Completed Design-1

Problem 1.py

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+# Time Complexity  : O(1) for all operations
+# Space Complexity : O(1) fixed size
+#Did this code successfully run on Leetcode : Yes
+#Any problem you faced while coding this : Yes: went back and forth to understand add funtion
+
+# We create a storage list of 1000 slots, each slot starts as None and only builds an inner list of 1000 False values when a key needs it
+# Every key maps to an exact position using key%1000 for outer slot and key//1000 for inner position
+# At that position, add sets True, remove sets False, contains returns whatever is there
+
+class MyHashSet:
+    def __init__(self):
+        self.hash1= 1000                                    #initializing the size of hash1 & hash2 to 1000
+        self.hash2 = 1000
+        self.storage = [None] *1000                         #initializing our storage with false
+
+    def primary(self,key:int):
+        return key % self.hash1                             #this is where we are deciding the index for the key in the storage array
+
+    def secondary(self,key:int):
+        return key// self.hash2                             #this is where we are deciding the index for the key in the secondary array
+
+    def add(self, key: int) -> None:
+        index = self.primary(key)                           #the key is passed in the primary function to check where we need to store the key
+        if self.storage[index] is None:                     #if the index is empty the we will create a new array under the index of size 1000
+            if index == 0:                                  #we will have a special case for index 0 because the 1M key will not be stored if we do not increase the size of secondary array to 1001
+                self.storage[index]= [False]*(self.hash2 +1)
+            else: 
+                self.storage[index]= [False]*self.hash2     #for all the other keys,the size of secondary array will be 1000 only
+        index2= self.secondary(key)                         #if the primary array is not empty then we will pass the key to secondary array and find it a place to store
+        self.storage[index][index2]=True                    #mark the place as true where the key is stored
+
+    def remove(self, key: int) -> None:  
+        index = self.primary(key)                           #we again pass the key to primary function to find the index of the key in the storage array
+        if self.storage[index] is None:                     #if it doesn not exist then we just return None
+            return 
+        index2 = self.secondary(key)                        #if the primary array is not empty then we will pass the key to secondary array and find it a place to remove  and return false
+        self.storage[index][index2]=False
+
+
+    def contains(self, key: int) -> bool:
+        index = self.primary(key)                           #we again pass the key to primary function to find the index of the key in the storage array
+        if self.storage[index] is None:                     #if it does not exist then we just return false
+             return False
+        index2 = self.secondary(key)                        #if the primary array is not empty then we will pass the key to secondary array and find it a place to check if the key is present or not and return true or false accordingly
+        return self.storage[index][index2]
+
+

Problem2.py

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+# Time Complexity: O(1) for all operations: push, pop, top and getMin all use
+# Space Complexity: O(n) both stacks grow by 1 on every push, giving us O(2n) which simplifies to O(n)
+
+# We maintain two stacks in parallel - the main stack stores all values while the min stack stores a snapshot of the current minimum at every level
+
+# On every push we compare the new value against the current min and update it, so we always know the minimum without ever having to search
+
+# Logic: On every pop we remove from both stacks simultaneously and the new minimum is automatically restored by looking at the top of the min stack
+
+class MinStack:
+
+    def __init__(self):                #here we are initializing 2 explty stacks
+      self.stack =[]                   #the first stack is the enter all the values
+      self.minstack=[]                 #the 2nd one is to maintain values that are min corresponding to the elements
+      self.min = float('inf')          #we need t set the value in the minstack to infinity so whatever number next enters will automatically be the new min value
+      self.minstack.append(self.min)   #we need to append infinity into min stack
+
+    def push(self, value: int) -> None:
+       self.min = min(value,self.min)   #it just returns the min value everytime
+       self.stack.append(value)         # here we are appending the values to the main stack
+       self.minstack.append(self.min)   #here we are appending the min values only
+
+    def pop(self) -> None:
+        self.stack.pop()               #we pop the element using pop funtion
+        self.minstack.pop()
+        self.min = self.minstack[-1]   #we need to update the new min value and it will be the topmost value in the table
+
+
+    def top(self) -> int:
+        return self.stack[-1]
+
+
+    def getMin(self) -> int:
+        return self.min
+
+
+
+# Your MinStack object will be instantiated and called as such:
+# obj = MinStack()
+# obj.push(value)
+# obj.pop()
+# param_3 = obj.top()
+# param_4 = obj.getMin()