2288. Apply Discount to Prices

Description

A sentence is a string of single-space separated words where each word can contain digits, lowercase letters, and the dollar sign '$'. A word represents a price if it is a sequence of digits preceded by a dollar sign.

For example, "$100", "$23", and "$6" represent prices while "100", "$", and "$1e5" do not.

You are given a string sentence representing a sentence and an integer discount. For each word representing a price, apply a discount of discount% on the price and update the word in the sentence. All updated prices should be represented with exactly two decimal places.

Return a string representing the modified sentence.

Note that all prices will contain at most 10 digits.

Example 1:

Input: sentence = "there are $1 $2 and 5$ candies in the shop", discount = 50
Output: "there are $0.50 $1.00 and 5$ candies in the shop"
Explanation: 
The words which represent prices are "$1" and "$2". 
- A 50% discount on "$1" yields "$0.50", so "$1" is replaced by "$0.50".
- A 50% discount on "$2" yields "$1". Since we need to have exactly 2 decimal places after a price, we replace "$2" with "$1.00".

Example 2:

Input: sentence = "1 2 $3 4 $5 $6 7 8$ $9 $10$", discount = 100
Output: "1 2 $0.00 4 $0.00 $0.00 7 8$ $0.00 $10$"
Explanation: 
Applying a 100% discount on any price will result in 0.
The words representing prices are "$3", "$5", "$6", and "$9".
Each of them is replaced by "$0.00".

Constraints:

1 <= sentence.length <= 10⁵
sentence consists of lowercase English letters, digits, ' ', and '$'.
sentence does not have leading or trailing spaces.
All words in sentence are separated by a single space.
All prices will be positive numbers without leading zeros.
All prices will have at most 10 digits.
0 <= discount <= 100

Solutions

Solution 1: Simulation

We can split the sentence into an array of words by spaces, then iterate through the array of words. For each word, if it represents a price, we update it to the price after applying the discount. Finally, we concatenate the updated array of words into a space-separated string.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the string sentence.

Python3JavaC++GoTypeScript

class Solution:
    def discountPrices(self, sentence: str, discount: int) -> str:
        ans = []
        for w in sentence.split():
            if w[0] == '$' and w[1:].isdigit():
                w = f'${int(w[1:]) * (1 - discount / 100):.2f}'
            ans.append(w)
        return ' '.join(ans)

class Solution {
    public String discountPrices(String sentence, int discount) {
        String[] words = sentence.split(" ");
        for (int i = 0; i < words.length; ++i) {
            if (check(words[i])) {
                double t = Long.parseLong(words[i].substring(1)) * (1 - discount / 100.0);
                words[i] = String.format("$%.2f", t);
            }
        }
        return String.join(" ", words);
    }

    private boolean check(String s) {
        if (s.charAt(0) != '$' || s.length() == 1) {
            return false;
        }
        for (int i = 1; i < s.length(); ++i) {
            if (!Character.isDigit(s.charAt(i))) {
                return false;
            }
        }
        return true;
    }
}

class Solution {
public:
    string discountPrices(string sentence, int discount) {
        istringstream is(sentence);
        string w;
        string ans;
        auto check = [](string s) {
            if (s[0] != '$' || s.size() == 1) {
                return false;
            }
            for (int i = 1; i < s.size(); ++i) {
                if (!isdigit(s[i])) {
                    return false;
                }
            }
            return true;
        };
        while (is >> w) {
            if (check(w)) {
                long long v = stoll(w.substr(1)) * (100 - discount);
                char t[20];
                sprintf(t, "$%lld.%02lld", v / 100, v % 100);
                ans += t;
            } else {
                ans += w;
            }
            ans += ' ';
        }
        ans.pop_back();
        return ans;
    }
};

func discountPrices(sentence string, discount int) string {
    words := strings.Split(sentence, " ")
    for i, w := range words {
        if w[0] == '$' {
            if v, err := strconv.Atoi(w[1:]); err == nil {
                words[i] = fmt.Sprintf("$%.2f", float64(v*(100-discount))/100)
            }
        }
    }
    return strings.Join(words, " ")
}

function discountPrices(sentence: string, discount: number): string {
    const sell = (100 - discount) / 100;
    const reg = new RegExp(/^(\$)(([1-9]\d*\.?\d*)|(0\.\d*))$/g);
    const words = sentence.split(' ').map(d => {
        if (!reg.test(d)) return d;
        return d.replace(reg, (s, $1, $2) => {
            return `$${(sell * $2).toFixed(2)}`;
        });
    });
    return words.join(' ');
}

2288. Apply Discount to Prices

Description

Solutions

Solution 1: Simulation

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