Leetcode 2451: Odd String Difference
You are given an array of strings where each string is of the same length. Each string can be converted into a difference array where each element represents the difference between two consecutive characters’ positions in the alphabet. All strings in the array have the same difference array except for one. Your task is to identify the string that has a unique difference array.
Problem
Approach
Steps
Complexity
Input: The input consists of an array `words` where each element is a string of lowercase English letters.
Example: words = ["xyz", "abb", "dcz"]
Constraints:
• 3 <= words.length <= 100
• 2 <= words[i].length <= 20
• words[i] consists of lowercase English letters
Output: Return the string that has a different difference array compared to all other strings in the input array.
Example: Output: "abb"
Constraints:
• There will be exactly one string with a unique difference array.
Goal: Find the string with the unique difference array.
Steps:
• 1. Compute the difference array for each string in the input array.
• 2. Store the difference arrays in a hash map where the key is the difference array and the value is a list of corresponding strings.
• 3. Identify the difference array that is unique by checking the size of the list for each key in the hash map.
• 4. Return the string corresponding to the unique difference array.
Goal: The problem requires comparing the difference arrays of strings in a time-efficient manner.
Steps:
• Words have the same length, making their difference arrays comparable.
• All words are valid lowercase English letters.
Assumptions:
• The difference array calculation is based on alphabetic positions starting from 'a' = 0 to 'z' = 25.
• The input array always has at least one string with a unique difference array.
• Input: words = ["xyz", "abb", "dcz"]
• Explanation: The difference array for 'xyz' is [1, 1], for 'abb' is [1, 1], and for 'dcz' is [1, -1]. The unique difference array is [1, -1], so 'dcz' is the string with a unique difference array.
• Input: words = ["abc", "def", "xyz"]
• Explanation: The difference array for 'abc' is [1, 1], for 'def' is [1, 1], and for 'xyz' is [1, 1]. No string has a unique difference array in this case.
Approach: We will calculate the difference array for each string and store it in a hash map. The string with a unique difference array will be found by checking which difference array is not repeated.
Observations:
• The problem is a comparison of difference arrays, which can be efficiently handled with a hash map.
• We can calculate the difference array in O(n) time for each string and use the hash map to store and track the frequency of each difference array.
Steps:
• 1. Create a helper function to calculate the difference array for a given string.
• 2. Initialize a hash map to store difference arrays as keys and lists of corresponding words as values.
• 3. Iterate through each string in the input array, compute its difference array, and store it in the hash map.
• 4. Once all difference arrays are stored, find the key with only one corresponding string, and return that string.
Empty Inputs:
• Input will not be empty as per the problem constraints.
Large Inputs:
• With a large number of words, the solution should remain efficient, as each word's difference array is computed in linear time.
Special Values:
• Special cases such as all words having the same difference array will not occur based on problem constraints.
Constraints:
• The time complexity of the solution should be manageable for the given constraints (up to 100 words of length 20).
class Solution {
string difference(string& s) {
string d;
for (int i = 0; i < s.size() - 1; i++) {
d += s[i + 1] - s[i];
}
return d;
}
public:
string oddString(vector<string>& words) {
int n = words[0].size();
unordered_map<string, vector<string>> mp;
for (auto &it : words) {
mp[difference(it)].push_back(it);
}
for (auto &it : mp) {
if (it.second.size() == 1) return it.second[0];
}
return "";
}
1 : Class Definition
class Solution {
Defines the `Solution` class, which contains the function implementations for solving the problem.
2 : Difference Function
string difference(string& s) {
This function calculates the difference between consecutive characters in the string `s` and returns a string representing the differences.
3 : Variable Initialization
string d;
Initializes an empty string `d` to store the differences between consecutive characters.
4 : Loop Setup
for (int i = 0; i < s.size() - 1; i++) {
Starts a loop to iterate over the string `s`, excluding the last character.
5 : Difference Calculation
d += s[i + 1] - s[i];
Calculates the difference between the ASCII values of consecutive characters `s[i + 1]` and `s[i]`, and appends the result to string `d`.
6 : Return Statement
return d;
Returns the string `d`, which contains the differences between consecutive characters of the input string `s`.
7 : Public Access Modifier
public:
Specifies that the following function `oddString` is publicly accessible from outside the class.
8 : Odd String Function
string oddString(vector<string>& words) {
This function finds the 'odd' string in the list of words, which differs in its pattern of consecutive character differences.
9 : Variable Initialization
int n = words[0].size();
Initializes `n` to the length of the first word in the `words` list.
10 : Map Initialization
unordered_map<string, vector<string>> mp;
Initializes an unordered map `mp` to store groups of words, where the key is the string of differences between consecutive characters.
11 : Map Population
for (auto &it : words) {
Iterates through each word `it` in the list `words`.
12 : Difference and Map Update
mp[difference(it)].push_back(it);
Calls the `difference` function for each word and stores the word in the map `mp` under the key corresponding to its character differences.
13 : Find Odd String
for (auto &it : mp) {
Iterates through each entry in the map `mp`, where each entry contains a list of words with the same character differences.
14 : Return Odd String
if (it.second.size() == 1) return it.second[0];
If a group in the map has only one word, it is the 'odd' string, so the function returns it.
15 : Return Empty String
return "";
If no odd string is found, the function returns an empty string.
Best Case: O(n * m)
Average Case: O(n * m)
Worst Case: O(n * m)
Description: For each word (n words), we calculate the difference array in O(m) time, where m is the length of the word.
Best Case: O(n)
Worst Case: O(n * m)
Description: The space complexity is determined by the number of words and the storage of their corresponding difference arrays.
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