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='')

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