Update and rename compare_2_tops.py to compare_two_tops.py

heaps/compare_2_tops.py

@@ -1,43 +0,0 @@
-#!/usr/bin/env python3
-# -*- coding: utf-8 -*-
-# author: bt3gl
-
-
-'''
-You are given an array of integers stones where 
-stones[i] is the weight of the ith stone.
-
-We are playing a game with the stones. On each turn, 
-we choose the heaviest two stones and smash them together. 
-Suppose the heaviest two stones have weights x and y with 
-x <= y. The result of this smash is:
-
-If x == y, both stones are destroyed, and
-If x != y, the stone of weight x is destroyed, and the stone 
-of weight y has new weight y - x.
-At the end of the game, there is at most one stone left.
-
-Return the weight of the last remaining stone. 
-'''
-
-def last_stone(stones) -> int:
-
-        for i in range(len(stones)):
-            stones[i] *= -1
-
-        heapq.heapify(stones)
-
-        while len(stones) > 1:
-
-            stone_1 = heapq.heappop(stones)
-            stone_2 = heapq.heappop(stones)
-
-            if stone_1 != stone_2:
-                heapq.heappush(stones, stone_1 - stone_2)
-
-        if stones:
-            return -heapq.heappop(stones)
-            
-        else:
-            return 0
-          

heaps/compare_two_tops.py

@@ -0,0 +1,25 @@
+#!/usr/bin/env python3
+# -*- coding: utf-8 -*-
+# author: bt3gl
+
+
+def compare_two_tops(array) -> int:
+
+        for i in range(len(array)):
+            array[i] *= -1
+
+        heapq.heapify(array)
+
+        while len(array) > 1:
+
+            val1 = heapq.heappop(array)
+            val2 = heapq.heappop(array)
+
+            if val1 != val2:
+                heapq.heappush(array, val1 - val2)
+
+        if array:
+            return -heapq.heappop(array)
+            
+        return 0
+