For me this was the most difficult type of problem with counting specific numbers whose digits satisfy some condition. After some failed attempts I finally cracked it. It requuires simple DP. In my case I used 3 dimensional arrays. The problem is tricky, so it needs carefull analysis.

What wrong in this Code?

n = int(input())
d = n
x = []
a = n

def check(a,d):
    if(d==0):
        pass
    elif(d==10):
        x.append(1)
    elif(a<10):
        x.append(a)
        d = ((d-a)/10)
        check(d,d)
    elif(a%10==0 and a==10):
        x.append(0)
        d = (d/10)
        check(d,d)
    elif(a%10==0):
        a=a/10;
        check(a,d)
    else:
        a=a%10;
        check(a,d)
check(a,d)
x.reverse()     

l = len(str(n))            

count = 0
for i in range(0,l-2):
        if((x[i]-x[i+1])==-1 or (x[i]-x[i+1])==1):
                count = count + 1
print(count)

My script gives correct output in IDLE but fails due to timeout for this problem. Any suggestion or help:

i = input() x=10**10

def prime(a3): n=a3+1 flag=0 count=2 for i in range(n): flag=0 while flag!=1: count=count+1 for i in range(2,count): if count%i==0: count=count+1 break else: continue flag=1 print count

def fun(x,i): if i<10*x: print 0 elif i==10*x: print 1 elif i>10*x: a=i/10*x
a1=str(int(a)) a2=list(a1) a3=len(a2)-1 prime(a3)

fun(x, i)

I submitted my solution. it works fine on my machine. Runtime errors shown on your results. why is that?

when do we get solutions of the problem?