Update Solution.py

Author: samarthswami1016

solution/3100-3199/3139.Minimum Cost to Equalize Array/Solution.py

@@ -1,35 +1,21 @@
 class Solution:
     MOD = 10**9 + 7
-
     def solve(self, k):
         sumDifferences = k * len(self.nums) - self.sumNums
-
         ones = max(2 * (k - self.minNums) - sumDifferences, 0)
         if (sumDifferences - ones) & 1 != 0:
             ones += 1
-
         return ones * self.cost1 + ((sumDifferences - ones) // 2) * self.cost2
-
-    def minCostToEqualizeArray(
-        self, nums: List[int], cost1: int, cost2: int
-    ) -> int:
+    def minCostToEqualizeArray(self, nums: List[int], cost1: int, cost2: int) -> int:
         cost2 = min(2 * cost1, cost2)
-
         self.nums = nums
         self.minNums = min(nums)
         self.sumNums = sum(nums)
         self.cost1 = cost1
         self.cost2 = cost2
-
         m = max(nums)
-
         sameParity = range(m, 10**18, 2)
         diffParity = range(m + 1, 10**18, 2)
-        i = bisect_left(
-            sameParity, 0, key=lambda i: self.solve(i + 2) - self.solve(i)
-        )
-        j = bisect_left(
-            diffParity, 0, key=lambda j: self.solve(j + 2) - self.solve(j)
-        )
-
+        i = bisect_left(sameParity, 0, key=lambda i: self.solve(i + 2) - self.solve(i))
+        j = bisect_left(diffParity, 0, key=lambda j: self.solve(j + 2) - self.solve(j))
         return min(self.solve(sameParity[i]), self.solve(diffParity[j])) % Solution.MOD