LOJ 1249 - Chocolate Thief

Problem statement

In this problem, each test case contains n number of lines. Each line contains: Name(n), length(l), breadth(b), height(h)

Denoting that the person with name (n) ate the cake of the length (l), breadth (b) and height (h). Each information is separated by a whitespace. Each person should eat the same amount (volume) of cake which means l*b*h should be the same. However, a single thief stole a portion from another victim, given that in all cases,

number of thief = number of victim = 1

We need to print out the name of the thief and victim in each testcase.

Example Output: Case 1: mun took chocolate from all

Approach

Basically we need to find out the name associated with higher than average volume and lower than average volume of consumed cake. Here, a map is used. It is a key-value pair. The key is the volume of the consumed cake, and the value is an Object. Which again contains two key-value pairs. The key is the name of the person and the value is the number of persons which consumed the exact amount of cake as the key of the map denotes. We are storing the name because as it is mentioned in the problem, there is only one pair of theif-victim in each test case. So, storing the names where frequency is 1 will give us the name of the thief and the victim.

If you want to learn more about javascript Maps, Here is the official documentation.

const tab = {
    [key]: // Amount of cake consumed
    {
        name, // Name of the person consuming the same amount of cake as the key
        frequency // Amount of people consuming the same amount fo cake 
    }
}

• If there is only one key, it means that everybody got the same volume of that cake
• Otherwise, we need to find out the minimum key and the maximum key, along with the associated name. Then print out the name of the victim and the thief

Javascript

const readline = require('readline')
const rl = readline.createInterface({
    input:process.stdin
})
const userInput = []
rl.on('line',(values)=>{
    userInput.push(values)
})


rl.on('close',()=>{
    const noOfCases = parseInt(userInput[0])
    let inputPtr = 1
    for(let i=0;i<noOfCases;i++){
        let victim = ''
        let thief = ''
        let students = parseInt(userInput[inputPtr])
        let tab = {}
        for(let i = inputPtr+1 ; i< students+inputPtr+1;i++){
            let temp = userInput[i].split(" ")
            const l = parseInt(temp[temp.length-1])
            const b = parseInt(temp[temp.length-2])
            const h = parseInt(temp[temp.length-3])
            let vol = l*b*h
            let name = temp[0]
            if(tab[vol]){
                tab[vol].name = name
                tab[vol].freq++
            }
            else{
                tab[vol] = {
                    name,
                    freq:1
                }
            }
        }
        const keys = Object.keys(tab)
        if(keys.length===1){
            console.log(`Case ${i+1}: no thief`)
        }
        else{
            let smallest = parseInt(keys[0])
            let largest = parseInt(keys[1])
            victim = tab[keys[0]].name
            thief = tab[keys[1]].name
            for(let key in tab){
                if(tab[key].freq===1){
                    if(parseInt(key)<smallest){
                        smallest = parseInt(key)
                        victim = tab[key].name
                    }
                    else if(parseInt(key)>largest){
                        largest = parseInt(key)
                        thief = tab[key].name
                    }
                }
            }
            console.log(`Case ${i+1}: ${thief} took chocolate from ${victim}`)
        }
        inputPtr = inputPtr + students + 1
    }
})

Tutorial source