maxheap.py

Algorithms/Dynamic Programming/rod_cutting_problem.c

@@ -0,0 +1,24 @@
+#include <limits.h> 
+#include <stdio.h> 
+
+int max(int a, int b) { return (a > b) ? a : b; } 
+
+int cutRod(int price[], int n) 
+{ 
+    if (n <= 0) 
+        return 0; 
+    int max_val = INT_MIN; 
+    for (int i = 0; i < n; i++) 
+        max_val = max(max_val, price[i] + cutRod(price, n - i - 1)); 
+
+    return max_val; 
+} 
+
+int main() 
+{ 
+    int arr[] = { 1, 5, 8, 9, 10, 17, 17, 20 }; 
+    int size = sizeof(arr) / sizeof(arr[0]); 
+    printf("Maximum Obtainable Value is %d", cutRod(arr, size)); 
+    getchar(); 
+    return 0; 
+}

Data Structures/Heap/Python/maxheap.py

@@ -0,0 +1,24 @@
+import heapq
+
+class Solution:
+
+    def kClosest(self, points: List[List[int]], K: int) -> List[List[int]]:
+
+        maxHeap = []
+
+        for (x, y) in points:
+            distance = math.sqrt(x*x + y*y)
+
+            if len(maxHeap) >= K:
+                if -1 * distance > maxHeap[0][0]:
+                    heapq.heappushpop(maxHeap, [-1 * distance, [x, y]])
+            else:
+                heapq.heappush(maxHeap, [-1 * distance, [x, y]])
+
+        resList = []
+
+        for _ in range(K):
+            resList.append(heapq.heappop(maxHeap)[1])
+
+        # return the list
+        return resList

Data Structures/Palindrome/py_palindrome_oneliner

@@ -0,0 +1,3 @@
+s=input()
+is_palindrome = lambda phrase: phrase == phrase[::-1]
+print(is_palindrome(s))