CareerCup

hope this helps...

public class CharCount
{

    public static void main(String[] args)
    {

        System.out.print("Enter String to find count of individual char : ");
        Scanner in = new Scanner(System.in);
        String string = in.next();

        char[] c = new char[string.length()];
        int[] count = new int[string.length()];
        int end = 0;

        boolean flag;

        for (int i = 0; i < string.length(); i++)
        {
            flag = true;
            for (int j = 0; j < end; j++)
            {
                if (string.charAt(i) == c[j])
                {
                    flag = false;
                    count[j]++;
                }
            }
            if (flag)
            {
                c[end] = string.charAt(i);
                count[end] = 1;
                end++;
            }
        }
        for (int i = 0; i < end; i++)
        {
//            System.out.println("char : " + c[i] + " count : " + count[i]);
            System.out.print(""+count[i]+c[i]);
        }
    }

}

#include <iostream>
#include <string>

using std::string;

int CalculateRepeats(int inputNum){
    string inString = std::to_string(inputNum);
    char curChar = ' ';											/// THIS MUST BE CHAR
    int count = 0;
    string outString = "";
    for(int i = 0; i<inString.length(); i++){
        if(inString[i]==curChar){count++;}
        else{
            if(count!=0) {outString += std::to_string(count) + curChar;}
            count = 1;
            curChar = inString[i];
        }
    }
    if(count!=0) {outString += std::to_string(count) + curChar;}
    int outNumber = atoi(outString.c_str());								/// CONVERSION STR TO INT
    return outNumber;
}

int main(){
    int testArg = 111222289;
    int testOut = CalculateRepeats(testArg);
    std::cout << testOut;
}

Of course the example is wrong. But the way to solve it :
1. Go over each digit keeping frequency and then read the frequency and the digit
2. If the frequency itself is more than 9, recurse.
Not solving it, because the problem [2] was meant, but not said.

Why is example wrong?

Input: 111222289
output: 31421819

Program should be return above output for that input

@whoknows, thanks a lot for downvoting my comment.
Now I know every downvote is loosing points by 5! It is a nice game!

I am thinking which sort of english language actually allows :
"three ones **three** fours" translated into "3142" .
Thus...

@PeyarTheriyaa
I had a similar solution. But I am wondering if there any better time and space complexity.

Thanks for sharing your thought

Here's my solution

#include <iostream>
#include <string>

using std::string;

int CalculateRepeats(int inputNum){
    string inString = std::to_string(inputNum);
    char curChar = ' ';											/// THIS MUST BE CHAR
    int count = 0;
    string outString = "";
    for(int i = 0; i<inString.length(); i++){
        if(inString[i]==curChar){count++;}
        else{
            if(count!=0) {outString += std::to_string(count) + curChar;}
            count = 1;
            curChar = inString[i];
        }
    }
    if(count!=0) {outString += std::to_string(count) + curChar;}
    int outNumber = atoi(outString.c_str());								/// CONVERSION STR TO INT
    return outNumber;
}

int main(){
    int testArg = 111222289;
    int testOut = CalculateRepeats(testArg);
    std::cout << testOut;
}

its actually n complexity... since we parse only once through the given string.. I'd say that 's about as low as one can go... since we need to parse at least once to look through the string