O(nlogn) time and O(n) space using sorting and set.

Author: 149ps

2231. Largest Number After Digit Swaps by Parity/2231. Largest Number After Digit Swaps by Parity.py

@@ -0,0 +1,49 @@
+"""
+You are given a positive integer num. You may swap any two digits of num that have the same parity (i.e. both odd digits or both even digits).
+
+Return the largest possible value of num after any number of swaps.
+
+
+
+Example 1:
+
+Input: num = 1234
+Output: 3412
+Explanation: Swap the digit 3 with the digit 1, this results in the number 3214.
+Swap the digit 2 with the digit 4, this results in the number 3412.
+Note that there may be other sequences of swaps but it can be shown that 3412 is the largest possible number.
+Also note that we may not swap the digit 4 with the digit 1 since they are of different parities.
+Example 2:
+
+Input: num = 65875
+Output: 87655
+Explanation: Swap the digit 8 with the digit 6, this results in the number 85675.
+Swap the first digit 5 with the digit 7, this results in the number 87655.
+Note that there may be other sequences of swaps but it can be shown that 87655 is the largest possible number.
+
+
+Constraints:
+
+1 <= num <= 109
+"""
+class Solution:
+    def largestInteger(self, num: int) -> int:
+        odd,even,odd_nums,even_nums = set(),set(),[],[]
+        for i,digit in enumerate(str(num)):
+            if int(digit) % 2 == 0:
+                even.add(i)
+                even_nums.append(int(digit))
+            else:
+                odd.add(i)
+                odd_nums.append(int(digit))
+        odd_nums.sort(reverse=True)
+        even_nums.sort(reverse=True)
+        result,o,e = [],0,0
+        for i in range(len(str(num))):
+            if i in odd:
+                result.append(str(odd_nums[o]))
+                o += 1
+            elif i in even:
+                result.append(str(even_nums[e]))
+                e += 1
+        return int(''.join(result))