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HackerRank Find Digits Problem Solution
In this Post, we will solve HackerRank Find Digits Problem Solution.
An integer d is a divisor of an integer n if the remainder of n÷d = 0.
Given an integer, for each digit that makes up the integer determine whether it is a divisor
Count the number of divisors occurring within the integer.
Example
n = 124
Check whether 1, 2 and 4 are divisors of 124. All 3 numbers divide evenly into 124 so
return 3.
n = 111
Check whether 1, 1, and 1 are divisors of 111. All 3 numbers divide evenly into 111 so
return 3.
n = 10
Check whether 1 and 0 are divisors of 10. 1 is, but 0 is not. Return 1.
Function Description
Complete the find Digits function in the editor below.
findDigits has the following parameter(s):
- int n: the value to analyze
Returns
- int: the number of digits in n that are divisors of n
Input Format
The first line is an integer, t, the number of test cases. The t subsequent lines each contain an integer, n.
Constraints
1≤t≤15
0 < n < 10 power 9
Sample Input
2
12
1012
Sample Output
2
3
Explanation
The number 12 is broken into two digits, 1 and 2. When 12 is divided by either of those two digits, the remainder is 0 so they are both divisors.
The number 1012 is broken into four digits, 1, 0, 1, and 2. 1012 is evenly divisible by its digits 1, 1, and 2, but it is not divisible by 0 as division by zero is undefined.
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Find Digits C Solution
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main() {
int t,i,j,buffnum;
char b, n[12];
unsigned long modnum, num, cnt=0;
scanf("%d",&t);
for(i=0;i<t;i++){
cnt = 0;
scanf("%s",n);
num=atoi(n);
for(j=0;j<strlen(n);j++){
buffnum = n[j] - '0';
if(buffnum != 0){
modnum=num%buffnum;
if(modnum == 0){
cnt += 1;
}
}
}
printf("%lu\n",cnt);
}
/* Enter your code here. Read input from STDIN. Print output to STDOUT */
return 0;
}
Find Digits C++ Solution
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
int main() {
/* Enter your code here. Read input from STDIN. Print output to STDOUT */
int n;
cin >> n;
for (int i = 0; i < n; i++)
{
int number;
cin >> number;
int sum = 0;
int loc = 0;
while (number / pow(10, loc))
{
int digit = static_cast<int>(number / pow(10, loc++)) % 10;
if (digit != 0 && number % digit == 0)
sum++;
}
cout << sum << endl;
}
return 0;
}
Find Digits C Sharp Solution
using System;
using System.Collections.Generic;
using System.IO;
using System.Linq;
class Solution
{
static void Main(String[] args)
{
int numTestCases = Convert.ToInt32(Console.ReadLine());
List<int> res = new List<int>();
for(int i = 1; i <= numTestCases; i++)
{
int count = 0;
string inputNumber = Console.ReadLine();
long n = Convert.ToInt64(inputNumber);
inputNumber = new string(inputNumber.Where(x => x != '0').ToArray());
foreach(char c in inputNumber)
{
if(n % Convert.ToInt32(c.ToString()) == 0)
{
count++;
}
}
res.Add(count);
}
foreach (int r in res)
{
Console.WriteLine(r);
}
Console.ReadLine();
}
}
Find Digits Java Solution
import java.io.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.function.*;
import java.util.regex.*;
import java.util.stream.*;
import static java.util.stream.Collectors.joining;
import static java.util.stream.Collectors.toList;
class Result {
/*
* Complete the 'findDigits' function below.
*
* The function is expected to return an INTEGER.
* The function accepts INTEGER n as parameter.
*/
public static int findDigits(int n) {
String m = String.valueOf(n);
int[] input = new int[m.length()];
int forRet = 0;
for(int i=0 ; i< m.length() ; i++){
input[i] = (int) (m.charAt(i));
input[i] -=48;
if(input[i] == 0){
continue;
}
if( n%input[i]==0){
forRet++;
}
}
return forRet;
}
}
public class Solution {
public static void main(String[] args) throws IOException {
BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));
BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH")));
int t = Integer.parseInt(bufferedReader.readLine().trim());
IntStream.range(0, t).forEach(tItr -> {
try {
int n = Integer.parseInt(bufferedReader.readLine().trim());
int result = Result.findDigits(n);
bufferedWriter.write(String.valueOf(result));
bufferedWriter.newLine();
} catch (IOException ex) {
throw new RuntimeException(ex);
}
});
bufferedReader.close();
bufferedWriter.close();
}
}
Find Digits JavaScript Solution
function processData(input) {
result = input.split("\n");
for(number in result){
if(result[++number] != result[result.length]){
console.log(getoutput(result[number]));
}
}
}
function getoutput(str){
var totaldivis = 0;
for(char in str ){
if(str % str[char] == 0 ){
totaldivis = totaldivis+1;
}
}
return totaldivis;
}
process.stdin.resume();
process.stdin.setEncoding("ascii");
_input = "";
process.stdin.on("data", function (input) {
_input += input;
});
process.stdin.on("end", function () {
processData(_input);
});
Find Digits Python Solution
T = int(input())
cases = [int(input()) for _ in range(T)]
for case in cases:
count = 0
rem = case
while rem > 0:
if rem % 10 != 0 and case % (rem % 10) == 0:
count += 1
rem //= 10
print(count)
other solutions