How to display duplicate characters in a string in C++? - c++

Question

I am working on some code for a class that requires me to output duplicates in a string. This string can have any ascii character but the output needs to show only the repeated character and the total number of times it repeats.
Here are some sample inputs and outputs
mom, m:2
taco, No duplicates
good job, o:3
tacocat, t:2 c:2 a:2
My code works for all but the last test case, the t:2 and a:2 appears twice, Now I have come to the conclusion that I need to store duplicated characters somewhere and run a check on that list to see if that duplicate has already been printed so I tried using a vector.
My method is to push the character into the vector as the duplicates are printed and if a character is already in the vector then it is skipped in the printing. But I have not been able to find a way to this. I tried to use the find() from #include<algorithm> but got a syntax error that I am unable to fix. Is there a function that I can apply for this? Or am I going about this in a bad way?
I found the implementation of find() here & I looked here but they don't match and it breaks my code completely when I try to apply it.
#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
using namespace std;
vector <char> alreadyprintedcharacters;
void findrepeats(string const&);
int main()
{
string input;
cout << "Enter the input : ";
getline(cin, input);
findrepeats(input);
return 0;
}
void findrepeats(string const &in)
{
int trackerOfDuplicates = 0;
int asciiArray[256];
char ch;
int charconv;
for (int i = 0; i < 256; i++) // creates my refference array for the comparison and sets all the values equal to zero
asciiArray[i] = 0;
for (unsigned int i = 0; i < in.length(); i++)
{
ch = in[i];
charconv = static_cast<int>(ch);
if (asciiArray[charconv] == 0)
{
asciiArray[charconv] = 1;
}
else if (asciiArray[charconv] > 0)
{
asciiArray[charconv] = asciiArray[charconv]++;
}
}
bool trip = false;
for (unsigned int i = 0; i < in.length(); i++)
{
char static alreadyprinted;
char ch = in[i];
if ((asciiArray[ch] > 1) && (ch != alreadyprinted) && (find(alreadyprintedcharacters.begin(), alreadyprintedcharacters.end(), ch)!= alreadyprintedcharacters.end()))// change reflected HERE
{
cout << in[i] << " : " << asciiArray[ch] << endl;//???? maybe a nested loop
trip = true;
alreadyprinted = ch;
alreadyprintedcharacters.push_back(alreadyprinted);
}
}
if (trip == false)
cout << "No repeated characters were found.\n";
}

Answer 1

Your code works fine for me (gives the correct output for tacocat) if you fix the error related to std::find:
std::find doesn't return a bool, it returns an iterator (in your case, a std::vector<char>::iterator). If you want to check if std::find found something, you should compare it to alreadyprintedcharacters.end(), because that's what std::find returns if it didn't find something.

Answer 2

You can create an integer array of 256 and initialize it to 0 at first. Then loop over characters in the string and increment each index that corresponds to that letter. In the end, you can print out letters that have values greater than 1. Just change your findrepeats function to the following:
void findrepeats(string const &in)
{
int asciiArray[256];
char ch;
int charconv;
bool foundAny = false;
for (int i = 0; i < 256; i++) asciiArray[i] = 0;
for (unsigned int i = 0; i < in.length(); i++)
{
ch = in[i];
charconv = static_cast<int>(ch);
asciiArray[charconv]++;
}
for (unsigned int i = 0; i < 256; i++)
{
char static alreadyprinted;
if (asciiArray[i] > 1)
{
foundAny = true;
cout << static_cast<char>(i) << " : " << asciiArray[i] << endl;
}
}
if (!foundAny)
cout << "No repeated characters were found.\n";
}

Answer 3

You have to make following changes in your code
change the loop body where you are updating the reference array for the comparison and sets all the values like this:
//your code
else if (asciiArray[charconv] > 0)
{
asciiArray[charconv] = asciiArray[charconv]++;
}
in the above code the value of asciiArray[charconv] doesn't change because it is a post increment asciiArray[charconv]++; , either change it to a pre increment ++asciiArray[charconv]; or write asciiArray[charconv] = asciiArray[charconv]+1;
Here is a link to this why it doesn't increment.
Also you can change the loop like this,more simplified:
for (unsigned int i = 0; i < in.length(); i++)
{
ch = in[i];
charconv = static_cast<int>(ch);
asciiArray[charconv]++;
}
change the type of found to std::vector<char>::iterator coz find returns an iterator to the first element in the range that compares equal to val & if no elements match, the function returns last.
std::vector<char>::iterator found = find(alreadyprintedcharacters.begin(), alreadyprintedcharacters.end(), ch);
Then your condition should be like
if((asciiArray[ch] > 1) && (ch!=alreadyprinted) && (found == alreadyprintedcharacters.end()))

Answer 4

I don't quite get why you need all of that code (given you stated you can't use std::map).
You declared an array of 256 and set each item to 0, which is OK:
for (int i = 0; i < 256; i++)
asciiArray[i] = 0;
Now the next step should be simple -- just go through the string, one character at a time, and increment the associated value in your array. You seem to start out this way, then go off on a tangent doing other things:
for (unsigned int i = 0; i < in.length(); i++)
{
ch = in[i]; // ok
asciiArray[ch]++;
We can set a boolean to true if we discover that the character count we just incremented is > 1:
bool dup = false;
for (unsigned int i = 0; i < in.length(); i++)
{
ch = in[i]; // ok
asciiArray[ch]++;
if ( asciiArray[ch] > 1 )
dup = true;
}
That is the entire loop to preprocess the string. Then you need a loop after this to print out the results.
As to printing, just go through your array only if there are duplicates, and you know this by just inspecting the dup value. If the array's value at character i is > 1, you print the information for that character, if not, skip to the next one.
I won't show the code for the last step, since this is homework.

Answer 5

Just met similar question last week, here is what I did, maybe not a best solution, but it did work well.
string str("aer08%&#&%$$gfdslh6FAKSFH");
vector<char> check;
vector<int> counter;
//subscript is the bridge between charcheck and count. counter[sbuscript] store the times that check[subscript] appeared
int subscript = 0;
bool charisincheck = false;
for (const auto cstr : str) //read every char in string
{
subscript = 0;
charisincheck = false;
for (const auto ccheck : check) // read every element in charcheck
{
if (cstr == ccheck)//check if the char get from the string had already existed in charcheck
{
charisincheck = true; //if exist, break the for loop
break;
}
subscript++;
}
if (charisincheck == true) //if the char in string, then the count +1
{
counter[subscript] += 1;
}
else //if not, add the new char to check, and also add a counter for this new char
{
check.push_back(cstr);
counter.push_back(1);
}
}
for (decltype(counter.size()) i = 0; i != counter.size(); i++)
{
cout << check[i] << ":" << counter[i] << endl;
}met

Answer 6

import java.util.*;
class dublicate{
public static void main(String arg[]){
Scanner sc =new Scanner(System.in);
String str=sc.nextLine();
int d[]=new int[256];
int count=0;
for(int i=0;i<256;i++){
d[i]=0;
}
for(int i=0;i<str.length();i++){
if(d[str.charAt(i)]==0)
for(int j=i+1;j<str.length();j++){
if(str.charAt(i)==str.charAt(j)){
d[str.charAt(i)]++;
}
}
}
for(char i=0;i<256;i++){
if(d[i]>0)
System.out.println(i+" :="+(d[i]+1));
}
}
}

Answer 7

//here simple code for duplicate characters in a string in C++
#include<iostream.h>
#include<conio.h>
#include<string.h>
void main(){
clrscr();
char str[100];
cin>>str;
int d[256];
int count=0;
for(int k=0;k<256;k++){
d[k]=0;
}
for(int i=0;i<strlen(str);i++){
if(d[str[i]]==0)
for(int j=i+1;j<strlen(str);j++){
if(str[i]==str[j]){
d[str[i]]++;
}
}
}
for(int c=0;c<256;c++){
if(d[c]>0)
cout<<(char)c<<" :="<<(d[c]+1)<<"\n";
}
getch();
}