# Find the homeless

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 -:

There are N Homeless people in the community and N houses in the community. It will be given in the array (people) , height of the person and in the array house capacity of the house is given.

Government decided to give homes for people on the basis of following conditions:

• Priority is given for the people from left to right of the array
• Each person is allotted to a house if and only if the capacity of house is greater than or equal to persons height
• Nearby empty Houses are alloted to the person( starting from extreme left)

You need to find the number of homeless people who have not allotted any home if the government follows the above conditions. So that government will have an idea for how many people they need to allot home for next time.

## Constraints:

• 1 <= N <= 10^3
• 1 <= people[i] <= 10^5
• 1 <= house[i] <= 10^5

## Input Format for Custom Testing:

• The first line contains an integer, N, denoting the number of  people and number of houses.
• Each line i of the N subsequent lines (where 0 <= i <= N) contains an integer describing peoplei.
• Each line i of the N subsequent lines (where 0 <= i <= N) contains an integer describing housei.

## Sample Test Cases

3
4
2
7
3
5
10

0

• ### Explanation

people=[4,2,7]
house=[3,5,10]
People[0] has more priority , from left to right order in houses 5 is the nearest one which fits for people[0]
people[1]=2 will fit in 3 which is nearer from left
people[2]=7 will fit in remaining house of capacity of 10
So no homeless people left so return 0

• Sample Input 2
3
3
8
5
1
9
4

2

• ### Explanation

people=[3,8,5]
house=[1,9,4]
people[0]=3 can fit in 9 which is nearest from left in array house
people[1]=8  cannot fit in any home which is left (i.e, 1 and 4)
people[2]=5 cannot fit in any home which is left (i.e, 1 and 4)
So return 2,which is number of homeless people

`N=int(input())people=[]house=[]#to read data of people and house arraysfor i in range(N): people.append(int(input()))for i in range(N): house.append(int(input()))count=0for i in range(N):    for j in range(N):        if(people[i]<house[j]):            count+=1             house[j]=-1            break print(N-count)`
`import java.util.*;class Main{   public static void main(String args[])  {    Scanner sc = new Scanner (System.in);        int n=sc.nextInt();    int people[]= new int[n];    int house[]= new int[n];          for(int i=0; i<n; i++)       people[i]=sc.nextInt();           for(int i=0; i<n; i++)       house[i]=sc.nextInt();           int count = 0;    for(int i=0; i<n; i++){                for(int j=0; j<n; j++){                        if(people[i]<house[j]){            count+=1 ;            house[j]=-1;            break ;            }        }    }     System.out.println(n-count);  } }`
`#include<bits/stdc++.h>using namespace std;int main(){        int n;    cin>>n;        int people[n], house[n];          for(int i=0; i<n; i++)       cin>>people[i];           for(int i=0; i<n; i++)       cin>>house[i];           int count = 0;    for(int i=0; i<n; i++){                for(int j=0; j<n; j++){                        if(people[i]<house[j]){            count+=1 ;            house[j]=-1;            break ;            }        }    }     cout<<n-count;  } `