Bank Compare | Code Vita 2018 | round 1

Bank Compare

 Problem Description

There are two banks; Bank A and Bank B. Their interest rates vary. You have received offers from both bank in terms of annual rate of interest, tenure and variations of rate of interest over the entire tenure.

You have to choose the offer which costs you least interest and reject the other.

Do the computation and make a wise choice.

The loan repayment happens at a monthly frequency and Equated Monthly Installment (EMI) is calculated using the formula given below :

EMI = loanAmount * monthlyInterestRate /

( 1 - 1 / (1 + monthlyInterestRate)^(numberOfYears * 12))

Constraints

1 <= P <= 1000000

1 <=T <= 50

1<= N1 <= 30

1<= N2 <= 30

Input Format

First line : P – principal (Loan Amount)

Second line : T – Total Tenure (in years).

Third Line : N1 is number of slabs of interest rates for a given period by Bank A. First slab starts from first year and second slab starts from end of first slab and so on.

Next N1 line will contain the interest rate and their period.

After N1 lines we will receive N2 viz. the number of slabs offered by second bank.

Next N2 lines are number of slabs of interest rates for a given period by Bank B. First slab starts from first year and second slab starts from end of first slab and so on.

The period and rate will be delimited by single white space.

Output

Your decision – either Bank A or Bank B.

Explanation

Example 1

Input

10000

20

3

5 9.5

10 9.6

5 8.5

3

10 6.9

5 8.5

5 7.9

Output

Bank B

Example 2

Input

500000

26

3

13 9.5

3 6.9

10 5.6

3

14 8.5

6 7.4

6 9.6

Output

Bank A

Program:

#include <stdio.h>

double power(double b,int a)

{

    int i;

    double pow=1;

    for(i=0;i<a;i++)

    {

        pow=pow*b;

    }

    return pow;

}

int main() {

double p,s,mi,sum,emi,j1,j,bank[5],sq;

int y,n,k,i,yrs,y1,l=0;

    scanf("%lf",&p);

scanf("%d",&y);

for(k=0;k<2;k++)

{

scanf("%d",&n);

sum=0;

for(i=0;i<n;i++)

{

    scanf("%d",&yrs);

    scanf("%lf",&s);

    mi=0;

    j=s/1200;

    j1=1+j;

    y1=yrs*12;

    sq=power(j1,y1);

    emi=p*(j/(1-(1/(sq))));

    mi=emi*y1;

    sum=sum+mi;

}

bank[l++]=sum;

}

if(bank[0]<bank[1])

printf("Bank A");

else

printf("Bank B");

return 0;

}

Output:

10000

20

3

5 9.5

10 9.6

5 8.5

3

10 6.9

5 8.5

5 7.9

Bank B

  

You can also run it in an online IDE: https://ide.geeksforgeeks.org/sPEYMvF71Y

If you have any doubts you can leave it in the comment section or contact me!!
Your feedback are welcomed so kindly leave your feedback below! Cheers!

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