LeetCode 2506: Count Pairs Of Similar Strings
LeetCode 2506 Solution Explanation
Explanation
To solve this problem, we can iterate over all pairs of strings in the words array and check if they are similar. We can do this by counting the frequency of characters in each string and comparing these frequencies. If the frequencies are the same, the strings are similar.
We will use a nested loop to compare each pair of strings. For each pair, we will maintain a set to store the unique characters present in both strings. If the size of the set is less than or equal to 26 (the number of lowercase English letters), we can conclude that the strings are similar.
The time complexity of this approach is O(n^2 * k), where n is the number of strings and k is the maximum length of a string. The space complexity is O(k) to store the set of unique characters.
LeetCode 2506 Solutions in Java, C++, Python
class Solution {
public int countPairs(String[] words) {
int count = 0;
for (int i = 0; i < words.length; i++) {
for (int j = i + 1; j < words.length; j++) {
Set<Character> charSet = new HashSet<>();
for (char c : words[i].toCharArray()) {
charSet.add(c);
}
for (char c : words[j].toCharArray()) {
charSet.add(c);
}
if (charSet.size() <= 26) {
count++;
}
}
}
return count;
}
}Interactive Code Editor for LeetCode 2506
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