Base Converter - Problem

Write functions to convert a number from any base (2-16) to any other base.

Your function should take three parameters:

number: The number to convert (as a string)
fromBase: The base of the input number (integer from 2 to 16)
toBase: The target base for conversion (integer from 2 to 16)

For bases greater than 10, use uppercase letters A-F to represent digits 10-15.

Note: Handle invalid inputs by returning an empty string.

Input & Output

Example 1 — Binary to Hexadecimal

$ Input: number = "1010", fromBase = 2, toBase = 16

› Output: "A"

💡 Note: Binary 1010 = 1×8 + 0×4 + 1×2 + 0×1 = 10 in decimal, then 10 in decimal = A in hexadecimal

Example 2 — Decimal to Binary

$ Input: number = "10", fromBase = 10, toBase = 2

› Output: "1010"

💡 Note: Decimal 10 divided by 2 repeatedly: 10÷2=5 R0, 5÷2=2 R1, 2÷2=1 R0, 1÷2=0 R1, reading remainders backwards gives 1010

Example 3 — Same Base

$ Input: number = "123", fromBase = 10, toBase = 10

› Output: "123"

💡 Note: Same base conversion returns the original number unchanged

Constraints

2 ≤ fromBase, toBase ≤ 16
1 ≤ number.length ≤ 30
number contains only valid digits for fromBase
Digits 0-9 for values 0-9, letters A-F for values 10-15

Visualization

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Asked in

G Google 12 a Amazon 8 M Microsoft 6

The key insight is to use decimal (base-10) as an intermediate conversion step, eliminating the need for complex direct base-to-base conversion tables. Best approach uses two-step conversion: source→decimal→target. Time: O(n), Space: O(1)

Common Approaches

✓ Direct Base-to-Base Conversion

⏱️ Time: O(n) Space: O(1)

Create separate conversion logic for every possible base pair combination. This requires handling 15×14 = 210 different conversion paths, making the code extremely complex and error-prone.

Two-Step Conversion via Decimal

⏱️ Time: O(n) Space: O(1)

Use decimal (base-10) as an intermediate step. First convert the input number from its source base to decimal, then convert the decimal result to the target base. This eliminates the need for complex direct conversions.

Direct Base-to-Base Conversion — Algorithm Steps

Create mapping tables for all base pairs
Apply direct conversion using the specific table
Handle edge cases for each combination

Visualization

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Step-by-Step Walkthrough

Input Validation

Check if all digits are valid for the source base

Direct Mapping

Apply complex conversion tables for the specific base pair

Result

Output the converted number

Code -

solution.c — C

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

char* solution(char* number, int fromBase, int toBase) {
    static char result[1000];
    result[0] = '\0';
    
    if (!number || strlen(number) == 0 || fromBase < 2 || fromBase > 16 || toBase < 2 || toBase > 16) {
        return result;
    }
    
    // Validate all digits are valid for fromBase
    for (int i = 0; i < strlen(number); i++) {
        char digit = number[i];
        if (isdigit(digit)) {
            if (digit - '0' >= fromBase) {
                return result;
            }
        } else {
            if (toupper(digit) - 'A' + 10 >= fromBase) {
                return result;
            }
        }
    }
    
    if (fromBase == toBase) {
        strcpy(result, number);
        return result;
    }
    
    // Convert to decimal first
    long long decimalVal = 0;
    for (int i = 0; i < strlen(number); i++) {
        char digit = number[i];
        decimalVal = decimalVal * fromBase;
        if (isdigit(digit)) {
            decimalVal += digit - '0';
        } else {
            decimalVal += toupper(digit) - 'A' + 10;
        }
    }
    
    // Convert from decimal to target base
    if (decimalVal == 0) {
        strcpy(result, "0");
        return result;
    }
    
    char temp[1000];
    int pos = 0;
    while (decimalVal > 0) {
        int remainder = decimalVal % toBase;
        if (remainder < 10) {
            temp[pos++] = '0' + remainder;
        } else {
            temp[pos++] = 'A' + remainder - 10;
        }
        decimalVal /= toBase;
    }
    
    // Reverse the result
    for (int i = 0; i < pos; i++) {
        result[i] = temp[pos - 1 - i];
    }
    result[pos] = '\0';
    
    return result;
}

int main() {
    char number[1000];
    int fromBase, toBase;
    
    fgets(number, sizeof(number), stdin);
    number[strcspn(number, "\n")] = 0;
    
    scanf("%d", &fromBase);
    scanf("%d", &toBase);
    
    char* result = solution(number, fromBase, toBase);
    printf("%s\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Still needs to process each digit once, but with complex lookup tables

✓ Linear Growth

Space Complexity

O(1)

No extra data structures besides conversion tables

✓ Linear Space

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Base Converter - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Direct Base-to-Base Conversion — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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