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Formatted question description: https://leetcode.ca/all/2451.html

# 2451. Odd String Difference

• Difficulty: Easy.
• Related Topics: .
• Similar Questions: Minimum Rounds to Complete All Tasks.

## Problem

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.

## Solution (Java, C++, Python)

• class Solution {
public String oddString(String[] words) {
Map<String, List<String>> cnt = new HashMap<>();
for (var w : words) {
List<String> d = new ArrayList<>();
for (int i = 0; i < w.length() - 1; ++i) {
}
cnt.computeIfAbsent(String.join(",", d), k -> new ArrayList<>()).add(w);
}
for (var v : cnt.values()) {
if (v.size() == 1) {
return v.get(0);
}
}
return "";
}
}

• class Solution {
public:
string oddString(vector<string>& words) {
unordered_map<string, vector<string>> cnt;
for (auto& w : words) {
string d;
for (int i = 0; i < w.size() - 1; ++i) {
d += (char) (w[i + 1] - w[i]);
d += ',';
}
cnt[d].emplace_back(w);
}
for (auto& [_, v] : cnt) {
if (v.size() == 1) {
return v[0];
}
}
return "";
}
};

• class Solution:
def oddString(self, words: List[str]) -> str:
cnt = defaultdict(list)
for w in words:
d = [str(ord(b) - ord(a)) for a, b in pairwise(w)]
cnt[','.join(d)].append(w)
return next(v[0] for v in cnt.values() if len(v) == 1)


• func oddString(words []string) string {
cnt := map[string][]string{}
for _, w := range words {
d := make([]byte, len(w)-1)
for i := 0; i < len(w)-1; i++ {
d[i] = w[i+1] - w[i]
}
t := string(d)
cnt[t] = append(cnt[t], w)
}
for _, v := range cnt {
if len(v) == 1 {
return v[0]
}
}
return ""
}

• function oddString(words: string[]): string {
const n = words[0].length;
const map = new Map<string, [boolean, number]>();
words.forEach((word, i) => {
const diff: number[] = [];
for (let j = 1; j < n; j++) {
diff.push(word[j].charCodeAt(0) - word[j - 1].charCodeAt(0));
}
const k = diff.join();
map.set(k, [!map.has(k), i]);
});
for (const [isOnly, i] of map.values()) {
if (isOnly) {
return words[i];
}
}
return '';
}


• use std::collections::HashMap;
impl Solution {
pub fn odd_string(words: Vec<String>) -> String {
let n = words[0].len();
let mut map: HashMap<String, (bool, usize)> = HashMap::new();
for (i, word) in words.iter().enumerate() {
let mut k = String::new();
for j in 1..n {
k.push_str(&(word.as_bytes()[j] - word.as_bytes()[j - 1]).to_string());
k.push(',');
}
let new_is_only = !map.contains_key(&k);
map.insert(k, (new_is_only, i));
}
for (is_only, i) in map.values() {
if *is_only {
return words[*i].clone();
}
}
String::new()
}
}



Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).