# Taum and B’day – HackerRank Solution

In this post, we will solve Taum and B’day HackerRank Solution. This problem (Taum and B’day) is a part of HackerRank Algorithms series.

Taum is planning to celebrate the birthday of his friend, Diksha. There are two types of gifts that Diksha wants from Taum: one is black and the other is white. To make her happy, Taum has to buyÂ bÂ black gifts andÂ wÂ white gifts.

• The cost of each black gift isÂ bcÂ units.
• The cost of every white gift isÂ wcÂ units.
• The cost to convert a black gift into white gift or vice versa isÂ zÂ units.

Determine the minimum cost of Diksha’s gifts.

Example
b = 3
w = 5
bc = 3
wc = 4
z = 1

He can buy a black gift forÂ 3Â and convert it to a white gift forÂ 1, making the total cost of each white giftÂ 4. That matches the cost of a white gift, so he can do that or just buy black gifts and white gifts. Either way, the overall cost isÂ 3 * 3 + 5 * 4 = 29.

Function Description

Complete the function taumBday in the editor below. It should return the minimal cost of obtaining the desired gifts.

taumBday has the following parameter(s):

• int b: the number of black gifts
• int w: the number of white gifts
• int bc: the cost of a black gift
• int wc: the cost of a white gift
• int z: the cost to convert one color gift to the other color

Returns

• int:Â the minimum cost to purchase the gifts

## Input Format

The first line will contain an integerÂ t, the number of test cases.

The nextÂ tÂ pairs of lines are as follows:
– The first line contains the values of integersÂ bÂ andÂ w.
– The next line contains the values of integersÂ bc,Â wc, andÂ z.

## Constraints

• 1 <= t <= 10
• 0 <= b, w, bc, wc, z <= 109

## Output Format

Â t lines, each containing an integer: the minimum amount of units Taum needs to spend on gifts.

Sample Input

``````STDIN   Function
-----   --------
5       t = 5
10 10   b = 10, w = 10
1 1 1   bc = 1, wc = 1, z = 1
5 9     b = 5, w = 5
2 3 4   bc = 2, wc = 3, z = 4
3 6     b = 3, w = 6
9 1 1   bc = 9, wc = 1, z = 1
7 7     b = 7, w = 7
4 2 1   bc = 4, wc = 2, z = 1
3 3     b = 3, w = 3
1 9 2   bc = 1, wc = 9, z = 2``````

Sample Output

``````20
37
12
35
12``````

Explanation

• Test Case #01:
Since black gifts cost the same as white, there is no benefit to converting the gifts. Taum will have to buy each gift forÂ 1Â unit. The cost of buying all gifts will be:Â b * bc + w * wc = 10 * 1 + 10 * 1 = 20.
• Test Case #02:
Again, he cannot decrease the cost of black or white gifts by converting colors.Â Â is too high. He will buy gifts at their original prices, so the cost of buying all gifts will be:Â b * bc + w * wc = 5 * 2 + 9 * 3 = 10 + 27 = 37.
• Test Case #03:
SinceÂ bc > wc + z, he will buyÂ b + w = 3 + 6 = 9Â white gifts at their original price ofÂ 1.Â b = 3Â of the gifts must be black, and the cost per conversion,Â z = 1. Total cost isÂ 9 * 1 + 3 * 1 = 12.
• Test Case #04:
Similarly, he will buyÂ w = 7Â white gifts at their original price,Â wc = 2. For black gifts, he will first buy white ones and color them to black, so that their cost will be reduced toÂ wc + z = 2 + 1 = 3. So cost of buying all gifts will be:Â 7 * 3 + 7 * 2 = 35.
• Test Case #05:Â He will buy black gifts at their original price,Â bc = 1. For white gifts, he will first black gifts worthÂ bc = 1Â unit and color them to white forÂ z = 2Â units. The cost for white gifts is reduced toÂ wc = bc + z = 2 + 1 = 3Â units. The cost of buying all gifts will be:Â 3 * 1 + 3 * 3 = 3 + 9 = 12.

## Solution – Taum and B’day – HackerRank Solution

### C++

```#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

int main() {
int cases;
scanf("%d", &cases);
while (cases--) {
int B, W, X, Y, Z;
scanf("%d %d %d %d %d", &B, &W, &X, &Y, &Z);
long long res = (long long)B * X + (long long)W * Y;
res = min(res, (long long)B * X + (long long)W * (X + Z));
res = min(res, (long long)B * (Y + Z) + (long long)W * Y);
printf("%lld\n", res);
}
return 0;
}
```

### Python

```#!/bin/python3

import os
import sys

#
# Complete the taumBday function below.
#
def taumBday(b, w, bc, wc, z):
res = 0

if bc <= wc:
res += bc*b
if bc + z <= wc:
res += (bc + z)*w
else:
res += wc*w
else:
res += wc*w
if wc + z <= bc:
res += (wc + z)*b
else:
res += bc*b

return res

if __name__ == '__main__':
fptr = open(os.environ['OUTPUT_PATH'], 'w')

t = int(input())

for t_itr in range(t):
bw = input().split()

b = int(bw[0])

w = int(bw[1])

bcWcz = input().split()

bc = int(bcWcz[0])

wc = int(bcWcz[1])

z = int(bcWcz[2])

result = taumBday(b, w, bc, wc, z)

fptr.write(str(result) + '\n')

fptr.close()
```

### Java

```import java.io.*;
import java.util.*;
public class Solution
{
public static void main(String[] args)
{
int m;
long b,w,x,y,z;
long v1,v2,v3,v4,s;
Scanner in = new Scanner(System.in);
m = in.nextInt();
while(m!=0)
{
b = in.nextLong();
w = in.nextLong();
x = in.nextLong();
y = in.nextLong();
z = in.nextLong();
v1=0L;v2=0L;v3=0L;v4=0L;s=0L;
v1 = (b*x)+(w*y);
v2 = (b*z)+(b*y) + (w*z)+(w*x);
if(v1<=v2)
s=v1;
else
s=v2;
v3 = (b*x) + (w*z)+(w*x);
if(v3<=s)
s=v3;
v4 = (b*z)+(b*y) + (w*y);;
if(v4<=s)
s=v4;
System.out.println(s);
m--;
}
}
}
```

Note: This problem (Taum and B’day) is generated by HackerRank but the solution is provided by CodingBroz. This tutorial is only for Educational and Learning purpose.