LeetCode in Kotlin
3849. Maximum Bitwise XOR After Rearrangement
Medium
You are given two binary strings s and t, each of length n.
You may rearrange the characters of t in any order, but s must remain unchanged.
Return a binary string of length n representing the maximum integer value obtainable by taking the bitwise XOR of s and rearranged t.
Example 1:
Input: s = “101”, t = “011”
Output: “110”
Explanation:
- One optimal rearrangement of
tis"011". - The bitwise XOR of
sand rearrangedtis"101" XOR "011" = "110", which is the maximum possible.
Example 2:
Input: s = “0110”, t = “1110”
Output: “1101”
Explanation:
- One optimal rearrangement of
tis"1011". - The bitwise XOR of
sand rearrangedtis"0110" XOR "1011" = "1101", which is the maximum possible.
Example 3:
Input: s = “0101”, t = “1001”
Output: “1111”
Explanation:
- One optimal rearrangement of
tis"1010". - The bitwise XOR of
sand rearrangedtis"0101" XOR "1010" = "1111", which is the maximum possible.
Constraints:
1 <= n == s.length == t.length <= 2 * 105s[i]andt[i]are either'0'or'1'.
Solution
class Solution {
fun maximumXor(s: String, t: String): String {
val cnt = IntArray(2)
for (c in t.toCharArray()) {
cnt[c.code - '0'.code]++
}
val ans = CharArray(s.length)
for (i in 0..<s.length) {
val x = s[i].code - '0'.code
if (cnt[x xor 1] > 0) {
cnt[x xor 1]--
ans[i] = '1'
} else {
cnt[x]--
ans[i] = '0'
}
}
return String(ans)
}
}