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  1. Prepare
  2. Algorithms
  3. Implementation
  4. Electronics Shop
  5. Discussions

Electronics Shop

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2391 Discussions

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  • c_canbul
     + 0 comments

    C#

    static int getMoneySpent(int[] keyboards, int[] drives, int b) {

         int temp =-1;
     foreach(var keyboard in keyboards){
        foreach(var drive in drives){
            if((keyboard + drive) > temp && (keyboard + drive) <=  b)
            temp = keyboard + drive;
        }


     }
     return temp;
}

  • bladoncgarland
     + 0 comments

    JS solution:

    function getMoneySpent(keyboards, drives, b) {
    /*
     * Write your code here.
     */
    keyboards.sort((a, b) => a - b)
    drives.sort((a, b) => b - a)
    
    let maxPrice = -1    
    let kIndex = 0
    let dIndex = 0
    
    while(kIndex < keyboards.length && dIndex < drives.length) {
        let combinedCost = keyboards[kIndex] + drives[dIndex]
        
        if (combinedCost > b) {
            dIndex++
            continue
        }
        
        kIndex++
        
        if (combinedCost > maxPrice) {
            maxPrice = combinedCost
        } 
    }
    
    return maxPrice
}

  • yadav940
     + 0 comments

    Here is my python 3 solution

    maxvalue=-1;
for keyboar in keyboards :
    for drive in drives :
        total = keyboar+drive
        if total< b and total > maxvalue  :
            maxvalue = total

return maxvalue

  • alban_tyrex
     + 0 comments

    Here is my c++ solution, you can find the video here : https://youtu.be/yC-TXToDbD0

    int getMoneySpent(vector<int> keyboards, vector<int> drives, int b) {
    int ans = -1;
    for(int k : keyboards){
        for(int d : drives){
            if(k + d > ans && k + d <= b) ans = k + d;
        }
    }
    return ans;
}

  • job_brquirino
     + 0 comments

    most simple solution i found

    let pricies = []
    for (const k of keyboards) {
        for (const d of drives) pricies.push(d + k)
    }
    const maxPrice = pricies.filter((p) => p <= b)
    return maxPrice.length > 0 ? Math.max(...maxPrice) : -1