Check if Word Equals Summation of Two Words

Check if Word Equals Summation of Two Words - Problem

String Easy

Imagine you're working on a secret code system where letters have special numerical meanings! In this fascinating problem, each letter from 'a' to 'j' represents a digit from 0 to 9 respectively.

Here's how the encoding works:

'a' → 0, 'b' → 1, 'c' → 2, ..., 'j' → 9
When you have a word like "acb", you convert each letter to its digit: a=0, c=2, b=1
Concatenate these digits to form "021", which becomes the integer 21

Your Mission: Given three encoded words (firstWord, secondWord, and targetWord), determine if the numerical values of the first two words add up to the numerical value of the target word.

Return true if firstWord + secondWord = targetWord, otherwise return false.

Input & Output

example_1.py — Basic Addition

$ Input: firstWord = "acb", secondWord = "cba", targetWord = "cdb"

› Output: true

💡 Note: Converting words to numbers: "acb" → 021 → 21, "cba" → 210 → 210, "cdb" → 231 → 231. Since 21 + 210 = 231, the answer is true.

example_2.py — Unequal Sum

$ Input: firstWord = "aaa", secondWord = "a", targetWord = "aab"

› Output: false

💡 Note: Converting words: "aaa" → 000 → 0, "a" → 0 → 0, "aab" → 001 → 1. Since 0 + 0 = 0 ≠ 1, the answer is false.

example_3.py — Single Characters

$ Input: firstWord = "a", secondWord = "a", targetWord = "b"

› Output: false

💡 Note: Converting: "a" → 0, "a" → 0, "b" → 1. Since 0 + 0 = 0 ≠ 1, the answer is false.

Constraints

1 ≤ firstWord.length, secondWord.length, targetWord.length ≤ 8
firstWord, secondWord, and targetWord consist of lowercase English letters from 'a' to 'j' inclusive
No string has leading zeros when converted to a number

Visualization

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Understanding the Visualization

Map Letters to Digits

Each letter a-j corresponds to digits 0-9 respectively

Build Numbers

Convert each word by concatenating its digit values

Verify Equation

Check if the first two numbers sum to the third

Key Takeaway

🎯 Key Insight: Convert letters directly to digits using character arithmetic (char - 'a'), eliminating the need for complex lookup tables or mappings.

Asked in

a Amazon 15 ⊞ Microsoft 12 G Google 8 f Meta 6

The optimal approach is to convert each word to its numerical value using character arithmetic (char - 'a') and then perform a simple addition check. This solution runs in O(n + m + k) time where n, m, k are the word lengths, with O(1) space complexity.

Common Approaches

Approach	Time	Space	Notes
✓ Direct String Conversion	O(n + m + k)	O(1)	Convert each word to its numerical value by processing each character

Direct String Conversion — Algorithm Steps

Create a helper function to convert a word to its numerical value
For each character in the word, convert it to its digit value (char - 'a')
Build the number by multiplying current result by 10 and adding the new digit
Convert all three words to their numerical values
Check if firstValue + secondValue equals targetValue

Visualization

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Step-by-Step Walkthrough

Convert First Word

Transform each character to its digit value and build the number

Convert Second Word

Repeat the process for the second word

Convert Target Word

Convert the target word to its numerical value

Check Sum

Verify if first + second equals target

Code -

solution.c — C

#include <stdio.h>
#include <string.h>
#include <stdbool.h>

int wordToNumber(const char* word) {
    int result = 0;
    int len = strlen(word);
    for (int i = 0; i < len; i++) {
        result = result * 10 + (word[i] - 'a');
    }
    return result;
}

bool isSumEqual(const char* firstWord, const char* secondWord, const char* targetWord) {
    return wordToNumber(firstWord) + wordToNumber(secondWord) == wordToNumber(targetWord);
}

int main() {
    char firstWord[10], secondWord[10], targetWord[10];
    scanf("%s %s %s", firstWord, secondWord, targetWord);
    printf("%s\n", isSumEqual(firstWord, secondWord, targetWord) ? "true" : "false");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n + m + k)

Where n, m, k are the lengths of firstWord, secondWord, and targetWord respectively

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables to store the numerical values

✓ Linear Space

28.4K Views

Medium Frequency

~8 min Avg. Time

892 Likes

Ln 1, Col 1

Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Direct String Conversion — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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