grid47 Exploring patterns and algorithms
Sep 15, 2024
6 min read
Solution to LeetCode 522: Longest Uncommon Subsequence II Problem
Given an array of strings strs, determine the length of the longest uncommon subsequence between them. An uncommon subsequence is a string that is a subsequence of one string but not the others. If no such subsequence exists, return -1.
Problem
Approach
Steps
Complexity
Input: The input consists of an array of strings strs, where each string contains only lowercase English letters.
Example: strs = ['hello', 'world', 'wonder']
Constraints:
• 2 <= strs.length <= 50
• 1 <= strs[i].length <= 10
• Each string consists of lowercase English letters.
Output: Return the length of the longest uncommon subsequence. If no uncommon subsequence exists, return -1.
Example: 6, -1
Constraints:
• The output should be an integer indicating the length of the longest uncommon subsequence.
Goal: The goal is to identify the longest subsequence that is unique to one string in the array, and not a subsequence of any other string.
Steps:
• 1. Iterate through each string in the array and check if it is a subsequence of any other string in the array.
• 2. If a string is not a subsequence of any other string, update the result with its length.
• 3. Return the length of the longest uncommon subsequence or -1 if none exist.
Goal: The input constraints define the limits on the size and content of the array and strings.
Steps:
• 2 <= strs.length <= 50
• 1 <= strs[i].length <= 10
• Each string is composed of lowercase English letters.
Assumptions:
• The input will always consist of valid strings containing only lowercase English letters.
• Input: strs = ['hello', 'world', 'wonder']
• Explanation: In this case, 'wonder' is the longest uncommon subsequence as it is not a subsequence of any other string.
• Input: strs = ['aaa', 'aaa', 'aa']
• Explanation: There is no uncommon subsequence as all strings share subsequences of each other.
Approach: The approach involves comparing each string with others to find the longest string that does not appear as a subsequence of any other string.
Observations:
• We need to determine the longest uncommon subsequence by comparing each string with others.
• A brute force approach can be used to check subsequences, but optimizations can be considered for larger inputs.
Steps:
• 1. Loop through each string in the array.
• 2. For each string, compare it against all other strings to check if it is a subsequence of any of them.
• 3. If it is not a subsequence of any other string, update the result with its length.
• 4. Return the length of the longest uncommon subsequence, or -1 if none are found.
Empty Inputs:
• An empty string array should return -1.
Large Inputs:
• The algorithm should be optimized to handle input arrays with 50 strings efficiently.
Special Values:
• Identical strings in the input will never produce an uncommon subsequence.
Constraints:
• The time complexity should be efficient for input sizes of up to 50 strings, each up to 10 characters long.
intfindLUSlength(vector<string>& strs) {
if(strs.empty()) return-1;
int rst =-1;
for(auto i =0; i < strs.size(); i++) {
int j =0;
for(; j < strs.size(); j++) {
if( i == j ) continue;
if(LCS(strs[i], strs[j])) break;
}
if (j == strs.size())
rst = max(rst, static_cast<int>(strs[i].size()));
}
return rst;
}
boolLCS(const string& a, const string& b) {
if(b.size() < a.size()) returnfalse;
int i =0;
for(autoch : b)
if (i < a.size() && a[i] == ch) i++;
return i == a.size();
}
1 : Function Definition
intfindLUSlength(vector<string>& strs) {
Defines the function `findLUSlength`, which takes a vector of strings and returns the length of the longest uncommon subsequence.
2 : Edge Case Handling
if(strs.empty()) return-1;
Handles the edge case where the input list of strings is empty, returning -1 as there is no subsequence.
3 : Variable Initialization
int rst =-1;
Initializes the result variable `rst` to -1, which will store the length of the longest uncommon subsequence.
4 : Outer Loop
for(auto i =0; i < strs.size(); i++) {
Iterates through each string in the `strs` vector to check against all other strings.
5 : Variable Initialization
int j =0;
Initializes a variable `j` that will be used in the inner loop to compare the current string with all other strings.
6 : Inner Loop
for(; j < strs.size(); j++) {
Starts an inner loop that compares the current string `strs[i]` with each other string `strs[j]`.
7 : Skipping Same String
if( i == j ) continue;
Skips comparison when the current string is being compared with itself.
8 : LCS Check
if(LCS(strs[i], strs[j])) break;
Checks if `strs[i]` is a subsequence of `strs[j]` using the helper function `LCS`. If it is, breaks out of the loop.
9 : Condition to Update Result
if (j == strs.size())
Checks if the inner loop completed without finding a common subsequence, meaning the string `strs[i]` is unique.
10 : Update Result
rst = max(rst, static_cast<int>(strs[i].size()));
Updates the result variable `rst` to store the maximum length of the longest uncommon subsequence.
11 : Return Statement
return rst;
Returns the length of the longest uncommon subsequence found.
12 : Helper Function Definition
boolLCS(const string& a, const string& b) {
Defines the helper function `LCS`, which checks if string `a` is a subsequence of string `b`.
13 : Subsequence Check
if(b.size() < a.size()) returnfalse;
Checks if string `b` is shorter than string `a`. If so, it cannot be a subsequence of `a`, so return false.
14 : Loop Through String b
int i =0;
Initializes a variable `i` to track the position in string `a`.
15 : Character Matching
for(autoch : b)
Iterates through each character `ch` in string `b`.
16 : Update Index i
if (i < a.size() && a[i] == ch) i++;
If the current character `ch` in `b` matches the character at index `i` in `a`, increment `i`.
17 : Return Statement
return i == a.size();
If the entire string `a` is matched in `b`, return true, indicating `a` is a subsequence of `b`.
Best Case: O(n^2 * m) where n is the number of strings and m is the average string length.
Average Case: O(n^2 * m) where n is the number of strings and m is the average string length.
Worst Case: O(n^2 * m) where n is the number of strings and m is the average string length.
Description: In the worst case, we compare each string with every other string, resulting in a quadratic time complexity.
Best Case: O(1), as no additional space is required for input.
Worst Case: O(n * m), where n is the number of strings and m is the length of each string.
Description: The space complexity is linear relative to the number of strings and their lengths.
