TCS Codevita | New ATM Design

In this blog we will discuss a problem asked in TCS Codevita 2019. Its a very interesting question which looks easy but it is quite tricky.

Before running through the solution i would recommend to try it by yourself.

Problem Description:

Automated Teller Machine (ATM) is an electronic device that enables people to withdraw cash from their bank account. Every ATM has a limit for number of currency notes (say N), it can give at a time.


A bank wants to design an ATM for school students. The unique feature of this ATM would be that it would always give maximum number of currency notes possible, to make the students happy. Available denomination of currency notes in the ATM are 100, 200, 500, 1000


Constraints:

N<100


Input Format:

First Line provides an integer, N


Second Line provides an integer denoting the amount you want to withdraw (in multiples of 100)


Third Line provides an integer denoting the available currency note of Rs 100 in the ATM


Fourth Line provides an integer denoting the available currency note of Rs 200 in the ATM


Fifth Line provides an integer denoting the available currency note of Rs 500 in the ATM


Sixth Line provides an integer denoting the available currency note of Rs 1000 in the ATM


Output:

One line containing the maximum number of currency note possible for the desired withdrawal amount. Output should be 0 (zero) if transaction is not possible, for example if sufficient fund is not available in the ATM.


Time Limit:

1


Explanation:

Example 1:


Input:

10
1300
10
10 
10
10

Output:

10

Explanation:

Here,

7 * 100 + 3 * 200 + 0 * 500 + 0 * 1000 hence maximum possible currency = 10.


Example 2:


Input:

5
1700
1
2
2
2

Output:

3

Explanation:

Here,

0 * 100 + 1 * 200 + 1 * 500 + 1 * 1000 hence maximum possible currency = 3.

Implementation using C:

#include<bits/stdc++.h>
using namespace std;
int main()
{
	int sum,sum1,sum2,sum3,amount,n,hundred,twohundred,fivehundred,thousand,i,j,k,l,count=0;
	cin>>n;
	cin>>amount;
	cin>>hundred;
	cin>>twohundred;
	cin>>fivehundred;
	cin>>thousand;
	for( i = 0; i <= hundred; i++)
	{
		sum= i*100;
		if(sum == amount && i<=n && i>count)
		count = i;
		if(sum < amount)
		for( j = 0; j<=twohundred; j++)
		{
		sum1=sum+j*200;
		if(sum1 == amount && (i+j) <= n && (j+i)>count)
		count= i + j;
		if(sum1 < amount)
		for(k = 0; k<=fivehundred; k++)
		{
		sum2= sum1 + k*500;
		if(sum2 == amount && (i+j+k)<=n && (i+j+k)>count)
	        count= i + j + k;
		if(sum2 < amount)
		for(l = 0; l<=thousand; l++)
		{
		sum3= sum2+ l*1000;
		if(sum3 == amount && (i+j+k+l)<=n && (i+j+k+l)>count)
		count= i + j + k + l;
		if(sum3 > amount)
		 l= thousand + 1;
		}
		}
		}
	}
	cout<<count;
	return 0;
}

Happy Coding!

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