Alien Dictionary

code:

 from collections import defaultdict

class Solution:

    def __init__(self):

        self.graph=defaultdict(list)

        

        

    def addEdge(self,u,v):

        self.graph[u].append(v)

        

    

    def topUtil(self,visited,i,stack):

        visited[i]=True

        

        for k in self.graph[i]:

            if visited[k]==False:

                self.topUtil(visited,k,stack)

        

        stack.append(i)    

        

        

        

    def findOrder(self,dict, N, k):

        

        for i in range(len(dict)-1):

            w1=dict[i]

            w2=dict[i+1]

            

            for j in range( min(len(w1),len(w2)) ):

                if w1[j]!=w2[j]:

                    #print(w1[j],w2[j])

                    self.addEdge(ord(w1[j])-97,ord(w2[j])-97)

                    break

                    

        #print(self.graph)

                

                    

    # top(self,k):

        visited=[False]*k

        stack=[]

        for i in range(k):

            if visited[i]==False:

                self.topUtil(visited,i,stack)

                

        ans=[]        

        while stack:

            ans.append(stack.pop())

        

        for l in range(len(ans)):

            

            ans[l]=chr(ans[l]+97)

            

        print(ans)

        

        

N = 5

K = 4

dict = ["baa","abcd","abca","cab","cad"]

g=Solution()       

g.findOrder(dict,N,K)

        

Find whether it is possible to finish all tasks or not from given dependencies Using Python

code:

from collections import defaultdict

class graph:

    def __init__(self,v):

        self.v=v

        self.graph=defaultdict(list)

        

    def addEdge(self,u,v):

        self.graph[u].append(v)

        

    

        

    def isCycleUtil(self,visited,i,recStack):

        visited[i]=True

        recStack[i]=True

        

        for j in self.graph[i]:

            if visited[j]==False:

                if self.isCycleUtil(visited,j,recStack)==True:

                    return True

            elif recStack[j]==True:

                return True

                

        recStack[i]=False

        return False

        

    def isCycle(self):

        visited=[False]*self.v

        recStack=[False]*self.v

        

        for i in range(self.v):

            if visited[i]==False:

                if self.isCycleUtil(visited,i,recStack)==True:

                    return True

                    

        return False

        

    

    def find(self):

        if self.isCycle()==True:

            return False

            

        inDegree=[0]*(self.v+1)

        

        for i in self.graph:

            for j in self.graph[i]:

                inDegree[j]+=1

               

        

        

        done=[False]*(self.v+1)

        q=[]

        

        for k in range(self.v+1):

            if inDegree[k]==0:

                q.append(k)

                done[k]=True

        

        print(inDegree)

        while q:

            t=q.pop(0)

            

            for node in self.graph[t]:

                

                done[node]=True

                inDegree[node]-=1

                if inDegree[node]==0:

                    q.append(node)

            

            

        

        for p in done:

            if p==False:

                

                return False

        return True

                

                

#Driver Code               

g=graph(4)

g.addEdge(1,0)

g.addEdge(2,1)

g.addEdge(3,2)

print(g.find())

                    

Rat in a Maze Solution

code:

 def pathUtil(maze,visited,sol,x,y,N,possible):

    #print(sol)

    if x<0 or y<0 or x>N-1 or y>N-1 or maze[x][y]==0 or visited[x][y]==1:

        return

    

    if x==N-1 and y==N-1 and maze[x][y]==1:

        possible.append(sol)

        return

    

    visited[x][y]=1

    

    pathUtil(maze,visited,sol+'D',x+1,y,N,possible)

    pathUtil(maze,visited,sol+'U',x-1,y,N,possible)

    pathUtil(maze,visited,sol+'R',x,y+1,N,possible)

    pathUtil(maze,visited,sol+'L',x,y-1,N,possible)

    

    visited[x][y]=0

    

def path(maze,N):

    visited=[[0 for i in range(N)] for j in range(N)]

    sol=''

    possible=[]

    pathUtil(maze,visited,sol,0,0,N,possible)

    

    print("possible path : ",possible)

    

N = 4

maze = [[1, 0, 0, 0],

         [1, 1, 0, 1], 

         [1, 1, 0, 0],

         [0, 1, 1, 1]]

         

path(maze,N)

Competitive Programming vs Development

Competitive programming vs development which would be good for our career this question arise in every IT Student. Every Would suggest you to go for competitive programming because it opens great opportunity in jobs. Their are many platform where you can practice your coding skills or do competitive programming like Hackerrank,Topcoder,codechef,codeforces..etc.. The development is also important but not necessary if you pro in Data structure and algorithm.

If you want to do startup or business in IT field then it will be ok to do more projects so you can understand how Software are developed but for job in Product Based company you have to crack competitive programming.

If you are in 1st or 2nd year then highly recommend you to try competitive Programming and learn Data Structure and algorithm , increase your programming level to advance or intermediate. If you are able to do this then you can continue this in 3rd or 4th year easily.

If you are start Competitive Programming in 3rd year or 4th year you just learn topics at least topics of data structure . Then focus on making one big project related to any trending technology.

Here you can get you project idea : https://www.codementor.io/projects

In My coding journey what mistake I have done I will explain you and you should not do that.. 

Best Platform For Competitive Programming

Actually Here many student get confuse that which competitive programming platform is best for practice and I have also wasted my 1st and 2nd year of college in this things but it is just wastage of time. Their is nothing like best platform through which if you practice on that you become pro in less time it is not like that rather than that programming is about how much you have practice it , you just have to solve more and more question ,start with easy level do  100-150 question of each topic related to Data structure and algorithm then switch into intermediate level and then into Advance level. Initially you can able to solve only one question in one day or more. but don't demotivate just continue your count of failure , Try to solve then see solution from editorials understand them.

Don't Confuse Too Much

I just recommend you to do practice in hackerrank And do competition in codechef ,You can use  Codechef rating into you resume to show case your Programming Skills.

Doing only Easy Problem

Here if you have done 50 question of easy level of each topic then it means you have done with easy level you have to switch into another level . It is necessary for you to understand when to switch to higher  level otherwise you cant progress you will be always into you comfort zone.

At The End I just Want to say Don't lose Focus your Time investment in CP will be worth for you in career. It will frustrating , depressing but don't give up.  

I will post Topic For Data structure and algorithm in future posts

Thank you.

Knapsack(0/1) Problem Solution Using Python

Given weights and values of n items, put these items in a knapsack of capacity W to get the maximum total value in the knapsack. In other words, given two integer arrays val[0..n-1] and wt[0..n-1] which represent values and weights associated with n items respectively. Also given an integer W which represents knapsack capacity, find out the maximum value subset of val[] such that sum of the weights of this subset is smaller than or equal to W. You cannot break an item, either pick the complete item or don’t pick it (0-1 property).

Code :

w=10 n=4 val=[10,40,30,50] wt=[5,4,6,3] row,col=(n,w) arr=[[0 for i in range(col+1)]for j in range(row+1)] for item in range(row): item=item+1 for capacity in range(col): capacity=capacity+1 maxWithCurr=0 maxWithoutCurr=arr[item-1][capacity] if capacity >= wt[item-1]: remaining=capacity-wt[item-1] maxWithCurr=val[item-1] maxWithCurr+=arr[item-1][remaining] arr[item][capacity]=max(maxWithCurr,maxWithoutCurr) print(arr[row][col]) print(arr)

Bit Stuffing in Computer Network Using Python

Bit Stuffing Mechanism

Here, the delimiting flag sequence generally contains six or more consecutive 1s. Most protocols use the 8-bit pattern 01111110 as flag. In order to differentiate the message from the flag in case of same sequence, a single bit is stuffed in the message. Whenever a 0 bit is followed by five consecutive 1bits in the message, an extra 0 bit is stuffed at the end of the five 1s. When the receiver receives the message, it removes the stuffed 0s after each sequence of five 1s. The un-stuffed message is then sent to the upper layers.

 

Code :

#Stuffing the original signal def stuff(sig): onec = 0 # one counter c = 0 # index counter one = [] # one indexes s = list(sig) for i in s: c += 1 if i == '0': onec = 0 else: onec += 1 if onec == 5: one.append(c) onec = 0 k = 0 # count extra index number for i in one: # print(i) s.insert(i+k,'0') k+=1 return s # destuffing the stuffed signal def destuff(sig): onec = 0 # one counter c = 0 # index counter one = [] # one indexes sig = list(sig) for i in sig: c += 1 if i == '0': onec = 0 else: onec += 1 if onec == 5: one.append(c) onec = 0 k = 0 # count extra index number for i in one: # print(i) sig.pop(i+k) k-=1 return sig # Driver Code sig = input("Enter the signal: ") print("Original Signal : ",sig) stf = stuff(sig) print("Stuffed Signal : ",end ="") print("".join([ x for x in stf])) dstf = destuff(stf) print("Destuffed Signal : ",end ="") print("".join([ x for x in dstf]))

Byte Stuffing in Computer Network Using Python

Byte Stuffing Mechanism

If the pattern of the flag byte is present in the message byte sequence, there should be a strategy so that the receiver does not consider the pattern as the end of the frame. Here, a special byte called the escape character (ESC) is stuffed before every byte in the message with the same pattern as the flag byte. If the ESC sequence is found in the message byte, then another ESC byte is stuffed before it.

Code :

st=input('Enter the Data :') flag=input('Flag :') esc=input('Esc :') def Sender(st,flag,esc): a=st.replace(esc,esc*2) y=a.replace(flag,esc+flag) return y print("Stuffed : "+Sender(st,flag,esc))