How to display duplicate characters in the string?

I have writ开发者_开发百科ten a program to display the duplicate character(s) in a string, but it displays the characters again if it comes more than 2 times. Any solution to find it exactly?

//to find duplicate characters in the string........

#include<iostream>
using namespace std;
int main()
{
    int i,j;
    char ar[100];
    cout<<"enter string:";
    cin.getline(ar,100);
    for(i=0;ar[i]!='\0';i++)
    {
        for(j=i+1;ar[j]!='\0';j++)
        {
            if(ar[i]==ar[j])
            {  
                cout<<ar[i]<<endl;
                break;
            }
        }
    }

    system("pause");
    return 0;
}

---

Another method is to sort the characters in the string, then examine the sorted string.

Duplicate characters will be easy to find, since they will be next to each other.

---

You should keep track of how many times each character appears:

int count[256]; // <-- assuming char is 8 bytes
for(i=0;i!=256;++i) 
{
    count[i] = 0; // <-- set all counts to be zero
}
for(i=0;ar[i]!='\0';i++)
{
    count[ar[i]] = count[ar[i]] + 1;

    // now you can check if count is 1, and if so then do whatever
}

---

here is the sample code to find duplicate chars in a string. Complete code will be available at http://java2novice.com/java-interview-programs/duplicate-string-character-count/

public void findDuplicateChars(String str){

    Map<Character, Integer> dupMap = new HashMap<Character, Integer>(); 
    char[] chrs = str.toCharArray();
    for(Character ch:chrs){
        if(dupMap.containsKey(ch)){
            dupMap.put(ch, dupMap.get(ch)+1);
        } else {
            dupMap.put(ch, 1);
        }
    }
    Set<Character> keys = dupMap.keySet();
    for(Character ch:keys){
        if(dupMap.get(ch) > 1){
            System.out.println(ch+"--->"+dupMap.get(ch));
        }
    }
}

    

---

I hope this would help:

#include <iostream>
using namespace std;

int main() {
    char str[50];
    cout << "Enter a string" << endl;
    gets(str);

    for(int i=0; str[i]!='\0'; i++)
    {
        for(int j=i+1; str[j]!='\0'; j++)
        {
            if(str[i]==str[j])
            cout << "Character " << str[i] << " is repeated" << endl;
        }
    }


    return 0;
}

---

C++ solution:

#include <iostream>
#include <set>

using namespace std;

set<char> findDuplicate(string s, int c[])
{
    for (int i = 0; i != 256; i++)
        c[i] = 0;

    set<char> set;

    for (int i = 0; s[i] != '\0'; i++)
    {
        c[s[i]] = c[s[i]] + 1;

        if (c[s[i]] > 1)
            set.insert(s[i]);
    }
    return set;
}

int main(int argc, char const *argv[])
{
    int c[256];
    string s = "test string";
    set<char> set = findDuplicate(s, c);
    for (auto it = set.begin(); it != set.end(); ++it)
    cout << *it << " ";
    return 0;
}

---

Below is the code works for [a-z] but it's very space efficient. Because it's using just two integers to find the solution. thanks.

public class AllDuplicatesInString
{
    static class BitSet
    {
        int justPresent, moreThanOnce;

        BitSet() 
        {
            justPresent = moreThanOnce = 0;
        }

        void set(int k)
        {
            if(isSetJustPresent(k))
            {
                k = k - 'a';
                moreThanOnce = moreThanOnce | (1<<k);
                return;
            }
            k = k - 'a';
            justPresent = justPresent | (1<<k);
        }
        boolean isSetJustPresent(int k)
        {
            k = k - 'a';
            return (justPresent & (1<<k))!=0;
        }
        boolean isSetMoreThanOnce(int k)
        {
            k = k - 'a';
            return (moreThanOnce & (1<<k))!=0;
        }
    }
    public static String duplicateChars(String str)
    {
        if(str==null || str.equals("")){
            throw new NullPointerException();
        }
        BitSet b = new BitSet();
        for(int i=0;i<str.length();i++){
            b.set(str.charAt(i));
        }
        StringBuilder stringBuilder = new StringBuilder();
        for(int i=0;i<26;i++){
            if(b.isSetMoreThanOnce(i+'a')){
                stringBuilder.append((char)(i+'a'));
            }
        }
        return stringBuilder.toString();
    }

    public static void main(String[] args)
    {
        String str = "aaaabbbbjjjjsfsfzcncnzcmcncmnczmjsdjs";
        System.out.println(duplicateChars(str));
    }
}

---

Here is the working code from GeeksForGeeks

// C program to count all duplicates from string using hashing 
# include <stdio.h> 
# include <stdlib.h> 
# define NO_OF_CHARS 256 

/* Fills count array with frequency of characters */
void fillCharCounts(char *str, int *count) 
{ 
   int i; 
   for (i = 0; *(str+i);  i++) 
      count[*(str+i)]++; 
} 

/* Print duplicates present in the passed string */
void printDups(char *str) 
{ 
  // Create an array of size 256 and fill count of every character in it 
  int *count = (int *)calloc(NO_OF_CHARS, sizeof(int)); 
  fillCharCounts(str, count); 

  // Print characters having count more than 0 
  int i; 
  for (i = 0; i < NO_OF_CHARS; i++) 
    if(count[i] > 1) 
        printf("%c,  count = %d \n", i,  count[i]); 

  free(count); 
} 

/* Driver program to test to pront printDups*/
int main() 
{ 
    char str[] = "test string"; 
    printDups(str); 
    getchar(); 
    return 0; 
} 

---

If you want to maintain the order, you can try out my code below. I have only made it for lower case tho.

The trick is to keep count of the frequency of each character and checking it each time during the loop.

#include <iostream>
#include <cstring>

using namespace std;
char* print_duplicates(const char str[], int n);

int main(){
    char * actual = new char(20);
    char * dup = new char(20);
    cout << "Enter the string: ";
    cin >> actual;
    int size = strlen(actual);
    dup = print_duplicates(actual,size);
    cout << "The duplicates are: " << dup << endl;
    return 0;
}

char* print_duplicates(const char str[], int n){
    int freq[26]{0}, index{0};
    char* duplicates = new char(20);
    for (int i=0;i<n;++i){
        for (int j=i+1;j<n;j++){
            if (freq[str[i]-'a'] == 0){
                if(str[i]==str[j]){
                    ++freq[str[i]-'a'];
                    duplicates[index++] = str[i];
                }
            }
        }
    }
    return duplicates;
}

---

/* Duplicates in string */

#include <iostream>
using namespace std;

int main() {
    int t; cin >> t;
    while(t--)
    {
        string str;
        cin >> str;
        int N = 128;
        int hsn[N] = {0};
        for(int i = 0; str[i] != '\0'; ++i)
        {
            hsn[str[i]] += 1;
        }
        
        for(int i = 0; i < N; ++i)
        {
            if(hsn[i] > 1)
            {
                cout << (char)i << "  "  << hsn[i] << endl;
            }
        }
    }
    return 0;
}