Leetcode 1657: Determine if Two Strings Are Close
Two strings are considered transformable into each other if you can convert one into the other by performing a series of operations. These operations include swapping any two characters or transforming all occurrences of one character into another.
Problem
Approach
Steps
Complexity
Input: You are given two strings, `word1` and `word2`. Each string consists of lowercase English letters.
Example: word1 = "abc", word2 = "bca"
Constraints:
• 1 <= word1.length, word2.length <= 10^5
• word1 and word2 contain only lowercase English letters
Output: Return true if `word1` and `word2` can be transformed into each other through the specified operations, otherwise return false.
Example: true
Constraints:
• The result will be either true or false.
Goal: Determine if `word1` and `word2` can be transformed into each other through a sequence of swaps and character transformations.
Steps:
• Check if the lengths of both strings are the same. If not, return false.
• Check if both strings contain the same set of unique characters. If they do not, return false.
• Compare the frequencies of characters in both strings. If the frequency distribution matches, return true, otherwise return false.
Goal: The constraints are designed to ensure that both strings are manageable in size and contain valid characters.
Steps:
• 1 <= word1.length, word2.length <= 10^5
• Both strings consist of lowercase English letters.
Assumptions:
• Both strings are valid lowercase English strings.
• The strings may have different lengths, but you must check their transformation potential.
• Input: word1 = "abc", word2 = "bca"
• Explanation: In this case, it is possible to transform `word1` into `word2` by swapping characters. After applying two swap operations, we get the string `bca`.
• Input: word1 = "a", word2 = "aa"
• Explanation: In this case, it's impossible to convert `word1` into `word2` since `word2` has more characters than `word1`, making it impossible to apply the given operations.
Approach: We will compare the character distributions and sets of the two strings to determine if they can be transformed into each other.
Observations:
• If two strings contain different characters, they can never be transformed into each other.
• The operations only allow swaps and transformations, which means the structure of the strings' characters must be identical.
• Start by checking if the lengths of the strings are the same. If not, return false immediately.
• Then, check the frequency of characters in both strings. If the frequencies match, they can be transformed into each other.
Steps:
• Compare the lengths of `word1` and `word2`. If they differ, return false.
• Create a map of character frequencies for both `word1` and `word2`.
• Compare the sets of characters in both strings. If they don't match, return false.
• Sort the frequency values of characters in both strings and compare them. If they match, return true; otherwise, return false.
Empty Inputs:
• If one string is empty and the other is not, return false.
Large Inputs:
• For large input strings, ensure that the solution efficiently compares characters without redundant operations.
Special Values:
• If both strings consist of the same repeated character, they should be transformable into each other.
Constraints:
• Be mindful of time complexity when processing large inputs.
bool closeStrings(string w1, string w2) {
if(w1.size() != w2.size()) return false;
int n = w1.size();
map<char, int> m1, m2;
int mask1 = 0, mask2 = 0;
for(int i = 0; i < n; i++) {
m1[w1[i]]++;
m2[w2[i]]++;
mask1 |= (1 << (w1[i] - 'a'));
mask2 |= (1 << (w2[i] - 'a'));
}
if(mask1 != mask2) return false;
vector<int> arr, nums;
for(auto it: m1) {
arr.push_back(it.second);
}
sort(arr.begin(), arr.end());
// for(int x: arr)
// cout << x << " ";
// cout << "\n";
for(auto it: m2) {
nums.push_back(it.second);
}
sort(nums.begin(), nums.end());
// for(int x: nums)
// cout << x << " ";
cout << "\n";
return nums == arr;
}
1 : Function Definition
bool closeStrings(string w1, string w2) {
Defines the function `closeStrings` that takes two strings as input and returns a boolean.
2 : Size Check
if(w1.size() != w2.size()) return false;
Checks if the two strings have the same length. If not, they cannot be close.
3 : Variable Initialization
int n = w1.size();
Initializes the variable `n` to store the size of the strings for later use.
4 : Map Initialization
map<char, int> m1, m2;
Initializes two maps to count the frequency of characters in each string.
5 : Variable Initialization
int mask1 = 0, mask2 = 0;
Initializes two integer masks to track the unique characters in the strings.
6 : Looping
for(int i = 0; i < n; i++) {
Loops through each character of the strings to update the maps and masks.
7 : Map Update
m1[w1[i]]++;
Increments the frequency of the current character in the first map.
8 : Map Update
m2[w2[i]]++;
Increments the frequency of the current character in the second map.
9 : Bit Manipulation
mask1 |= (1 << (w1[i] - 'a'));
Updates the bitmask for the first string to track its unique characters.
10 : Bit Manipulation
mask2 |= (1 << (w2[i] - 'a'));
Updates the bitmask for the second string to track its unique characters.
11 : Condition Check
if(mask1 != mask2) return false;
Checks if the sets of unique characters in the two strings are different. If so, they cannot be close.
12 : Vector Initialization
vector<int> arr, nums;
Initializes two vectors to store the frequency counts from the maps.
13 : Map Iteration
for(auto it: m1) {
Iterates through the first map to extract character frequencies.
14 : Vector Update
arr.push_back(it.second);
Appends the frequency of a character from the first map to the first vector.
15 : Sorting
sort(arr.begin(), arr.end());
Sorts the frequencies in the first vector to prepare for comparison.
16 : Map Iteration
for(auto it: m2) {
Iterates through the second map to extract character frequencies.
17 : Vector Update
nums.push_back(it.second);
Appends the frequency of a character from the second map to the second vector.
18 : Sorting
sort(nums.begin(), nums.end());
Sorts the frequencies in the second vector to prepare for comparison.
19 : Return
return nums == arr;
Compares the sorted frequency vectors. If they are equal, the strings are close.
Best Case: O(n)
Average Case: O(n log n)
Worst Case: O(n log n)
Description: The best-case time complexity occurs when the strings are already identical or trivially transformable. The worst case arises from sorting the frequency distributions of characters.
Best Case: O(n)
Worst Case: O(n)
Description: The space complexity is O(n) due to the storage of character frequencies and other intermediate data structures.
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