to all the programmers here.
JavaScript version <script type="text/javascript"> <!-- var str="Abracadabra"; //source Talabi Olabode var times=str.match(/a/gi).length;// gives the frequency for A var Bleep=str.match(/r/gi).length;// gives the frequency for R
document.write("The number of times "A" appears is:"+times+"<br />'); document.write("The number of times "R" appears is:"+Bleep); //--> </script>

edicied:
Count the number of Duplicates

In any Programing Language, Write a function that will return the count of distinct case-insensitive alphabetic characters and numeric digits that occur more than once in the input string. The input string can be assumed to contain only alphabets (both uppercase and lowercase) and numeric digits.

Example

"abcde" -> 0 # no characters repeats more than once
"aabbcde" -> 2 # 'a' and 'b'
"aabBcde" -> 2 # 'a' occurs twice and 'b' twice (bandB)
"aA11" -> 2 # 'a' and '1'

Java

public class Result {
     private int numDist;
     private List<String> charsRepeated = new ArrayList<String>();

    //Getters and setters skipped, but assume they have been defined.
}

---Separate file.

public class Naira {
     public static int numDistinctChars(String alphaNumWord) {
          Result res = new Result();
          String[] chars = alphaNumWord.toCharArray();
          Set<String> distinctChars = new TreeSet<String>();
          for (String char : chars) {
              if (!distinctChars.add(char)) {
                  res.getCharsRepeated().add(char);
             }         
          }
         res.setNumDist(distinctChars.size());
     }
}

//From the consumer.

.... {
          .....
          Result res = Naira.numDistinctChars("dsds323212" ) ;
         .... 
       }

import java.util.HashMap;
import java.util.Map;
import java.util.Set;
 
public class Details {
 
  public void countDupChars(String str){
 
   
    Map<Character, Integer> map = new HashMap<Character, Integer>(); 

    char[] chars = str.toCharArray();

    for(Character ch:chars){
      if(map.containsKey(ch)){
         map.put(ch, map.get(ch)+1);
      } else {
         map.put(ch, 1);
        }
    }
 
    Set<Character> keys = map.keySet();
 

    for(Character ch:keys){
      if(map.get(ch) > 1){
        System.out.println("Char "+ch+" "+map.get(ch));
      }
    }
  }
 
  public static void main(String a[]){
    Details obj = new Details();
    obj.countDupChars("hello";
  
  }
}

solution in Python(2)
<code>
def counter(word):
numberList = [] # empty list for numbers
alphaList = [] # empty list for alphabets

#sorting num and alpha from word into resp. list
for ch in word:
if ch.isdigit():
numberList.append(ch)
elif ch.isalpha():
alphaList.append(ch.lower()) #converting to lower case since matching is case-INsensitive

#counting and print only num/alpha occurring more than once
for x in numberList:
count = numberList.count(x)
if count >1:
print "{} --> {} ".format(x,count)
while numberList.count(x) > 1 : #weeding tested num
numberList.remove(x)

for i in alphaList:
count = alphaList.count(i)and
if count > 1:
print" {} --> {}".format(i, count)
while alphaList.count(i) > 1: #weeding tested alpha
alphaList.remove(i)
</code>

call function with your string as argument e.g
counter("pEpper62532622")
That's all... can attachment or visit link below if code is not well formatted...

https://gist.github.com/nny326/93d34e5d63b023d17aa5fa4534a9fb4a#file-duplicatecounter-py