LeetCode in Kotlin
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
class Solution {
fun similarPairs(words: Array<String>): Int {
val len = words.size
if (len == 1) {
return 0
}
val alPh = Array(len) { ByteArray(26) }
val map: MutableMap<String, Int> = HashMap()
for (i in 0 until len) {
val word = words[i]
for (c in word.toCharArray()) {
val idx = c.code - 'a'.code
if (alPh[i][idx].toInt() == 0) {
alPh[i][idx]++
}
}
val s = String(alPh[i])
if (map.containsKey(s)) {
map[s] = map[s]!! + 1
} else {
map[s] = 1
}
}
var pairs = 0
for ((_, freq) in map) {
if (freq > 1) {
pairs += freq * (freq - 1) / 2
}
}
return pairs
}
}