Check if Word Equals Summation of Two Words

Check if Word Equals Summation of Two Words - Problem

String Easy

The letter value of a letter is its position in the alphabet starting from 0 (i.e. 'a' → 0, 'b' → 1, 'c' → 2, etc.).

The numerical value of some string of lowercase English letters s is the concatenation of the letter values of each letter in s, which is then converted into an integer.

For example, if s = "acb", we concatenate each letter's letter value, resulting in "021". After converting it, we get 21.

You are given three strings firstWord, secondWord, and targetWord, each consisting of lowercase English letters 'a' through 'j' inclusive.

Return true if the summation of the numerical values of firstWord and secondWord equals the numerical value of targetWord, or false otherwise.

Input & Output

Example 1 — Basic Case

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

› Output: true

💡 Note: acb → "021" → 21, cba → "210" → 210, cdb → "231" → 231. Since 21 + 210 = 231, return true.

Example 2 — False Case

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

› Output: false

💡 Note: aaa → "000" → 0, a → "0" → 0, aab → "001" → 1. Since 0 + 0 = 0 ≠ 1, return false.

Example 3 — Single Characters

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

› Output: false

💡 Note: a → "0" → 0, b → "1" → 1, c → "2" → 2. Since 0 + 1 = 1 ≠ 2, return false.

Constraints

1 ≤ firstWord.length, secondWord.length, targetWord.length ≤ 8
firstWord, secondWord, and targetWord consist of lowercase English letters from 'a' to 'j'

Visualization

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Understanding the Visualization

Input

Three words: firstWord, secondWord, targetWord

Convert

Convert each word to numerical value using letter positions

Check Sum

Return true if first + second equals target

Key Takeaway

🎯 Key Insight: Convert letters to digit positions, concatenate to form numbers, then verify if sum equation holds

Asked in

G Google 15 M Microsoft 12

The key insight is to convert each word to its numerical value by treating letter positions as digits, then check if the sum equals the target. The most efficient approach converts letters to numbers in O(n) time and O(1) space.

Common Approaches

Approach	Time	Space	Notes
✓ Mathematical Conversion	O(n)	O(1)	Build numbers mathematically without string concatenation
Direct Conversion and Comparison	O(n)	O(1)	Convert each word to its numerical value and check if first + second equals target

Mathematical Conversion — Algorithm Steps

Step 1: For each character, calculate its digit value (char - 'a')
Step 2: Build number by: result = result * 10^(digits in current value) + current_value
Step 3: Compare the three computed values

Visualization

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Step-by-Step Walkthrough

Process Each Letter

Convert letter to digit value

Build Number

Multiply by 10/100 and add digit

Compare Results

Check if sum equals target

Code -

solution.c — C

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

long long getNumericValue(char* word) {
    long long result = 0;
    
    for (int i = 0; i < strlen(word); i++) {
        int digit = word[i] - 'a';
        if (digit >= 10) {
            result = result * 100 + digit;
        } else {
            result = result * 10 + digit;
        }
    }
    
    return result;
}

int solution(char* firstWord, char* secondWord, char* targetWord) {
    long long firstValue = getNumericValue(firstWord);
    long long secondValue = getNumericValue(secondWord);
    long long targetValue = getNumericValue(targetWord);
    
    return firstValue + secondValue == targetValue;
}

int main() {
    char firstWord[100], secondWord[100], targetWord[100];
    
    fgets(firstWord, sizeof(firstWord), stdin);
    firstWord[strcspn(firstWord, "\n")] = 0;
    
    fgets(secondWord, sizeof(secondWord), stdin);
    secondWord[strcspn(secondWord, "\n")] = 0;
    
    fgets(targetWord, sizeof(targetWord), stdin);
    targetWord[strcspn(targetWord, "\n")] = 0;
    
    int result = solution(firstWord, secondWord, targetWord);
    printf(result ? "true\n" : "false\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Still need to process each character once

✓ Linear Growth

Space Complexity

O(1)

Only using integer variables, no string building

✓ Linear Space

25.0K Views

Medium Frequency

~15 min Avg. Time

890 Likes

Ln 1, Col 1

Smart Actions

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Input & Output

Constraints

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Related Problems

Common Approaches

Mathematical Conversion — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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