Apply Discount to Prices

Problem

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.

Examples

Example 1

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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

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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 <= 10^5
• 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

Solution

Method 1 – String Parsing and Formatting

Intuition

The key idea is to identify words that represent prices (start with $ and followed by digits), apply the discount, and format the result to two decimal places. We process each word, check if it’s a price, and update it accordingly.

Approach

1. Split the sentence into words.
2. For each word:

• Check if it starts with $ and the rest are digits.
• If yes, parse the number, apply the discount, and format to two decimals.
• Otherwise, keep the word unchanged.

3. Join the processed words back into a sentence.

Code

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class Solution {
public:
   string discountPrices(string sentence, int discount) {
      stringstream ss(sentence);
      string word, ans;
      while (ss >> word) {
        if (word.size() > 1 && word[0] == '$' && all_of(word.begin() + 1, word.end(), ::isdigit)) {
           long long val = stoll(word.substr(1));
           double price = val * (1 - discount / 100.0);
           char buf[20];
           sprintf(buf, "$%.2f", price + 1e-8);
           ans += string(buf) + " ";
        } else {
           ans += word + " ";
        }
      }
      if (!ans.empty()) ans.pop_back();
      return ans;
   }
};

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class Solution {
   public String discountPrices(String sentence, int discount) {
      String[] arr = sentence.split(" ");
      StringBuilder ans = new StringBuilder();
      for (String w : arr) {
        if (w.length() > 1 && w.charAt(0) == '$' && w.substring(1).chars().allMatch(Character::isDigit)) {
           long val = Long.parseLong(w.substring(1));
           double price = val * (1 - discount / 100.0);
           ans.append(String.format("$%.2f", price)).append(" ");
        } else {
           ans.append(w).append(" ");
        }
      }
      return ans.toString().trim();
   }
}

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

Complexity

• ⏰ Time complexity: O(N) where N is the length of the sentence (each word is processed once).
• 🧺 Space complexity: O(N) for storing the result.