Todays leetcode problem solution

LC_TODAaY.py.txt

@@ -0,0 +1,29 @@
+class Solution:
+    def smallestDistancePair(self, nums: List[int], k: int) -> int:
+        list_size = len(nums)
+
+        # Find the maximum element in the list
+        max_element = max(nums)
+
+        # Initialize a bucket list to store counts of each distance
+        distance_bucket = [0] * (max_element + 1)
+
+        # Populate the bucket list with counts of each distance
+        for i in range(list_size):
+            for j in range(i + 1, list_size):
+                # Compute the distance between nums[i] and nums[j]
+                distance = abs(nums[i] - nums[j])
+
+                # Increment the count for this distance in the bucket
+                distance_bucket[distance] += 1
+
+        # Find the k-th smallest distance
+        for dist in range(max_element + 1):
+            # Reduce k by the number of pairs with the current distance
+            k -= distance_bucket[dist]
+
+            # If k is less than or equal to 0, return the current distance
+            if k <= 0:
+                return dist
+
+        return -1  # Return -1 if no distance found, should not reach here