Question

A top-secret message containing letters from A-Z is encoded to numbers using the following mapping: A -> 1 B- 2 Z-> 26 You are a SPY. You have to determine the total number of ways that message can be decoded. Note: An empty digit sequence is considered to have one decoding. It may be assumed that the input contains valid digits from 0 to 9 and If there are leading 0's, extra trailing 0's and two or more consecutive 0's then it is an invalid string. Example: Given encoded message "123", it could be decoded as "ABC" (1 2 3) or "LC" (12 3) or "AW"(1 23) So total ways are three.

A top-secret message containing letters from A-Z is encoded to numbers using the following mapping: A -> 1 B- 2 Z-> 26 You are a SPY. You have to determine the total number of ways that message can be decoded. Note: An empty digit sequence is considered to have one decoding. It may be assumed that the input contains valid digits from 0 to 9 and If there are leading 0's, extra trailing 0's and two or more consecutive 0's then it is an invalid string. Example: Given encoded message "123", it could be decoded as "ABC" (1 2 3) or "LC" (12 3) or "AW"(1 23) So total ways are three.

requirements
write c++ code for the above programs use only 1D arrays with loops and if else statement don't use any other way as it would be useful

A top-secret message containing letters from A-Z is encoded to numbers using the following mapping: A -> 1 B- 2 Z-> 26 You are a SPY. You have to determine the total number of ways that message can be decoded. Note: An empty digit sequence is considered to have one decoding. It may be assumed that the input contains valid digits from 0 to 9 and If there are leading 0's, extra trailing 0's and two or more consecutive 0's then it is an invalid string. Example: Given encoded message "123", it could be decoded as "ABC" (1 2 3) or "LC" (12 3) or "AW"(1 23) So total ways are three.

A top-secret message containing letters from A-Z is encoded to numbers using the following mapping: A -> 1 B- 2 Z-> 26 You are a SPY. You have to determine the total number of ways that message can be decoded. Note: An empty digit sequence is considered to have one decoding. It may be assumed that the input contains valid digits from 0 to 9 and If there are leading 0's, extra trailing 0's and two or more consecutive 0's then it is an invalid string. Example: Given encoded message "123", it could be decoded as "ABC" (1 2 3) or "LC" (12 3) or "AW"(1 23) So total ways are three.

Answer all four parts with comments for better understanding

Answer 1

There are only two questions in Pictures provided above out of which one question i.e spy and decoding is repeated three times. So effectively only two questions:

Question 1. Spy and Decoding

This question has been solved using dynamic programming most of the logic has been discussed in the code with the help of comments I hope that you find it useful.

CODE :

#include <bits/stdc++.h> //including the necessary libraries
using namespace std;

//Decoding Function which return number of possible decodings
int decoding(string sequence){
    int n = sequence.size(); // calculating the size of the digit sequence
    int dp[n+1]; //dp table in which dp[i] denotes the number of possible decodings of the sequence upto length i
    dp[0] = 1; //base case
    dp[1] = 1; //base case
    if(sequence[0] == '0') return 0; //Invalid case as if any sequence starts with 0 then it is invalid Eg: 01122

    for(int i = 2;i <= n; i++){
        //initializing the dp value to be 0
        dp[i] = 0;

        //if the previous digit was greater than 0 then it would also contribute upto next length also
        if(sequence[i-1] > '0') dp[i] = dp[i-1];

        //Here we handle the two digit case as we lookup behind upto two indexes in the sequence
        //If the digit 2 places behind is 1 then we can have any digit at the next position as
        //all 2 digits numbers having 1 at it's ten's place would be valid Eg: 10,11,12,13 ... 19
        //But if the digit 2 places behind is 2 then we only have 7 choices namely 20,21,22,23,24,25,26
        // Hence we check only upto 6 by using the < 7 condition.
        if(sequence[i-2] == '2' && sequence[i-1] < '7' || sequence[i-2] == '1') dp[i] += dp[i-2];
    }
    return dp[n]; //returning the answer for the length n

}

//DRIVER CODE//

int main(){
    string digit_sequence1 = "123"; //Valid with answer 3
    string digit_sequence2 = "204"; //Valid with answer 1
    string digit_sequence3 = "603"; //InValid with answer 0

    cout<<"The Number of Possible Decodings for digit_sequence1 are : "<<decoding(digit_sequence1)<<endl;
    cout<<"The Number of Possible Decodings for digit_sequence2 are : "<<decoding(digit_sequence2)<<endl;
    cout<<"The Number of Possible Decodings for digit_sequence3 are : "<<decoding(digit_sequence3)<<endl;
}

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Screenshots:

Decoding Function :

Driver Code :

Compiling & Running:

Output :

/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/-/

Question 2: Pattern Finding

Code:

#include <bits/stdc++.h> //including the necessary libraries
using namespace std;

int main(){
    string sentence; //declaraing sentence variable
    cin>>sentence;   //taking input for the sentence
  
    string pattern; //declaraing pattern variable
    cin>>pattern;    //taking input for the pattern

    int s_length = sentence.length(); //calculating length of sentence
    int pat_length = pattern.length();//calculating length of pattern

    //running the loop on sentence from index 0 to (sentence length - pattern length)
    //as it is only feasible to run the loop upto (sentence length - pattern length)
    //because after this any substring would be of less length then pattern length
    //Eg: sentence : elephant pattern : ant ---- here we will run the loop only till
    //a of ant in elephant as after this all sub strings will have length shorter than
    //length of pattern i.e ant here
    for(int i=0;i<s_length-pat_length+1;i++){
        bool flag = true; //using a flag varible to check if the pattern has been found
        int j=0; //iterator variable

        //searching for the pattern in the sentence
        for(j=0;j<pat_length;j++){
            if(sentence[i+j] == pattern[j]) continue;
            else{
                flag = false; //setting the flag varible false if pattern doesn't matches at any place
                break; //breaking the loop
            }
        }
        //NOTE : Here we are printing the index based on 1-based indexing
        //Eg : ELEPHANT == 12345678
        //checking if the pattern has been found or not
        if(flag) cout<<i+1<<"--"<<i+j<<endl; //printing the index of pattern in the sentence
    }
}

Screenshots:

Code:

Compiling :

Running and Output :

Happy Coding !