Leetcode 2840: Check if Strings Can be Made Equal With Operations II
You are given two strings s1 and s2, both of length n, consisting of lowercase English letters. You can perform the following operation on any of the two strings: choose any two indices i and j such that i < j and the difference j - i is even, then swap the two characters at those indices in the string. Return true if you can make s1 and s2 equal, and false otherwise.
Problem
Approach
Steps
Complexity
Input: The input consists of two strings, s1 and s2, each containing n lowercase English letters.
Example: s1 = 'aceb', s2 = 'caeb'
Constraints:
• s1 and s2 have the same length n.
• 1 <= n <= 10^5.
• s1 and s2 consist only of lowercase English letters.
Output: Return true if it's possible to make the two strings equal using the described operations, otherwise return false.
Example: true
Constraints:
• The output should be a boolean value indicating whether the strings can be made equal or not.
Goal: The goal is to determine if s1 can be transformed into s2 using the allowed swap operations.
Steps:
• For each character in s1 and s2, group them by their positions modulo 2 (even or odd indexed positions).
• Compare the counts of characters in corresponding groups for s1 and s2.
• If the counts match for both even and odd indexed positions, return true; otherwise, return false.
Goal: The constraints on the input are as follows:
Steps:
• Both s1 and s2 must have the same length n.
• s1 and s2 consist only of lowercase English letters.
• 1 <= n <= 10^5.
Assumptions:
• The problem assumes that swapping characters is the only allowed operation.
• The strings can have repeated characters.
• Input: s1 = 'aceb', s2 = 'caeb'
• Explanation: We can swap characters at indices 0 and 2 and then swap characters at indices 1 and 3 to make s1 equal to s2.
• Input: s1 = 'abe', s2 = 'bea'
• Explanation: It's not possible to transform s1 into s2 since there is no valid swap that can achieve this.
Approach: The approach involves checking if we can group characters of s1 and s2 based on their even and odd indexed positions, and verify that the counts of each character match.
Observations:
• The problem involves swapping characters at specific indices, so checking the counts of characters in different indexed groups is key.
• By comparing character frequencies in the even and odd indexed groups of both strings, we can determine if it's possible to make the strings equal.
Steps:
• Create two frequency arrays for each string: one for characters at even indices and one for characters at odd indices.
• Compare the frequency arrays for both strings. If the arrays match, return true; otherwise, return false.
Empty Inputs:
• Empty strings are not allowed since the minimum length is 1.
Large Inputs:
• The solution should efficiently handle large strings with up to 10^5 characters.
Special Values:
• The strings may consist of identical characters or all distinct characters.
Constraints:
• Ensure the solution works efficiently within the given constraint of 1 <= n <= 10^5.
bool checkStrings(string s1, string s2) {
int map[2][26] = {0}; // Initialize a 2D array with all elements set to 0
for (int i = 0; i < s1.length(); i++) {
map[i % 2][s1[i] - 'a']++;
map[i % 2][s2[i] - 'a']--;
}
for (int i = 0; i < 26; i++) {
if (map[0][i] != 0 || map[1][i] != 0) return false;
}
return true;}
1 : Initialization
bool checkStrings(string s1, string s2) {
Start of the function to check if two strings are identical by swapping characters.
2 : Array Setup
int map[2][26] = {0}; // Initialize a 2D array with all elements set to 0
Creates a 2D array to track character frequencies for both strings (two rows, 26 columns for each alphabet letter).
3 : Loop Through Strings
for (int i = 0; i < s1.length(); i++) {
Iterates through the characters of both strings simultaneously.
4 : Increment First String Frequency
map[i % 2][s1[i] - 'a']++;
Increments the frequency of the character in the first string (mapped to one of the rows in the array).
5 : Decrement Second String Frequency
map[i % 2][s2[i] - 'a']--;
Decrements the frequency of the character in the second string (mapped to the other row in the array).
6 : Final Check
for (int i = 0; i < 26; i++) {
Iterates through all the alphabet letters to check if the frequencies match.
7 : Check Frequency
if (map[0][i] != 0 || map[1][i] != 0) return false;
If any character's frequency is non-zero in either row, return false, meaning the strings cannot be made equal.
8 : End Function
return true;}
Returns true if all character frequencies match, meaning the strings can be made equal by swapping characters.
Best Case: O(n)
Average Case: O(n)
Worst Case: O(n)
Description: The time complexity is linear, as we are processing the string once to count the character frequencies.
Best Case: O(1)
Worst Case: O(n)
Description: The space complexity is linear, as we need to store the character frequencies for even and odd indexed positions.
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