# 1363 Largest Multiple of Three

Given an array of digits `digits`, return the largest multiple of three that can be formed by concatenating some of the given digits in any order. If there is no answer return an empty string.

Since the answer may not fit in an integer data type, return the answer as a string. Note that the returning answer must not contain unnecessary leading zeros.

Example 1:

``````Input: digits = [8,1,9]
Output: "981"
``````

Example 2:

``````Input: digits = [8,6,7,1,0]
Output: "8760"
``````

Example 3:

``````Input: digits = 
Output: ""
``````

Constraints:

• `1 <= digits.length <= 104`
• `0 <= digits[i] <= 9`
 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 `````` ``````class Solution: def largestMultipleOfThree(self, digits: List[int]) -> str: total = sum(digits) count = collections.Counter(digits) digits.sort(reverse=True) def f(i): if count[i]: digits.remove(i) count[i] -= 1 if not digits: return '' if not any(digits): return '0' if sum(digits) % 3 == 0: return ''.join(map(str, digits)) ''' If total % 3 == 0, done If total % 3 == 1 and there is one of 1,4,7 in digits: try to remove one of 1,4,7 If total % 3 == 2 and we have one of 2,5,8 in digits: try to remove one of 2,5,8 If total % 3 == 2: try to remove two of 1,4,7 If total % 3 == 1: try to remove two of 2,5,8 ''' if total % 3 == 0: return f(-1) if total % 3 == 1 and count + count + count: return f(1) or f(4) or f(7) if total % 3 == 2 and count + count + count: return f(2) or f(5) or f(8) if total % 3 == 2: return f(1) or f(1) or f(4) or f(4) or f(7) or f(7) return f(2) or f(2) or f(5) or f(5) or f(8) or f(8) ''' DP ''' class Solution: def largestMultipleOfThree(self, digits: List[int]) -> str: # dp[i] is the largest number which dp[i] % 3 = i dp = [-1, -1, -1] for a in sorted(digits, reverse=True): for x in dp[:] + : y = x * 10 + a dp[y % 3] = max(dp[y % 3], y) return str(dp) if dp >= 0 else ""``````