##### Welcome to Subscribe On Youtube

Formatted question description: https://leetcode.ca/all/2288.html

# 2288. Apply Discount to Prices

• Difficulty: Medium.
• Related Topics: String.
• Similar Questions: Multiply Strings, Apply Discount Every n Orders.

## Problem

A sentence is a string of single-space separated words where each word can contain digits, lowercase letters, and the dollar sign ’

## Solution (Java, C++, Python)

• 


Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n). . 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 89 $10$", 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 ’

## Solution (Java, C++, Python)

class Solution {
public String discountPrices(String sentence, int discount) {
String[] words = sentence.split(" ");
StringBuilder sb = new StringBuilder();
for (String word : words) {
sb.append(applyDiscount(word, discount));
sb.append(" ");
}
sb.deleteCharAt(sb.length() - 1);
return sb.toString();
}

private String applyDiscount(String s, int discount) {
if (s.charAt(0) == '$' && s.length() > 1) { long price = 0; for (int i = 1; i < s.length(); i++) { if (!Character.isDigit(s.charAt(i))) { // Error case. We could also use Long.parseLong() here. return s; } price *= 10; price += (s.charAt(i) - '0') * (100 - discount); } String stringPrice = String.valueOf(price); if (price < 10) { return "$0.0" + stringPrice;
}
if (price < 100) {
return "$0." + stringPrice; } return "$"
+ stringPrice.substring(0, stringPrice.length() - 2)
+ "."
+ stringPrice.substring(stringPrice.length() - 2);
}
return s;
}
}


Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n). .

• 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 (Java, C++, Python)



• Todo

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

############

# 2288. Apply Discount to Prices
# https://leetcode.com/problems/apply-discount-to-prices/

class Solution:
def discountPrices(self, sentence: str, discount: int) -> str:

def getDiscount(x):
# print(x)
if discount == 100: return "0.00"
res = float(x) * float(1 - discount / 100)

return "{:.2f}".format(res)

res = []

for x in sentence.split(" "):
if x != "$": res.append(x) else: count = 0 for c in x: if c == "$":
count += 1
elif c in string.ascii_lowercase:
count += 1

if len(x) > 1 and x == "\$" and count == 1:
res.append(x + str(getDiscount(x[1:])))
else:
res.append(x)

return " ".join(res)



Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).