Number Parser - Problem

Parse strings to numerical values (integers and floats) with comprehensive error handling. Your solution should support various input formats including:

Standard numbers: "123", "-45", "3.14"
Formatted numbers: "1,000", "1,234.56"
Currency format: "$5.99", "$1,234"
Scientific notation: "1e3", "2.5e-2"

Return the parsed number as a float. If the string cannot be parsed, return null.

Important: Handle edge cases like empty strings, invalid characters, and malformed numbers gracefully.

Input & Output

Example 1 — Currency Format

$ Input: s = "$1,234.56"

› Output: 1234.56

💡 Note: Remove the dollar sign and comma, then parse '1234.56' to get 1234.56

Example 2 — Scientific Notation

$ Input: s = "1e3"

› Output: 1000.0

💡 Note: Scientific notation: 1 × 10³ = 1000.0

Example 3 — Invalid Input

$ Input: s = "abc123"

› Output: null

💡 Note: Contains invalid characters 'abc' before the number, cannot be parsed

Constraints

1 ≤ s.length ≤ 100
s contains only digits, letters, and symbols: $ , . + - e E

Visualization

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Asked in

G Google 25 a Amazon 20 M Microsoft 18 A Apple 12

The key insight is to clean formatting characters first, then use built-in parsing functions. The cleanup and parse approach removes currency symbols and commas, then leverages tested parsing logic. Time: O(n), Space: O(n)

Common Approaches

✓ Character-by-Character Parsing

⏱️ Time: O(n) Space: O(1)

Iterate through each character, handling digits, decimal points, signs, and special formats manually. Build the final number by processing each valid character.

Cleanup and Built-in Parsing

⏱️ Time: O(n) Space: O(n)

Remove formatting characters (currency symbols, commas) and use language built-in parsing functions with proper error handling.

Character-by-Character Parsing — Algorithm Steps

Iterate through each character in the string
Handle special cases (currency symbols, commas, scientific notation)
Build the numeric result character by character
Return null if any invalid character is encountered

Visualization

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Step-by-Step Walkthrough

Initialize

Set up variables for result, sign, decimal tracking

Process

Examine each character: digit, decimal, comma, currency

Build

Accumulate the final numeric value

Code -

solution.c — C

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

double* solution(char* s) {
    if (s == NULL || strlen(s) == 0) {
        return NULL;
    }
    
    // Trim whitespace
    while (*s == ' ' || *s == '\t') s++;
    int len = strlen(s);
    while (len > 0 && (s[len-1] == ' ' || s[len-1] == '\t' || s[len-1] == '\n')) {
        s[len-1] = '\0';
        len--;
    }
    
    if (len == 0) return NULL;
    
    double result = 0.0;
    int sign = 1;
    int decimal_places = 0;
    int has_decimal = 0;
    int i = 0;
    
    // Handle currency symbol
    if (s[0] == '$') {
        i = 1;
    }
    
    // Handle sign
    if (i < len && (s[i] == '+' || s[i] == '-')) {
        sign = s[i] == '-' ? -1 : 1;
        i++;
    }
    
    // Process digits
    while (i < len) {
        char c = s[i];
        if (isdigit(c)) {
            int digit = c - '0';
            if (has_decimal) {
                decimal_places++;
                result += digit / pow(10, decimal_places);
            } else {
                result = result * 10 + digit;
            }
        } else if (c == '.' && !has_decimal) {
            has_decimal = 1;
        } else if (c == ',') {
            // Skip comma
        } else if ((c == 'e' || c == 'E') && i + 1 < len) {
            char* exp_part = s + i + 1;
            int exp = atoi(exp_part);
            if (exp != 0 || exp_part[0] == '0') {
                result *= pow(10, exp);
                break;
            } else {
                return NULL;
            }
        } else {
            return NULL;
        }
        i++;
    }
    
    double* res = (double*)malloc(sizeof(double));
    *res = result * sign;
    return res;
}

int main() {
    char s[1000];
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;
    
    double* result = solution(s);
    if (result == NULL) {
        printf("null\n");
    } else {
        printf("%g\n", *result);
        free(result);
    }
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through all characters in the string

✓ Linear Growth

Space Complexity

O(1)

Only using variables for current state and result

✓ Linear Space

24.5K Views

Medium Frequency

~15 min Avg. Time

890 Likes

Ln 1, Col 1

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Number Parser - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Character-by-Character Parsing — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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