Class Solution
Easy
You are given an array of equal-length strings words
. Assume that the length of each string is n
.
Each string words[i]
can be converted into a difference integer array difference[i]
of length n - 1
where difference[i][j] = words[i][j+1] - words[i][j]
where 0 <= 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'
is 0
, 'b'
is 1
, and 'z'
is 25
.
- 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.
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Constructor Summary
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Method Summary
Methods inherited from class java.lang.Object
clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait
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Constructor Details
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Solution
public Solution()
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Method Details
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oddString
public java.lang.String oddString(java.lang.String[] w)
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