HackerRank Taum and B’day problem solution

In this HackerRank Taum and B’day problem, you need to determine the minimum cost of Disksha’s gifts.

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.

Hackerrank Taum and B’day problem solution in Python programming.

tcs = int(input())
for i in range(tcs):
    b, w = map(int, input().split(' '))
    x, y, z = map(int, input().split(' '))
    if x + z < y:
        print(b*x+w*(x+z))
    elif y + z < x:
        print(w*y+b*(y+z))
    else:
        print(b*x+w*y)

Taum and B’day problem solution in Java Programming.

import java.io.*;
import java.util.*;

public class Solution 
{
    private static long cost( long B,long W,long X,long Y,long Z )
    {
        long cost = 0;
        
        if( X < Y )
        {
            cost = cost + ( B * X );
            if( Y < Z + X )
            {
                cost = cost + ( W * Y );
            }
            else
            {
                cost = cost + ( W * ( Z + X ) );
            }
        }
        else
        {
            cost = cost + ( W * Y );
            if( X < Z + Y )
            {
                cost = cost + ( B * X );
            }
            else
            {
                cost = cost + ( B * ( Z + Y ) );
            }   
        }
        return cost;
    }

    public static void main(String[] args) 
    {
        Scanner scan = new Scanner( System.in );
        int testCases = scan.nextInt();
        
        for( int i = 0;i < testCases;i++ )
        {
            System.out.println( cost( scan.nextLong(),scan.nextLong(),scan.nextLong(),scan.nextLong(),scan.nextLong() ) );
        }
    }
}

Problem solution in C++ programming.

#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("%lldn", res);
  }
  return 0;
}

Problem solution in C programming.

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {

    unsigned long long int B, W, X, Y, Z, black, white, BtoW, WtoB, sum, T;

scanf("%llu", &T);
 
while(T--)
    {
    scanf("%llu", &B);
    scanf("%llu", &W);
    scanf("%llu", &X);
    scanf("%llu", &Y);
    scanf("%llu", &Z);
    
    black = B * X;
    white = W * Y;
    BtoW = W * (Z + X);
    WtoB = B * (Z + Y);
    
    //no transfer
    sum = 0;
    sum = black + white;
    
    if ( black + BtoW < sum)
        sum = black + BtoW;
    if ( white + WtoB < sum)
        sum = white + WtoB;
    
    printf("%llun", sum);
    }
}

Problem solution in JavaScript programming.

function processData(input) {
    var testCases = input.split('n');
    testCases.shift();
    while (testCases.length) {
        var presents = testCases.shift().split(' ').map(function(n) { return new BigNum(n); });
        var prices = testCases.shift().split(' ').map(function(n) { return parseInt(n, 10); });
        var blackPresentCost = new BigNum(Math.min(prices[0], prices[1] + prices[2]) + "").multiplyBy(presents[0]);
        var whitePresentCost = new BigNum(Math.min(prices[1], prices[0] + prices[2]) + "").multiplyBy(presents[1]);
        console.log(blackPresentCost.add(whitePresentCost).toString());
    }
}

var BigNum = function(num) {
    if (typeof num === 'string') {
        this.parts = [];
        while (num.length) {
            this.parts.push(parseInt(num.slice(-5),10));
            num = num.substring(0,num.length-5)
        }    
    } else {
        this.parts = num;
    }
    
}

BigNum.prototype.multiplyBy = function(bigNum) {
    var newParts = [];
    for (var i = 0, l1 = this.parts.length; i < l1; ++i) {
        for (var j = 0, l2 = bigNum.parts.length; j < l2; ++j) {
            var newPartIndex = i + j;
            newParts[newPartIndex] = this.parts[i] * bigNum.parts[j] + (newParts[newPartIndex] || 0);
        }
    }
    this._fixOverflows(newParts);
    return new BigNum(newParts);
};

BigNum.prototype.add = function(bigNum) {
    var newParts = [];
    var length = Math.max(this.parts.length, bigNum.parts.length);
    for (var i = 0; i < length; ++i) {
        newParts[i] = (this.parts[i] || 0) + (bigNum.parts[i] || 0);
    }
    this._fixOverflows(newParts);
    return new BigNum(newParts);
}

BigNum.prototype._fixOverflows = function(parts) {
    for (var k = 0; k < parts.length; ++k) {
        var currentPart = parts[k].toString();
        if (currentPart.length > 5) {
            var overflowLength = currentPart.length - 5;
            var overflow = parseInt(currentPart.substr(0, overflowLength), 10);
            parts[k] = parseInt(currentPart.substr(overflowLength), 10)
            parts[k + 1] = overflow + (parts[k + 1] || 0);
        }
    }
}

BigNum.prototype.toString = function() {
    var fullNumber = this.parts.map(function(num) {
        num = num.toString();
        paddingAmount = 5 - num.length;
        for (var i = 0; i < paddingAmount; ++i) {
            num = '0' + num;
        }
        return num
    }).reverse().join('');
    while( fullNumber.charAt(0) === '0' && fullNumber.length !== 1) {
        fullNumber = fullNumber.substring(1);
    }
    return fullNumber;
};

process.stdin.resume();
process.stdin.setEncoding("ascii");
_input = "";
process.stdin.on("data", function (input) {
    _input += input;
});

process.stdin.on("end", function () {
   processData(_input);
});