Add No_1363: Largest Multiple of Three

Author: brianchiang-tw

No_1363_Largest Multiple of Three/No_1363_Largest Multiple of Three_description.md

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+'''
+
+Description:
+
+Given an integer array of digits, return the largest multiple of three that can be formed by concatenating some of the given digits in any order.
+
+Since the answer may not fit in an integer data type, return the answer as a string.
+
+If there is no answer return an empty string.
+
+ 
+
+Example 1:
+
+Input: digits = [8,1,9]
+Output: "981"
+Example 2:
+
+Input: digits = [8,6,7,1,0]
+Output: "8760"
+Example 3:
+
+Input: digits = [1]
+Output: ""
+Example 4:
+
+Input: digits = [0,0,0,0,0,0]
+Output: "0"
+ 
+
+Constraints:
+
+1 <= digits.length <= 10^4
+0 <= digits[i] <= 9
+The returning answer must not contain unnecessary leading zeros.
+
+'''

No_1363_Largest Multiple of Three/largest_multiple_of_three.py

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+'''
+
+Description:
+
+Given an integer array of digits, return the largest multiple of three that can be formed by concatenating some of the given digits in any order.
+
+Since the answer may not fit in an integer data type, return the answer as a string.
+
+If there is no answer return an empty string.
+
+ 
+
+Example 1:
+
+Input: digits = [8,1,9]
+Output: "981"
+Example 2:
+
+Input: digits = [8,6,7,1,0]
+Output: "8760"
+Example 3:
+
+Input: digits = [1]
+Output: ""
+Example 4:
+
+Input: digits = [0,0,0,0,0,0]
+Output: "0"
+ 
+
+Constraints:
+
+1 <= digits.length <= 10^4
+0 <= digits[i] <= 9
+The returning answer must not contain unnecessary leading zeros.
+
+'''
+
+
+
+from collections import defaultdict
+from functools import reduce
+from typing import List
+
+class Solution:
+    def largestMultipleOfThree(self, digits: List[int]) -> str:
+        
+        digits.sort(reverse = True)
+        
+        remainder_dict = defaultdict( list)
+        summation = 0
+        
+        for number in digits:
+            
+            summation += number
+            remainder_dict[number % 3].append( number )
+
+            
+        if summation == 0:
+            # Quick response for all-zero digits case
+            return "0"
+            
+        ## Redundant remainder removal for modulo 3 system            
+        redundant = summation % 3     
+        if redundant != 0:
+            
+            if remainder_dict[redundant]:
+                remainder_dict[redundant].pop()
+            else:
+                
+                if len(remainder_dict[3-redundant]) >= 2:
+                    remainder_dict[3-redundant].pop()
+                    remainder_dict[3-redundant].pop()
+                else:
+                    return ""
+
+        candidates = remainder_dict[0] + remainder_dict[1] + remainder_dict[2]
+        
+        if not candidates:
+            # Quck response if no digits remain after removal of redundant remainder
+            return ""
+        
+        # Keep digits in descending order
+        candidates.sort( reverse = True )
+        
+        # Make the number
+        func = lambda x, y: 10*x + y
+        value = reduce( func, candidates)
+        
+        # Convert to string form
+        return str(value)
+
+
+
+# n :　the length input array, digits.
+
+## Time Complexity: O( n log n)
+#
+# The overhead in time is the cost of sorting, which is of O( n log n ) in Python by timSort.
+
+## Space Complexity: O( n )
+#
+# The overhead in space is the storage for dictionary, remainder_dict, which is of O( n )
+
+
+
+from collections import namedtuple
+TestEntry = namedtuple('TestEntry', 'sequence')
+
+def test_bench():
+
+    test_data = [
+                    TestEntry( sequence = [8,1,9]),
+                    TestEntry( sequence = [8,6,7,1,0]),
+                    TestEntry( sequence = [1]),
+                    TestEntry( sequence = [0,0,0,0,0,0]),
+                    TestEntry( sequence = [9,8,6,8,6]),
+
+                ]
+
+    # expected output:
+    # Note: the third output is empty string '', which means no solution by definition.
+    '''
+    981
+    8760
+
+    0
+    966
+    '''
+
+    for t in test_data:
+
+        print( Solution().largestMultipleOfThree( t.sequence) )
+
+    return
+        
+
+
+if __name__ == '__main__':
+
+    test_bench()
+