Python Interview Questions


 #q1(prime number)

def prime(x,y):
    s=[]
    for i in range(x,y+1):
        
        for n in range(2,i):
            
            if i%n==0:
                break
        else:
            s.append(i)
    return s
            
print(prime(100,200))
     
 #q2(sort element accending order)
l=[10,20,30]
def sort(l):
    return sorted(l)
x=sort([90,100,80])
print(x[::-1])



#q3(sort without using sort)
l=[90,100,80]
def sort(l):
    k=[]
    for i in range(len(l)):
        k.append(min(l))
        l.remove(min(l))
    return k
sort(l)
   


#q4(fibonachi series)
def fib(z):
    x=0
    y=1
    p=[]
    while y<=z:
        p.append(y)
        x,y=y,y+x
    return p
fib(50)
        

#q5(reverese list)
def rev(l):
    new_list = l[::-1]
    return new_list

l=[40,50,70
print(rev(l)) 


#q6(check string palindrom)
def pal(s):
    x=s[::-1]
    if s==x:
        return 'palindrom'
    else:
        return 'notpalindrom'
pal('121')



#q7(set of duplicate)
l=[1,2,3,1,2,4,5]
def dup(l):
    k=[]
    for i in l:
        if l.count(i)>1:
            if i not in k:
                k.append(i)
             
    return k  
dup(l)
       
#q8(count no of occourance)
l=[1,2,3,4,1,2,3,6,7]
def countn(l):
    s={}
    for i in l:
        s[i]=l.count(i)
    return s
countn(l)


#q9(count no of word)
l='hi ram'
def word(l):
    x=l.split(' ')
     
    return x,len(x)
word(l)


 #10(find element present in list or not)
l=[1,2,3,4,5,6,8]
i=int(input('enter no for find:-'))
def find(l):
    for a in range(len(l)):
        if l[a]==i:
            return a,i
    print('nt')
       
find(l)



#11(find all num inside string)
l='a1s2d3f45'

def fi(d):
    s=['0','1','2','3','4','5','6','7','8','9']
    k=[]
    for i in l:
        if i in s:
            k.append(i)
    return k
fi(l)



#12(delete rept string valu)
l='satyam'
s=list(l)
for i in s:
    if l.count(i)>1:
         s.remove(i)
print(s)



#13(removing duplicate from string)
l='satyam'
def dup(l):
    #s=list(l)
    p=[]
    for i in l:
        if i not in p:
            p.append(i)
    return p
dup(l)




#14(k-1)
l=[1,2,3]
def k1(l):
    s=[]
    if len(l)<4:
        for i in range(len(l)):
            s.append(max(l))
            l.remove(max(l))
    
    else:
        for i in range(4):
        
            s.append(max(l))
            l.remove(max(l))
         
    x=sum(s)
    return x
k1(l)



#14(k-2)
l=[1,1,2,10,3,1,12]
def k2(l):
    s=sum(l)*2
    x=max(l)
    l.remove(x)




      

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