Solutions for Maximize the number of toys that can be purchased

    • @
    • @Zafir, submitted in Python 3, achived 100% score
    # Complete the maximumToys function below.
    def solve(prices, k):
        prices.sort()
        count = 0
        for i in prices:
            if (i <= k):
                count += 1
                k -= i
            else:
                break
        return count
    
    
    
    • @
    • @Zafir, submitted in Python 3, achived 100% score
    # Complete the maximumToys function below.
    def solve(prices, k):
        prices.sort()
        count = 0
        for i in prices:
            if (i <= k):
                count += 1
                k -= i
            else:
                break
        return count
    
    
    
    • @admin
    • @admin, submitted in Python 2, achived 100% score
    def solve(prices, k):
        prices.sort()
        count = 0
        for i in prices:
            if (i <= k):
                count += 1
                k -= i
            else:
                break
        return count
    • @admin
    • @admin, submitted in Python 2, achived 100% score
    def solve(prices, k):
        prices.sort()
        count = 0
        for i in prices:
            if (i <= k):
                count += 1
                k -= i
            else:
                break
        return count
    • @MatthewIdete
    • @alexander, submitted in Python 2, achived 100% score
    def solve(prices, k):
        prices.sort()
        count = 0
        for i in prices:
            if (i <= k):
                count += 1
                k -= i
            else:
                break
        return count
    • @JavierVow
    • @jacob, submitted in JavaScript, achived 5% score
    function solve(prices,k)
    {
        prices.sort();
        count=0;
        for (let i=0;i<prices;i++)
        {
            if (i<=k)
            {
                count+=1;
                k-=i;
            }
            else
            {
                break;
            }
        }
        return count;
    }
    • @JavierVow
    • @jacob, submitted in Python 3, achived 100% score
    # Complete the maximumToys function below.
    def solve(prices, k):
        prices.sort()
        count = 0
        for i in prices:
            if (i <= k):
                count += 1
                k -= i
            else:
                break
        return count
    
    
    
    • @JavierVow
    • @jacob, submitted in Python 2, achived 100% score
    def solve(prices, k):
        prices.sort()
        count = 0
        for i in prices:
            if (i <= k):
                count += 1
                k -= i
            else:
                break
        return count
    
    • @MatthewIdete
    • @alexander, submitted in Python 2, achived 100% score
    def solve(prices, k):
        prices.sort()
        count = 0
        for i in prices:
            if (i <= k):
                count += 1
                k -= i
            else:
                break
        return count
    • @JavierVow
    • @jacob, submitted in JavaScript, achived 5% score
    function solve(prices,k)
    {
        prices.sort();
        count=0;
        for (let i=0;i<prices;i++)
        {
            if (i<=k)
            {
                count+=1;
                k-=i;
            }
            else
            {
                break;
            }
        }
        return count;
    }