 # TCS NQT HOUSES IN VILLAGE CODING PROBLEM

TCS NQT HOUSES IN VILLAGE PROBLEM

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There are ‘N’ number of houses in a village. Each house requires a voltage ‘V’ volts of electricity to power up their home appliances. In this context, the houses are arranged in the form of a matrix (row* columns) where ‘N’ is the order of the matrix. The amount of voltage supplied to each house is calculated as current(i) * voltage(r) where ‘i’ is the row number and ‘r’ is the column number. When the voltage supplied to the house (i * r) matches the value ‘V’, the appliances in the house can work. The task here is to find the number of houses whose voltage value(i*r) is the same as ‘V’. If none of the house’s value matches display a message “NO POWER”.

**Input format**

– First line: N, a non-negative integer.

– Second line: V, a non-negative integer.

**Example 1:**

– Input:

– 5     -> Value of N i.e. rows, columns=5

– 15    -> Value of V

– Output :

– 2     -> number of houses whose i*r value matches

– Explanation:

– From the inputs given above, i=5, r=5, the below table represents values (i*r):

“`

+=====+===+====+====+====+====+

| i/j | 1 |  2 |  3 |  4 |  5 |

+—–+—+—-+—-+—-+—-+

|  1  | 1 |  2 |  3 |  4 |  5 |

+—–+—+—-+—-+—-+—-+

|  2  | 2 |  4 |  6 |  8 | 10 |

+—–+—+—-+—-+—-+—-+

|  3  | 3 |  6 |  9 | 12 | 15 |

+—–+—+—-+—-+—-+—-+

|  4  | 4 |  8 | 12 | 16 | 20 |

+—–+—+—-+—-+—-+—-+

|  5  | 5 | 10 | 15 | 20 | 25 |

+=====+===+====+====+====+====+

“`

– From the table above, we can deduce that the value v=15 occurs only in the houses (3,5) (5,3). Hence, the output is 2.

**Example 2:**

– Input:

– 3     -> Value of N i.e. rows, columns = 3

– 10    -> Value of V

– Output :

– NO POWER

– Explanation:

– From the inputs given above, i = 3, r = 3, the below table represents values (i*r):

“`

+=====+===+====+====+

| i/j | 1 |  2 |  3 |

+—–+—+—-+—-+

|  1  | 1 |  2 |  3 |

+—–+—+—-+—-+

|  2  | 2 |  4 |  6 |

+—–+—+—-+—-+

|  3  | 3 |  6 |  9 |

+=====+===+====+====+

“`

– From the table above, we can deduce that the value v=10 doesn’t match with the voltage in any of the houses. Hence, the output is NO POWER

**Constraints:**

– 0 < N <= 50

– 0 < V <= 250

Program in C

``````#include <stdio.h>

int main() {
int n, v, count = 0;
scanf("%d%d", &n, &v);
for(int i = 1 ; i <= n ; i++)
for(int j = 1 ; j <= n ; j++)
if( i * j == v )
count++;
if(count == 0)
printf("NO POWER");
else
printf("%d", count);
return 0;
}
``````

Program in C++

``````#include <iostream>

int main() {
int n, v, count = 0;
std::cin >> n >> v;
for(int i = 1 ; i <= n ; i++)
for(int j = 1 ; j <= n ; j++)
if( i * j == v )
count++;
if(count == 0)
printf("NO POWER");
else
std::cout << count;
return 0;
}
``````

Program in Java

``````import java.util.*;
class Main
{
public static void main(String[] args)
{
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int v = in.nextInt();
in.close();
int count = 0;
for(int i = 1 ; i <= n ; i++)
for(int j = 1 ; j <= n ; j++)
if( i * j == v )
count++;
if(count == 0)
System.out.print("NO POWER");
else
System.out.print(count);
}
}
``````

Program in Python

``````if __name__ == '__main__':
n = int(input())
v = int(input())
count = 0
for i in range(1, n + 1):
for j in range(1, n + 1):
if i * j == v:
count += 1
if count == 0:
print("NO POWER", end='')
else:
print(count, end='')
``````