Apply Discount to Prices - Practice Coding Problems

Apply Discount to Prices - Problem

String Medium

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: All prices will contain at most 10 digits.

Input & Output

Example 1 — Basic Case

$ Input: sentence = "there are $1 $2 and 5$ candies", discount = 10

› Output: "there are $0.90 $1.80 and 5$ candies"

💡 Note: $1 becomes $0.90 (90% of $1), $2 becomes $1.80 (90% of $2). 5$ is invalid ($ at end) so unchanged.

Example 2 — No Valid Prices

$ Input: sentence = "1 2 3 4 5 6 7 8 9 10", discount = 50

› Output: "1 2 3 4 5 6 7 8 9 10"

💡 Note: No words start with $, so sentence remains unchanged.

Example 3 — Edge Cases

$ Input: sentence = "$ $1e5 $100", discount = 20

› Output: "$ $1e5 $80.00"

💡 Note: $ (only $) and $1e5 (contains 'e') are invalid. Only $100 → $80.00 (80% of $100).

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
0 ≤ discount ≤ 100

Visualization

Tap to expand

Understanding the Visualization

Input

Sentence with mixed words and potential prices

Process

Identify valid prices and apply discount formula

Output

Transformed sentence with discounted prices

Key Takeaway

🎯 Key Insight: A valid price must start with $ and contain only digits after - apply discount formula and format to 2 decimal places

Asked in

G Google 15 a Amazon 12 M Microsoft 8

The key insight is to identify valid prices ($ followed by digits only) and apply the discount formula: new_price = original_price × (100 - discount) / 100. Best approach uses string splitting and validation. Time: O(n), Space: O(n)

Common Approaches

Approach	Time	Space	Notes
✓ String Processing with Word Iteration	O(n)	O(n)	Split sentence into words, check each word if it's a valid price, apply discount

String Processing with Word Iteration — Algorithm Steps

Split sentence into words
For each word, validate if it's a price format
If valid price, parse number and apply discount formula
Format result to 2 decimal places and reconstruct sentence

Visualization

Tap to expand

Step-by-Step Walkthrough

Split Words

Break sentence into individual words

Validate Price

Check if word starts with $ and has only digits after

Apply Discount

Calculate discounted price and format to 2 decimals

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>

int isValidPrice(char* s) {
    int len = strlen(s);
    if (len == 0) return 0;
    for (int i = 0; i < len; i++) {
        if (!isdigit(s[i])) return 0;
    }
    return 1;
}

char* solution(char* sentence, int discount) {
    char* result = malloc(10000);
    result[0] = '\0';
    
    char* copy = malloc(strlen(sentence) + 1);
    strcpy(copy, sentence);
    
    char* word = strtok(copy, " ");
    int first = 1;
    
    while (word != NULL) {
        if (!first) {
            strcat(result, " ");
        }
        first = 0;
        
        if (strlen(word) > 1 && word[0] == '$' && isValidPrice(word + 1)) {
            // Valid price format
            long long price = atoll(word + 1);
            double discountedPrice = price * (100 - discount) / 100.0;
            char formattedPrice[50];
            sprintf(formattedPrice, "$%.2f", discountedPrice);
            strcat(result, formattedPrice);
        } else {
            strcat(result, word);
        }
        
        word = strtok(NULL, " ");
    }
    
    free(copy);
    return result;
}

int main() {
    char sentence[10000];
    fgets(sentence, sizeof(sentence), stdin);
    sentence[strcspn(sentence, "\n")] = 0;
    
    int discount;
    scanf("%d", &discount);
    
    char* result = solution(sentence, discount);
    printf("%s\n", result);
    free(result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through all characters in the sentence to split and process words

✓ Linear Growth

Space Complexity

O(n)

Store array of words and build result string

⚡ Linearithmic Space

32.3K Views

Medium Frequency

~15 min Avg. Time

850 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

String Processing with Word Iteration — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler