Completed Design 1 Homework

design-1.py

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+#Time Complexity : The time complexity is O(1) for all the operations ie add, remove and contains
+#Space Complexity : The space complexity is O(M + K*N) M is the initial buck K is the number 
+# of secondary bucket and N is number of primary bucket initialized
+#Did this code successfully run on Leetcode : YES
+#Any problem you faced while coding this : Faced error in some of the syntax and was confused in 
+# writing the space complexity
+
+
+#Your code here along with comments explaining your approach
+
+class MyHashSet(object):
+
+    def __init__(self):
+        self.initialBucket=1000
+        self.secondaryBucket=1000
+        self.storage=[None]*self.initialBucket
+
+    def hashfunc1(self,key):
+        return key % self.initialBucket
+
+    def hashfunc2(self,key):
+        return key // self.secondaryBucket  
+
+    def add(self, key):
+        """
+        :type key: int
+        :rtype: None
+        """
+        bucket=self.hashfunc1(key)
+        internalBucket=self.hashfunc2(key)
+        if self.storage[bucket]==None:
+            if bucket==0:
+                self.storage[bucket]=[False]*(self.secondaryBucket+1)
+            else:
+                self.storage[bucket]=[False]*self.secondaryBucket
+        self.storage[bucket][internalBucket]=True
+
+
+    def remove(self, key):
+        """
+        :type key: int
+        :rtype: None
+        """
+        bucket=self.hashfunc1(key)
+        internalBucket=self.hashfunc2(key)
+        if self.storage[bucket]==None:
+            return
+        self.storage[bucket][internalBucket]=False
+
+
+    def contains(self, key):
+        """
+        :type key: int
+        :rtype: bool
+        """
+        bucket=self.hashfunc1(key)
+        internalBucket=self.hashfunc2(key)
+        if self.storage[bucket]==None:
+            return False
+        return self.storage[bucket][internalBucket]
+
+
+
+# Your MyHashSet object will be instantiated and called as such:
+# obj = MyHashSet()
+# obj.add(key)
+# obj.remove(key)
+# param_3 = obj.contains(key)

designminheap.py

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+#Time Complexity : The time complexity is O(1) for all the operations 
+#Space Complexity : The space complexity is O(N) where N is no of elements pushed in stack
+#Did this code successfully run on Leetcode : YES
+
+class MinStack(object):
+
+    def __init__(self):
+        self.st=[]
+        self.min=float('inf')
+
+    def push(self, val):
+        """
+        :type val: int
+        :rtype: None
+        """
+        if val <= self.min:
+            self.st.append(self.min)
+            self.min = val
+
+        self.st.append(val)
+
+    def pop(self):
+        """
+        :rtype: None
+        """
+        if not self.st:
+            return
+        if self.st.pop() == self.min:
+            self.min = self.st.pop()
+
+
+    def top(self):
+        """
+        :rtype: int
+        """
+        return self.st[-1] if self.st else None
+
+
+    def getMin(self):
+        """
+        :rtype: int
+        """
+        return self.min
+
+
+
+# Your MinStack object will be instantiated and called as such:
+# obj = MinStack()
+# obj.push(val)
+# obj.pop()
+# param_3 = obj.top()
+# param_4 = obj.getMin()