# 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 buy b black gifts and w white gifts.

• The cost of each black gift is bc units.
• The cost of every white gift is wc units.
• The cost to convert a black gift into white gift or vice versa is z 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 for 3 and convert it to a white gift for 1, making the total cost of each white gift 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 is 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: the minimum cost to purchase the gifts

## Input Format

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

The next t pairs of lines are as follows:
– The first line contains the values of integers b and w.
– The next line contains the values of integers bcwc, and z.

## Constraints

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

## Output Format

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 for 1 unit. The cost of buying all gifts will be: 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.  is too high. He will buy gifts at their original prices, so the cost of buying all gifts will be: b * bc + w * wc = 5 * 2 + 9 * 3 = 10 + 27 = 37.
• Test Case #03:
Since bc > wc + z, he will buy b + w = 3 + 6 = 9 white gifts at their original price of 1b = 3 of the gifts must be black, and the cost per conversion, z = 1. Total cost is 9 * 1 + 3 * 1 = 12.
• Test Case #04:
Similarly, he will buy w = 7 white gifts at their original price, wc = 2. For black gifts, he will first buy white ones and color them to black, so that their cost will be reduced to wc + z = 2 + 1 = 3. So cost of buying all gifts will be: 7 * 3 + 7 * 2 = 35.
• Test Case #05: He will buy black gifts at their original price, bc = 1. For white gifts, he will first black gifts worth bc = 1 unit and color them to white for z = 2 units. The cost for white gifts is reduced to wc = bc + z = 2 + 1 = 3 units. The cost of buying all gifts will be: 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)

w = int(bw)

bcWcz = input().split()

bc = int(bcWcz)

wc = int(bcWcz)

z = int(bcWcz)

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.