LeetCode-in-Java
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
@SuppressWarnings("java:S135")
public class Solution {
public boolean isItPossible(String word1, String word2) {
int[] count1 = count(word1);
int[] count2 = count(word2);
int d = count1[26] - count2[26];
int[] zip1 = zip(count1, count2);
int[] zip2 = zip(count2, count1);
for (int i = 0; i < 26; i++) {
int d1 = zip1[i];
if (d1 == -1) {
continue;
}
for (int j = 0; j < 26; j++) {
int 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 int[] zip(int[] c1, int[] c2) {
int[] zip = new int[26];
for (int i = 0; i < 26; i++) {
int 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 int[] count(String word) {
int[] count = new int[27];
int len = word.length();
for (int i = 0; i < len; i++) {
count[word.charAt(i) - 'a']++;
}
for (int i = 0; i < 26; i++) {
if (count[i] > 0) {
count[26]++;
}
}
return count;
}
}