Get Minimum Element From The Stack || Program 34 || Competitive Coding || Learning Monkey || скачать в хорошем качестве

Get Minimum Element From The Stack || Program 34 || Competitive Coding || Learning Monkey ||

Get Minimum Element From The Stack In this class, We discuss Get Minimum Element From The Stack. Readers can prepare a full competitive coding course to crack product development companies. Click Here. The reader should have basic coding skills to work with competitive coding. Click here. Question: Given N elements. Our task is to implement a stack that provides a minimum element in the stack in O(1) time. The time complexity of push and pop functions is O(1). We need to write the function push, pop, and get minimum. The push function will push the element in the stack. The pop function will pop the element from the stack. The get minimum function has to display the minimum element from the stack. Examples and step-by-step explanations are provided in the video. Code: class stack: def __init__(self): self.s=[] self.minEle=None def push(self,x): if len(self.s)==0: self.s.append(x) self.minEle=x elif x lt self.minEle: temp=(2*x)-self.minEle self.s.append(temp) self.minEle=x else: self.s.append(x) def pop(self): k=0 if len(self.s)==0: return -1 else: z=self.s.pop() if z lt self.minEle: k=self.minEle self.minEle=((2*k)-z) else: k=z return k def getMin(self): if len(self.s)==0: return -1 else: return self.minEle ob=stack() while(True): x=int(input("enter your option 1. Push 2. Pop 3. quit")) if(x==1): y=int(input("enter the element to insert")) ob.push(y) print("element inserted") elif(x==2): z=ob.pop() print(z) elif(x==3): z=ob.getMin() print(z) else: print("wrong option") Link for playlists:    / @wisdomerscse   Link for our website: https://learningmonkey.in Follow us on Facebook @   / learningmonkey   Follow us on Instagram @   / learningmonkey1   Follow us on Twitter @   / _learningmonkey   Mail us @ learningmonkey01@gmail.com