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Author: hogan-tech

3512-minimum-operations-to-make-array-sum-divisible-by-k/3512-minimum-operations-to-make-array-sum-divisible-by-k.py

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+# time complexity: O(n)
+# space complexity: O(n)
+from typing import List
+
+
+class Solution:
+    def minOperations(self, nums: List[int], k: int) -> int:
+        total = sum(nums)
+        remainder = total % k
+        if remainder == 0:
+            return 0
+
+        mods = []
+        for num in nums:
+            mods.append(num % k)
+        mods.sort(reverse=True)
+
+        result = 0
+        for m in mods:
+            if remainder == 0:
+                break
+            take = min(remainder, m)
+            result += take
+            remainder -= take
+        return result
+
+
+nums = [3, 9, 7]
+k = 5
+print(Solution().minOperations(nums, k))
+nums = [4, 1, 3]
+k = 4
+print(Solution().minOperations(nums, k))
+nums = [3, 2]
+k = 6
+print(Solution().minOperations(nums, k))