CareerCup

public int howmany(int[] given, int k) {
        int total =0;

        for (int n : given) {
            total = n== 0 && k == n ? total +1: total;
            while (n > 0) {
                total = n %10 == k ? total +1: total;

                n /= 10;
            }
        }

        return total;
    }

import java.util.Arrays;
import java.util.List;

/**
 *
 * @author vensi
 */
public class CountDigitList {

    public static void main(String[] args) {
        // Conventional way
        List<Integer> l = Arrays.asList(0, 100, 20, 44, 56, 64, 79);
        int k = 4;
        int totalCount = 0;
        for (int i : l) {
             totalCount+=countDigitInNumber(k,i);
        }
        System.out.println(totalCount);
        
        //Using java 8 streams and Strings
        String s = l.parallelStream().map(a->a.toString()).reduce("",(a,b)->a.concat(b));
        System.out.println(s.length() - s.replace(k+"", "").length());
    }

    private static int countDigitInNumber(int digit, int number) {
        int count = 0;
        do {
            int remainder = number % 10;
            number = number / 10;
            if (digit == remainder) {
                count++;
            }
        } while (number > 0);
        return count;
    }
}

import java.util.*;
public class TestDemo
{
	public static void main(String args[])
	{
		Scanner input = new Scanner(System.in);
		int n = input.nextInt(); //total numbers in the list L
		int k = input.nextInt(); //the value to count
		String[] array = new String[n]; //list of numbers goes here
		String totalString = " ";
		for(int i = 0;i<n;i++)
		{
			array[i]=input.next();
			totalString = totalString + array[i];
		}
		//now you have a string for all the list numbers
		char[] arrayOfTotalString = totalString.toCharArray();
		int count = 0;
		for(int i = 0;i<totalString.length();i++)
		{
			if(k==Character.getNumericValue(arrayOfTotalString[i]))
			{
				count++;
			}
		}
		System.out.println(count);  //final count
	}
}

}

import java.util.*;
public class TestDemo
{
public static void main(String args[])
{
Scanner input = new Scanner(System.in);
int n = input.nextInt(); //total numbers in the list L
int k = input.nextInt(); //the value to count
String[] array = new String[n]; //list of numbers goes here
String totalString = " ";
for(int i = 0;i<n;i++)
{
array[i]=input.next();
totalString = totalString + array[i];
}
//now you have a string for all the list numbers
char[] arrayOfTotalString = totalString.toCharArray();
int count = 0;
for(int i = 0;i<totalString.length();i++)
{
if(k==Character.getNumericValue(arrayOfTotalString[i]))
{
count++;
}
}
System.out.println(count); //final count
}
}

#include<iostream>
using namespace std;
int main()
{
int t,n,i;
cin>>t;
while(t--){
cin>>n;
int a[n],c=0,k;
for(i=0;i<n;i++)
cin>>a[i];
cin>>k;
for(i=0;i<n;i++){
while(a[i]){
if(a[i]%10==k)
c++;
a[i]/=10;
}
}
cout<<c<<endl;
}
}

public class AppTest
{

	public static void main(String[] args)
	{
		int[] k = {0, 3, 9};
		int[] l = {0, 45, 3900, 9};
		for (int ek : k)
			System.out.println(ek + " : " + howmany(l, ek));
	}

	public static int howmany(int[] L, int k)
	{
		int total = 0;
		for (int l: L) {
			if (l == 0 && l == k)
				total++;
			if (l > 0) {
				String strL = String.valueOf(l);
				for (int i = 0; i < strL.length(); i++) {
					if (Character.getNumericValue(strL.charAt(i)) == k)
						total++;
				}
			}
		}
		return total;
	}
}

In python3

import re

L = list(map(str,input().split()))

k = int(input())

string = "".join(L)

getNo = re.findall(str(k),string)

print(len(getNo))

First time submitter, but here goes nothing.

int frequencyCount (int[] list, int k)
{
int[] countlist = new int[10];
	for (int i = 0; i < list.length; i++)
	{
	int newK = list[i];
		while(newK > 9)
		{
		countlist[newK % 10]++;
		newK = newK  / 10;
		}
	countlist[newK]++;
	}
return countlist[k];
}

Unfortunately I have only relatively slow O(list.length * digits in each list member) solution to the problem. Inside my algorithm there is a for loop that for every number in a collection calls another function countDigit(int n, int digit) that counts how many times digit appears in that number. countDigit is recursive function with O(digits in number) complexity.

import java.util.*;

public class Example{
    
    public static int countDigit(int n, int digit)
    {
        if(n == digit)return 1;
        if(n == 0 && n != digit)return 0;
        if(n % 10 == digit)
        {
            return 1 + countDigit(n/10, digit);
        }else
        {
            return 0 + countDigit(n/10, digit);
        }
        //for example n = 3242 digit = 2;
        //1. 3242 % 10 == 2 so return 1 + countDigit(3242/10, 2)
        //2. 324 % 10 != 2 so return 0 + countDigit(324/10, 2)
        //3. 32 % 10 == 2 so return 1 + countDigit(32/10, 2);
        //4. 3 % 10 != 2 so return 0 + countDigit(3/10, 2);
        //5. 0 == base case so return 0;
        //countDigit = cD
        //cD(3242, 2) = 1 + cD(324, 2); cd(3242, 2) = 1 + 0 + cD(32, 2);
        //cd(3242, 2) = 1 + 0 + 1 + cd(3, 2); cd(3242, 2) = 1+0+1+cD(0, 2) = 1+0+1+0 = 2;
    }
    
    public static int howMany(List<Integer> list, int digit)
    {
        int counter = 0;
        for(int number : list)
        {
            counter += countDigit(number, digit);
        }
        return counter;
    }

     public static void main(String []args){
        List<Integer> list = new ArrayList<Integer>();
        list.add(2);
        list.add(12);
        list.add(245);
        list.add(34);
        list.add(764);
        list.add(2227); 
        list.add(324262);
        
        int number = 2;
        
        System.out.println("In list there are " + howMany(list,  number) + " digits equals " + number);
        //In list there are 9 digits equals 2
     }
}

ll = new Array(10).fill(0);

function add(aa){
ll[aa]++;
}

function spl(a){
s=a.split("");
// document.write(s[0]+" "+ll[1]+" ");
s.forEach(add);
}

var lists = ["12","1"];

lists.forEach(spl);

document.write(ll);

Flattening the number list to string and iterating over it.
In Python:

flat_L = ''.join(str, L)
k_occurrences = [0] * 10
for num in flat_L:
  k_occurrences[int(num)] += 1

Here L is the initial list with integer values;
k_occurrences - list where index represent the k number; and value represents the occurrences