# HR issues

In this article we will see InfyTQ Sample Coding question. You will find solution of InfyTQ Coding question in different programming language on this page.

## Problem statement -:

Shovon is an HR in a renowned company and he is assigning people to work. Now he is assigning people work in a fashion where if he assigns somework a work of cost 2, the next person will be strictly getting a job with cost equal or more than 2. Given that Shovon’s company has infinite work and a number of employees, how many distributions can be possible. The cost of jobs can go 0 to 9.

## Function Description:

Complete the special_numbers function in the editor below. It has the following parameter(s):

Parameters:

 Name Type Description N Integer The number of depts. arr[ ] Integer array The number of  employees in each dept..

### Return:

The function must return an INTEGER denoting the sum of answers for all distinct distributions.

### Constraints:

• 1 <= n <= 100
• 1 <= arr[i] <= 200

## Sample Cases:

2
4
1

725
• ### Description

The ans if m = 1 is 1o, which is all numbers from 0 to 9
The ans for m = 2 is 55
The answer for m = 3 is 220
The answer for m = 4 is 715
So fun(4) + fun(1) = 725
`#include <bits/stdc++.h>using namespace std; int func(int s,int p,int n){  if(p==n-1) return 1;  int ans=0;  for(int i=s;i<=9;i++)  ans+=func(i,p+1,n);   return ans;}  int main(){  int n,a, ans=0;  cin>>n;  vector<int> v(n);  for(int i=0;i<n;i++)   {    cin>>a;    for(int i=0;i<=9;i++)    ans+=func(i,0,a);  }  cout<<ans;}`
`n=int(input())a1=[]for i in range(n): a1.append(int(input()))dp=[0]*201dp[1]=10dp[2]=55a=[1]*10i=3while i<201: s=0 for i1 in range(10):   s+=a[i1]   a[i1]=s   dp[i]+=(s*(10-i1))   dp[i]=dp[i]%((10**9)+7) i+=1s1=0for i in a1: s1+=dp[i] s1=s1%((10**9)+7)print(s1)`
`import java.util.*;class Main{  static int func(int s,int p,int n)  {    if(p==n-1) return 1;        int ans=0;        for(int i=s;i<=9;i++)    ans += func(i,p+1,n);     return ans;  }  public static void main (String[]args)  {    Scanner sc = new Scanner (System.in);    int n = sc.nextInt ();    int  ans=0;        for(int i=0; i<n; i++){                int a = sc.nextInt ();        for(int j=0; j<=9; j++)           ans+=func(j,0,a);            }        System.out.println (ans);       }}`