NewStats: 3,265,279 , 8,186,227 topics. Date: Saturday, 14 June 2025 at 08:28 AM 156029

young02:
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

Bro this might work but it's quite inefficient.why looping thrice...

efosky1246:


Bro this might work but it's quite inefficient.why looping thrice...

quite inefficient... see as u talk am self...
No be for FUN? tsk...

If point-and-kill is what u prefer, have it!

def counter(word):
word = list(word.lower())
for i in word:
if (i.isalpha()) or (i.isdigit()):
count = word.count(i)
if (count > 1):
print "{} ={}" %(i,count)
while(word.count(i) > 1):
word.remove(i)

that's all....
code attached...

edicied:

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";
  
  }
}

Better than the previous java solution, but if all the character is not in the same case, your solution would fail for entry with same character but different cases I.e lower and upper cases which are supposed to count as same.