3759

Author: vikahaze

explanations/3759/en.md

@@ -29,9 +29,6 @@ We sort the array first, which allows us to efficiently determine how many eleme
 5. If the number of greater elements is at least `k`, add the count of current value to the result.
 6. Move to the next distinct value and repeat.
 
-  * **Brute Force:** For each element, count how many other elements are strictly greater by comparing with all other elements. This takes O(n^2) time.
-  * **Optimized Strategy:** Sort the array first. Then for each distinct value, use binary search to find the first element strictly greater than it. The number of greater elements is `n - upper_bound_index`. This reduces the time complexity to O(n log n).
-
 **2.1 Initialization & Example Setup:**
 
 Let's use the example `nums = [3, 1, 2]`, `k = 1` to trace through the solution.

solutions/3759/01.py

@@ -0,0 +1,39 @@
+from typing import List
+
+
+class Solution:
+    def countElements(self, nums: List[int], k: int) -> int:
+        # Sort the array
+        sorted_nums = sorted(nums)
+        n = len(sorted_nums)
+
+        res = 0
+
+        # For each distinct value, count how many elements are strictly greater
+        i = 0
+        while i < n:
+            # Count occurrences of current value
+            count = 1
+            while i + count < n and sorted_nums[i] == sorted_nums[i + count]:
+                count += 1
+
+            # Find the first index with value strictly greater than current
+            # Use binary search to find upper bound
+            left, right = i + count, n
+            while left < right:
+                mid = (left + right) // 2
+                if sorted_nums[mid] > sorted_nums[i]:
+                    right = mid
+                else:
+                    left = mid + 1
+
+            # Number of elements strictly greater
+            greater_count = n - left
+
+            # If at least k elements are greater, add count to result
+            if greater_count >= k:
+                res += count
+
+            i += count
+
+        return res