LeetCode-in-Java
2506. Count Pairs Of Similar Strings
Easy
You are given a 0-indexed string array words.
Two strings are similar if they consist of the same characters.
- For example,
"abca"and"cba"are similar since both consist of characters'a','b', and'c'. - However,
"abacba"and"bcfd"are not similar since they do not consist of the same characters.
Return the number of pairs (i, j) such that 0 <= i < j <= word.length - 1 and the two strings words[i] and words[j] are similar.
Example 1:
Input: words = [“aba”,”aabb”,”abcd”,”bac”,”aabc”]
Output: 2
Explanation: There are 2 pairs that satisfy the conditions:
-
i = 0 and j = 1 : both words[0] and words[1] only consist of characters ‘a’ and ‘b’.
-
i = 3 and j = 4 : both words[3] and words[4] only consist of characters ‘a’, ‘b’, and ‘c’.
Example 2:
Input: words = [“aabb”,”ab”,”ba”]
Output: 3
Explanation: There are 3 pairs that satisfy the conditions:
-
i = 0 and j = 1 : both words[0] and words[1] only consist of characters ‘a’ and ‘b’.
-
i = 0 and j = 2 : both words[0] and words[2] only consist of characters ‘a’ and ‘b’.
-
i = 1 and j = 2 : both words[1] and words[2] only consist of characters ‘a’ and ‘b’.
Example 3:
Input: words = [“nba”,”cba”,”dba”]
Output: 0
Explanation: Since there does not exist any pair that satisfies the conditions, we return 0.
Constraints:
1 <= words.length <= 1001 <= words[i].length <= 100words[i]consist of only lowercase English letters.
Solution
import java.util.HashMap;
import java.util.Map;
public class Solution {
public int similarPairs(String[] words) {
int len = words.length;
if (len == 1) {
return 0;
}
byte[][] alPh = new byte[len][26];
Map<String, Integer> map = new HashMap<>();
for (int i = 0; i < len; i++) {
String word = words[i];
for (char c : word.toCharArray()) {
int idx = c - 'a';
if (alPh[i][idx] == 0) {
alPh[i][idx]++;
}
}
String s = new String(alPh[i]);
if (map.containsKey(s)) {
map.put(s, map.get(s) + 1);
} else {
map.put(s, 1);
}
}
int pairs = 0;
for (Map.Entry<String, Integer> entry : map.entrySet()) {
int freq = entry.getValue();
if (freq > 1) {
pairs += (freq * (freq - 1)) / 2;
}
}
return pairs;
}
}