Class Solution
- java.lang.Object
-
- g2401_2500.s2451_odd_string_difference.Solution
-
public class Solution extends Object
2451 - Odd String Difference.Easy
You are given an array of equal-length strings
words
. Assume that the length of each string isn
.Each string
words[i]
can be converted into a difference integer arraydifference[i]
of lengthn - 1
wheredifference[i][j] = words[i][j+1] - words[i][j]
where0 <= j <= n - 2
. Note that the difference between two letters is the difference between their positions in the alphabet i.e. the position of'a'
is0
,'b'
is1
, and'z'
is25
.- For example, for the string
"acb"
, the difference integer array is[2 - 0, 1 - 2] = [2, -1]
.
All the strings in words have the same difference integer array, except one. You should find that string.
Return the string in
words
that has different difference integer array.Example 1:
Input: words = [“adc”,“wzy”,“abc”]
Output: “abc”
Explanation:
-
The difference integer array of “adc” is [3 - 0, 2 - 3] = [3, -1].
-
The difference integer array of “wzy” is [25 - 22, 24 - 25]= [3, -1].
-
The difference integer array of “abc” is [1 - 0, 2 - 1] = [1, 1].
The odd array out is [1, 1], so we return the corresponding string, “abc”.
Example 2:
Input: words = [“aaa”,“bob”,“ccc”,“ddd”]
Output: “bob”
Explanation: All the integer arrays are [0, 0] except for “bob”, which corresponds to [13, -13].
Constraints:
3 <= words.length <= 100
n == words[i].length
2 <= n <= 20
words[i]
consists of lowercase English letters.
- For example, for the string
-
-
Constructor Summary
Constructors Constructor Description Solution()
-