Problem
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 get21
.
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.**
Example 1:
Input: firstWord = "acb", secondWord = "cba", targetWord = "cdb"
Output: true
Explanation:
The numerical value of firstWord is "acb" -> "021" -> 21.
The numerical value of secondWord is "cba" -> "210" -> 210.
The numerical value of targetWord is "cdb" -> "231" -> 231.
We return true because 21 + 210 == 231.
Example 2:
Input: firstWord = "aaa", secondWord = "a", targetWord = "aab"
Output: false
Explanation:
The numerical value of firstWord is "aaa" -> "000" -> 0.
The numerical value of secondWord is "a" -> "0" -> 0.
The numerical value of targetWord is "aab" -> "001" -> 1.
We return false because 0 + 0 != 1.
Example 3:
Input: firstWord = "aaa", secondWord = "a", targetWord = "aaaa"
Output: true
Explanation:
The numerical value of firstWord is "aaa" -> "000" -> 0.
The numerical value of secondWord is "a" -> "0" -> 0.
The numerical value of targetWord is "aaaa" -> "0000" -> 0.
We return true because 0 + 0 == 0.
Constraints:
1 <= firstWord.length,
secondWord.length,
targetWord.length <= 8
firstWord
,secondWord
, andtargetWord
consist of lowercase English letters from'a'
to'j'
inclusive.
Solution (Java)
class Solution {
public boolean isSumEqual(String firstWord, String secondWord, String targetWord) {
StringBuilder sb = new StringBuilder();
int a = getSum(firstWord, sb);
sb.setLength(0);
int b = getSum(secondWord, sb);
sb.setLength(0);
int c = getSum(targetWord, sb);
return a + b == c;
}
private int getSum(String firstWord, StringBuilder sb) {
for (char c : firstWord.toCharArray()) {
sb.append(c - 'a');
}
return Integer.parseInt(sb.toString());
}
}
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).