Runtime: 69 ms (Top 5.88%) | Memory: 14 MB (Top 23.53%)

Author: AnasImloul

scripts/algorithms/E/Equal Rational Numbers/Equal Rational Numbers.py

@@ -1,3 +1,4 @@
+// Runtime: 69 ms (Top 5.88%) | Memory: 14 MB (Top 23.53%)
 class Solution:
     # inspired from:
     # https://coolconversion.com/math/recurring-decimals-as-a-fraction/
@@ -9,16 +10,16 @@ def isRationalEqual(self, s: str, t: str) -> bool:
         # intuition:
         # write each numbes as fraction: num / den
         # then compare the two fractions.
-        
+
         num1, den1 = self.toFraction(s)
         num2, den2 = self.toFraction(t)
-        
+
         return den1 * num2 == den2 * num1
-    
+
     def toFraction(self, s: str) -> typing.Tuple[int, int]:
         if "." not in s:
             return int(s), 1
-        
+
         intp, frac = s.split(".")
         # decimal dot, but no repeating part:
         # xyz.abc = xyzabc / 1000
@@ -27,22 +28,22 @@ def toFraction(self, s: str) -> typing.Tuple[int, int]:
             num = int(intp) * (10 ** len(frac)) + ifrac
             den = 10 ** len(frac)
             return num, den
-        
+
         # this is for cases like a.b(c)
-        # let n = a.b(c) 
+        # let n = a.b(c)
         # then, 10^(len(b+c)) * n = abc.(c)
-        # and  10^(len(b)) * n = ab.(c)
+        # and 10^(len(b)) * n = ab.(c)
         # subtract the two, and solve for n:
         # n = (abc - ab) / (10^len(b + c) - 10^len(b))
         frac, repfrac = frac.split("(")
         repfrac = repfrac[:-1]
-        
+
         iintp = int(intp)
         ifrac = int(frac) if len(frac) > 0 else 0
         irep = int(repfrac)
-        
+
         return (
             (iintp * (10 ** (len(frac + repfrac))) + ifrac * 10 ** len(repfrac) + irep) - (iintp * 10 ** len(frac) + ifrac),
             (10** len(frac+repfrac) - 10 **len(frac))
         )
-	```
+    ```