Time: 602 ms (58.72%), Space: 32.3 MB (79.07%) - LeetHub

Author: hogan-tech

3622-maximum-frequency-of-an-element-after-performing-operations-i/3622-maximum-frequency-of-an-element-after-performing-operations-i.py

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+# time complexity: O(max(nlogn, klogn))
+# space complexity: O(n)
+from bisect import bisect_left, bisect_right
+from typing import List
+
+
+class Solution:
+    def maxFrequency(self, nums: List[int], k: int, numOperations: int) -> int:
+        nums.sort()
+        result = 0
+        numCount = {}
+        lastNumIdx = 0
+        for i in range(len(nums)):
+            if nums[i] != nums[lastNumIdx]:
+                numCount[nums[lastNumIdx]] = i - lastNumIdx
+                result = max(result, i - lastNumIdx)
+                lastNumIdx = i
+
+        numCount[nums[lastNumIdx]] = len(nums) - lastNumIdx
+        result = max(result, len(nums) - lastNumIdx)
+
+        for i in range(nums[0], nums[-1] + 1):
+            left = bisect_left(nums, i - k)
+            right = bisect_right(nums, i + k) - 1
+            if i in numCount:
+                temp = min(right - left + 1, numCount[i] + numOperations)
+            else:
+                temp = min(right - left + 1, numOperations)
+            result = max(result, temp)
+
+        return result
+
+
+nums = [1, 4, 5]
+k = 2
+numOperations = 2
+print(Solution().maxFrequency(nums, k, numOperations))
+nums = [5, 16, 20, 20, 20, 24, 24]
+k = 5
+numOperations = 4
+print(Solution().maxFrequency(nums, k, numOperations))
+nums = [2, 70, 73]
+k = 39
+numOperations = 2
+print(Solution().maxFrequency(nums, k, numOperations))