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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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

✓ Direct Conversion and Comparison

⏱️ Time: O(n) Space: O(1)

For each word, iterate through characters, convert each letter to its position value (a=0, b=1, etc.), concatenate these values as strings, then convert to integer. Finally check if the sum equals target.

Mathematical Conversion

⏱️ Time: O(n) Space: O(1)

Instead of concatenating strings, build the numerical value mathematically by multiplying previous result by appropriate power of 10 and adding the current digit value. This avoids string operations entirely.

Direct Conversion and Comparison — Algorithm Steps

Step 1: Create helper function to convert word to numerical value
Step 2: Convert firstWord, secondWord, and targetWord to their numerical values
Step 3: Check if firstValue + secondValue equals targetValue

Visualization

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Step-by-Step Walkthrough

Convert Letters

Convert each letter to its position value

Concatenate

Join all position values as string

Compare Sum

Check if first + second equals target

Code -

solution.c — C

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

int getNumericValue(char* word) {
    char result[1000] = "";
    char temp[10];
    
    for (int i = 0; i < strlen(word); i++) {
        sprintf(temp, "%d", word[i] - 'a');
        strcat(result, temp);
    }
    
    return atoi(result);
}

int solution(char* firstWord, char* secondWord, char* targetWord) {
    int firstValue = getNumericValue(firstWord);
    int secondValue = getNumericValue(secondWord);
    int 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)

Linear scan through all characters in the three words

✓ Linear Growth

Space Complexity

O(1)

Only using variables to store converted values

✓ Linear Space

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Medium Frequency

~15 min Avg. Time

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Ln 1, Col 1

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Check if Word Equals Summation of Two Words - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Direct Conversion and Comparison — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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