Problem Description

You need to determine if two words add up to equal a third word, where each word is converted to a number based on letter positions.

Letter Value System:

• Each letter has a value based on its position in the alphabet, starting from 0
• 'a' → 0, 'b' → 1, 'c' → 2, and so on

Numerical Value Calculation: To get the numerical value of a word:

1. Convert each letter to its letter value
2. Concatenate these values as a string
3. Convert the concatenated string to an integer

Example:

• For the word "acb":
  • 'a' → 0, 'c' → 2, 'b' → 1
  • Concatenate: "021"
  • Convert to integer: 21

Task: Given three strings firstWord, secondWord, and targetWord (all containing only lowercase letters 'a' through 'j'), determine if:

numerical_value(firstWord) + numerical_value(secondWord) = numerical_value(targetWord)

Return true if the equation holds, false otherwise.

Intuition

The problem is essentially asking us to perform arithmetic on words that have been encoded as numbers. The key insight is that we need to convert each word into its numerical representation before we can check if the addition equation holds.

Think of each word as a special number system where instead of digits 0-9, we use letters with their alphabetical positions. When we see a word like "acb", we need to:

1. Map each letter to its position value (a=0, c=2, b=1)
2. Treat these values as digits placed side by side (021)
3. Convert this to a regular integer (21)

Once we have this conversion process figured out, the solution becomes straightforward - convert all three words to their numerical values and check if firstWord + secondWord = targetWord.

The conversion process itself follows a pattern similar to how we normally convert strings to integers. For each character in the string:

• Get its letter value by subtracting 'a' from it (since 'a' maps to 0)
• Build the final number by multiplying the current result by 10 and adding the new digit

For example, converting "acb" step by step:

• Start with ans = 0
• Process 'a': ans = 0 * 10 + 0 = 0
• Process 'c': ans = 0 * 10 + 2 = 2
• Process 'b': ans = 2 * 10 + 1 = 21

This approach naturally handles the concatenation and conversion in one pass through each string.

Solution Approach

The solution implements a helper function f(s) that converts a string to its numerical value. Here's how the implementation works:

Helper Function f(s):

1. Initialize ans = 0 to store the result and a = ord("a") to get the ASCII value of 'a'
2. For each character c in the string:
   • Convert the character to its ASCII value using ord(c)
   • Calculate the letter value: x = c - a (this gives us 0 for 'a', 1 for 'b', etc.)
   • Build the number using the formula: ans = ans * 10 + x
3. Return the final numerical value

Main Logic: The main function simply calls f() on all three input strings and checks if the equation holds:

return f(firstWord) + f(secondWord) == f(targetWord)

Example Walkthrough: Let's trace through firstWord = "acb", secondWord = "cdb", targetWord = "cfd":

For firstWord = "acb":

• 'a': x = 0, ans = 0 * 10 + 0 = 0
• 'c': x = 2, ans = 0 * 10 + 2 = 2
• 'b': x = 1, ans = 2 * 10 + 1 = 21

For secondWord = "cdb":

• 'c': x = 2, ans = 0 * 10 + 2 = 2
• 'd': x = 3, ans = 2 * 10 + 3 = 23
• 'b': x = 1, ans = 23 * 10 + 1 = 231

For targetWord = "cfd":

• 'c': x = 2, ans = 0 * 10 + 2 = 2
• 'f': x = 5, ans = 2 * 10 + 5 = 25
• 'd': x = 3, ans = 25 * 10 + 3 = 253

Check: 21 + 231 = 252 ≠ 253, so return false

The time complexity is O(n) where n is the total length of all three strings, and the space complexity is O(1).

Example Walkthrough

Let's work through a simple example where the equation holds true.

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

Step 1: Convert firstWord = "aaa" to its numerical value

• Start with ans = 0
• Process 'a' (1st): letter value = 0, ans = 0 * 10 + 0 = 0
• Process 'a' (2nd): letter value = 0, ans = 0 * 10 + 0 = 0
• Process 'a' (3rd): letter value = 0, ans = 0 * 10 + 0 = 0
• Result: 0

Step 2: Convert secondWord = "a" to its numerical value

• Start with ans = 0
• Process 'a': letter value = 0, ans = 0 * 10 + 0 = 0
• Result: 0

Step 3: Convert targetWord = "aab" to its numerical value

• Start with ans = 0
• Process 'a' (1st): letter value = 0, ans = 0 * 10 + 0 = 0
• Process 'a' (2nd): letter value = 0, ans = 0 * 10 + 0 = 0
• Process 'b': letter value = 1, ans = 0 * 10 + 1 = 1
• Result: 1

Step 4: Check if the equation holds

• firstWord + secondWord = 0 + 0 = 0
• targetWord = 1
• Since 0 ≠ 1, return false

Let's try another example where the equation is true:

Input: firstWord = "j", secondWord = "j", targetWord = "bi"

Converting each word:

• "j": letter value = 9, result = 9
• "j": letter value = 9, result = 9
• "bi":
  • 'b': letter value = 1, ans = 0 * 10 + 1 = 1
  • 'i': letter value = 8, ans = 1 * 10 + 8 = 18
  • Result: 18

Verification: 9 + 9 = 18, which equals 18, so return true

Solution Implementation

Time and Space Complexity

The time complexity is O(L), where L is the sum of the lengths of all three strings (firstWord, secondWord, and targetWord). This is because the function f iterates through each character of a string exactly once, performing constant-time operations for each character. Since f is called three times (once for each string), and each call processes every character in its respective string, the total time is proportional to the sum of all string lengths.

The space complexity is O(1). The algorithm uses only a fixed amount of extra space regardless of input size. The helper function f maintains only two integer variables (ans and a), and the map(ord, s) operation creates an iterator that doesn't consume additional space proportional to the string length. No auxiliary data structures that scale with input size are created.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Leading Zeros Confusion

A common misconception is thinking that leading zeros might cause issues or need special handling. For example, if a word starts with 'a' (which maps to 0), like "aab" → "001" → 1, developers might worry about the leading zero disappearing.

Why it's not a problem: The algorithm naturally handles this correctly. When we build the number using result = result * 10 + digit_value, starting with 'a' (0) simply keeps the result at 0 until we encounter a non-'a' character. The mathematical operations handle this automatically without any special cases needed.

2. Integer Overflow Concerns

Since we're potentially creating large numbers from long strings, developers might worry about integer overflow, especially in languages with fixed integer sizes.

Solution: In Python, integers have arbitrary precision, so overflow isn't an issue. However, if implementing in languages like Java or C++, consider:

• Using long or long long data types
• Adding overflow checks if the input strings can be very long
• The problem constraint mentions only letters 'a' through 'j', which limits the maximum digit value to 9, making the resulting numbers standard decimal integers

3. Incorrect ASCII Arithmetic

A subtle but critical error occurs when developers try to directly use character values without proper conversion:

Wrong approach:

def word_to_number(word):
    result = ""
    for char in word:
        result += str(ord(char) - ord('a'))
    return int(result)

Why it's inefficient: This creates an intermediate string representation, which is unnecessary and less efficient than building the number directly through arithmetic operations.

4. Misunderstanding the Base System

Some developers might think this is a base-26 number system since there are 26 letters in the alphabet.

Clarification: Despite having 26 possible letter values (0-25), we're still working in base-10. Each letter position contributes its value as a decimal digit. The problem specifically limits input to letters 'a' through 'j' (values 0-9), making it exactly like decimal numbers where each position can only hold digits 0-9.

5. Edge Case: Empty Strings

The code doesn't explicitly handle empty strings, which would return 0 from the helper function.

Consideration: While mathematically correct (empty string → 0), verify whether the problem constraints allow empty strings. If they do, the current implementation handles them correctly. If not, add validation:

if not word:
    return 0  # or raise an exception based on requirements