Runtime: 4515 ms (Top 5.02%) | Memory: 38.7 MB (Top 5.74%)

Author: AnasImloul

scripts/algorithms/V/Verbal Arithmetic Puzzle/Verbal Arithmetic Puzzle.py

@@ -1,24 +1,25 @@
+# Runtime: 4515 ms (Top 5.02%) | Memory: 38.7 MB (Top 5.74%)
 from itertools import permutations
 class Solution:
     def isSolvable(self, words: List[str], result: str) -> bool:
         # we need begins set to track start letters if word is longer than one to not have them as 0
         begins = {result[0]} if len(result) > 1 else set()
-        
+
         # the idea to have counter is to sum repeted letters based on where they are in the word
-        
+
         c = Counter()
         for w in words:
             if len(w) > 1:
                 begins.add(w[0])
             for i,l in enumerate(w[::-1]):
                 c[l] += 10**i
-        
+
         # we subtract "result" letters from the counter
         for i,l in enumerate(result[::-1]):
             c[l] -= 10**i
-            
+
         # we get at the end something like that Counter({'S': 1000, 'E': 91, 'R': 10, 'D': 1, 'Y': -1, 'N': -90, 'O': -900, 'M': -9000})
-        
+
         # lets split problem into 2 let's check positive letters and negative
         pos = []
         neg = []
@@ -30,28 +31,28 @@ def isSolvable(self, words: List[str], result: str) -> bool:
 
         if not neg and not pos:
             return True
-        
+
         if not neg or not pos:
             # False if begins else True this one is needed for words with one letter like words = ["A","B"], result = "A"
             return False if begins else True
-        
+
         NN = len(neg)
         NP = len(pos)
-        
+
         # we need this one to process permutaion with the smaller number first
         # that would optimize speed
         if NN > NP:
             neg,pos = pos,neg
             NN,NP = NP, NN
-        
+
         # we create dict with list of sets of permutations
         # keys sum of digit for letter * coefficient
         nd = defaultdict(list)
         for used in permutations(list(range(10)), NN):
             if 0 in used and neg[used.index(0)][0] in begins:
                 continue
             nd[sum(v * used[i] for i,(l,v) in enumerate(neg))].append(set(used))
-        
+
         # same for the other part
         # but while we are doing that we check if overlap of digits is missing
         for used in permutations(list(range(10)), NP):
@@ -64,4 +65,4 @@ def isSolvable(self, words: List[str], result: str) -> bool:
                     if not s1 & s2:
                         return True
 
-        return False
+        return False