LeetCode in Kotlin
2531. Make Number of Distinct Characters Equal
Medium
You are given two 0-indexed strings word1 and word2.
A move consists of choosing two indices i and j such that 0 <= i < word1.length and 0 <= j < word2.length and swapping word1[i] with word2[j].
Return true if it is possible to get the number of distinct characters in word1 and word2 to be equal with exactly one move. Return false otherwise.
Example 1:
Input: word1 = “ac”, word2 = “b”
Output: false
Explanation: Any pair of swaps would yield two distinct characters in the first string, and one in the second string.
Example 2:
Input: word1 = “abcc”, word2 = “aab”
Output: true
Explanation: We swap index 2 of the first string with index 0 of the second string. The resulting strings are word1 = “abac” and word2 = “cab”, which both have 3 distinct characters.
Example 3:
Input: word1 = “abcde”, word2 = “fghij”
Output: true
Explanation: Both resulting strings will have 5 distinct characters, regardless of which indices we swap.
Constraints:
1 <= word1.length, word2.length <= 105word1andword2consist of only lowercase English letters.
Solution
class Solution {
fun isItPossible(word1: String, word2: String): Boolean {
val count1 = count(word1)
val count2 = count(word2)
val d = count1[26] - count2[26]
val zip1 = zip(count1, count2)
val zip2 = zip(count2, count1)
for (i in 0..25) {
val d1 = zip1[i]
if (d1 == -1) {
continue
}
for (j in 0..25) {
val d2 = zip2[j]
if (d2 == -1) {
continue
}
if (i == j) {
if (d == 0) {
return true
}
continue
}
if (d - d1 + d2 == 0) {
return true
}
}
}
return false
}
private fun zip(c1: IntArray, c2: IntArray): IntArray {
val zip = IntArray(26)
for (i in 0..25) {
var d = 0
if (c1[i] == 0) {
d = -1
} else {
if (c2[i] == 0) {
d++
}
if (c1[i] == 1) {
d++
}
}
zip[i] = d
}
return zip
}
private fun count(word: String): IntArray {
val count = IntArray(27)
val len = word.length
for (i in 0 until len) {
count[word[i].code - 'a'.code]++
}
for (i in 0..25) {
if (count[i] > 0) {
count[26]++
}
}
return count
}
}