feat(0789_kth-largest-element-in-a-stream): add py3 soln

0789_kth-largest-element-in-a-stream/README.md

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+Design a class to find the `kth` largest element in a stream. Note that it is the `kth` largest element in the sorted order, not the `kth` distinct element.
+
+Implement `KthLargest` class:
+
+*   `KthLargest(int k, int[] nums)` Initializes the object with the integer `k` and the stream of integers `nums`.
+*   `int add(int val)` Appends the integer `val` to the stream and returns the element representing the `kth` largest element in the stream.
+
+**Example 1:**
+
+    Input
+    ["KthLargest", "add", "add", "add", "add", "add"]
+    [[3, [4, 5, 8, 2]], [3], [5], [10], [9], [4]]
+    Output
+    [null, 4, 5, 5, 8, 8]
+    
+    Explanation
+    KthLargest kthLargest = new KthLargest(3, [4, 5, 8, 2]);
+    kthLargest.add(3);   // return 4
+    kthLargest.add(5);   // return 5
+    kthLargest.add(10);  // return 5
+    kthLargest.add(9);   // return 8
+    kthLargest.add(4);   // return 8
+    
+
+**Constraints:**
+
+*   `1 <= k <= 104`
+*   `0 <= nums.length <= 104`
+*   `-104 <= nums[i] <= 104`
+*   `-104 <= val <= 104`
+*   At most `104` calls will be made to `add`.
+*   It is guaranteed that there will be at least `k` elements in the array when you search for the `kth` element.

0789_kth-largest-element-in-a-stream/python3/naive_approach_tle.py

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+# Naive approach, will TLE
+# Time for each add: O(k)
+class KthLargest:
+
+    def __init__(self, k: int, nums: List[int]):
+        self.k = k
+        
+        len_nums = len(nums)
+        
+        # Only keep max k elements, rest are useless
+        if len_nums >= k:
+            self.nums = sorted(nums)[len_nums - k:]
+        # If we're provided less than k elements, we
+        # will it up with least possible number
+        else:
+            self.nums = [
+                *([float("-inf")] * (k - len_nums)),
+                *sorted(nums)
+            ]
+
+    def add(self, val: int) -> int:
+        # Find the right place for val in the list
+        # of length k
+        
+        i = self.k - 1
+        while i >= 0 and val < self.nums[i]:
+            i -= 1
+        
+        # If val is big enough to fit in the list, shift
+        # existing elements to left
+        j = 0
+        while j < i:
+            self.nums[j] = self.nums[j + 1]
+            j += 1
+        
+        if i >= 0:
+            self.nums[i] = val
+        
+        # kth largest element is always at index 0
+        return self.nums[0]
+
+
+# Your KthLargest object will be instantiated and called as such:
+# obj = KthLargest(k, nums)
+# param_1 = obj.add(val)

0789_kth-largest-element-in-a-stream/python3/solution.py

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+import heapq
+
+class KthLargest:
+
+    def __init__(self, k: int, nums: List[int]):
+        self.k = k
+        self.heap = nums
+        heapq.heapify(self.heap)
+        
+        # Keep removing elements until heap size is k
+        while len(self.heap) > k:
+            heapq.heappop(self.heap)
+        
+
+    def add(self, val: int) -> int:
+        heapq.heappush(self.heap, val)
+        
+        if len(self.heap) > self.k:
+            heapq.heappop(self.heap)
+        
+        return self.heap[0]
+
+
+# Your KthLargest object will be instantiated and called as such:
+# obj = KthLargest(k, nums)
+# param_1 = obj.add(val)