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Vpropel VIT | POD | 18/05/2020 | Cryptography-II | 08
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//Dp | |
#include <iostream> | |
#include<string> | |
using namespace std; | |
unsigned long int MOD=1000000007; | |
unsigned long int counter(string digits, unsigned long int n) | |
{ | |
unsigned long int count[n+1]={0}; | |
count[0] = 1; | |
if(digits[0]=='*') | |
count[1]=9; | |
else | |
count[1] = 1; | |
for (int i = 2; i <= n; i++) | |
{ | |
if(digits[i-2]=='*' && digits[i-1]!='*') | |
{ | |
count[i]=count[i-1]; | |
char varnum='1'; | |
while(varnum<='9') | |
{ | |
if((varnum<='2' && digits[i-1]<='6')||varnum=='1') | |
count[i]=(count[i]+count[i-2])%MOD; | |
varnum++; | |
} | |
} | |
else if(digits[i-1]=='*' && digits[i-2]!='*') | |
{ | |
count[i]=(9*count[i-1])%MOD; | |
char varnum='1'; | |
while(varnum<='9') | |
{ | |
if((varnum<='6' && digits[i-2]<='2')||digits[i-2]=='1') | |
count[i]=(count[i]+count[i-2])%MOD; | |
varnum++; | |
} | |
} | |
else if(digits[i-1]=='*' && digits[i-2]=='*') | |
{ | |
count[i]=(9*count[i-1])%MOD; | |
char v1='1'; | |
char v2; | |
for(v1;v1<='9';v1++) | |
for(v2='1';v2<='9';v2++) | |
{ | |
if((v1<='2'&& v2<='6')||v1=='1') | |
count[i]=(count[i]+count[i-2])%MOD;; | |
} | |
} | |
else | |
{ | |
count[i] = count[i-1]; | |
if((digits[i-2] <='2' && digits[i-1] <='6')||digits[i-2]=='1') | |
count[i]=(count[i]+count[i-2])%MOD;; | |
} | |
} | |
/* | |
// To see how Table is formed and DP applied :) | |
for(int i=0;i<=n;i++) | |
cout<<digits[i]<<"\t"; | |
cout<<endl; | |
for(int i=0;i<=n;i++) | |
cout<<count[i]<<'\t'; | |
*/ | |
return (unsigned long int)count[n]; | |
} | |
int main() | |
{ | |
string digits; | |
cin>>digits; | |
unsigned long int n = digits.length(); | |
cout<< counter(digits,n); | |
return 0; | |
} | |
// ### Thank YOu :) | |
// @cs_jawanda | |
// .Chiranjeet Singh Jawanda. | |
// VIT VELLORE |
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//Dp | |
import java.util.Scanner; | |
public class Main | |
{ | |
public static void counter(char []digits,int n) | |
{ | |
final long MOD=1000000007; | |
long count[]=new long[n+1]; | |
count[0] = 1; | |
if(digits[0]=='*') | |
count[1]=9; | |
else | |
count[1] = 1; | |
for (int i = 2; i <= n; i++) | |
{ | |
if(digits[i-2]=='*' && digits[i-1]!='*') | |
{ | |
count[i]=count[i-1]; | |
count[i]=(count[i]+count[i-2]*(digits[i-1]<='6'?2:1))%MOD; | |
} | |
else if(digits[i-1]=='*' && digits[i-2]!='*') | |
{ | |
count[i]=(9*count[i-1])%MOD; | |
char varnum='1'; | |
while(varnum<='9') | |
{ | |
if((varnum<='6' && digits[i-2]<='2')||digits[i-2]=='1') | |
count[i]=(count[i]+count[i-2])%MOD; | |
varnum++; | |
} | |
} | |
else if(digits[i-1]=='*' && digits[i-2]=='*') | |
{ | |
count[i]=(9*count[i-1])%MOD; | |
count[i]=(count[i]+count[i-2]*15)%MOD; | |
} | |
else | |
{ | |
count[i] = count[i-1]; | |
if((digits[i-2] <='2' && digits[i-1] <='6')||digits[i-2]=='1') | |
count[i]=(count[i]+count[i-2])%MOD;; | |
} | |
} | |
System.out.print(count[n]); | |
} | |
public static void main(String []args) | |
{ | |
Scanner s=new Scanner(System.in); | |
String str=s.next(); | |
char digits[]=str.toCharArray(); | |
counter(digits,digits.length); | |
} | |
} | |
// ### Thank YOu :) | |
// @cs_jawanda | |
// .Chiranjeet Singh Jawanda. | |
// VIT VELLORE |
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//Recursion using Memoisation | |
import java.util.Scanner; | |
class Solve | |
{ | |
long M=1000000007; | |
public int numDecodings(String s) | |
{ | |
Integer[]memo=new Integer[s.length()]; | |
return ways(s,s.length()-1,memo); | |
} | |
public int ways(String s,int i,Integer[] memo) | |
{ | |
if(i<0) return 1; | |
if(memo[i]!=null) return memo[i]; | |
if(s.charAt(i)=='*') | |
{ | |
long res=9*ways(s,i-1,memo); | |
if(i>0 && s.charAt(i-1)=='1') | |
res=(res+9*ways(s,i-2,memo))%M; | |
else if(i>0 && s.charAt(i-1)=='2') | |
res=(res+6*ways(s,i-2,memo))%M; | |
else if(i>0 && s.charAt(i-1)=='*') | |
res=(res+15*ways(s,i-2,memo))%M; | |
memo[i]=(int)res; | |
return memo[i]; | |
} | |
long res=s.charAt(i)!='0'?ways(s,i-1,memo):0; | |
if(i>0 && (s.charAt(i-1)=='1' ||(s.charAt(i-1)=='2' && s.charAt(i)<='6'))) | |
res=(res + ways(s,i-2,memo))%M; | |
else if(i>0 && s.charAt(i-1)=='*') | |
res=(res+(s.charAt(i)<='6'?2:1)*ways(s,i-2,memo))%M; | |
memo[i]=(int)res; | |
return memo[i]; | |
} | |
} | |
public class Practice | |
{ | |
public static void main(String []args) throws Exception | |
{ | |
Scanner s=new Scanner(System.in); | |
String str=s.next(); s.close(); | |
Solve obj=new Solve(); | |
System.out.println(obj.numDecodings(str)); | |
} | |
} |
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Cryptography-II | |
A message containing letters from A-Z is being encoded to numbers using the following mapping way:'A' -> 1,'B' -> 2,'Z' -> 26 | |
Beyond that, now the encoded string can also contain the character '*', which can be treated as one of the numbers from 1 to 9. | |
Given the encoded message containing digits and the character '*', return the total number of ways to decode it. | |
For example, 1* has got 18 decoding possibilities as shown below | |
(i) 1* - 1 may be decoded as A and * can take values from 1 to 9 | |
(ii) 1* - Keep 1 as 1 and * can take values from 1 to 9 then characters corressponding to 11 to 19 will be deocoded | |
Input Format | |
First line contains the encoded string | |
Output Format | |
Print the decoded string | |
Also, since the answer may be very large, you should return the output mod 10^9 + 7. |
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https://leetcode.com/articles/decode-ways-ii/