Hacker rank Solution For Day 6: Let's Review

Problem:- Write a Hacker rank Solution For Day 6: Let's Review or Hacker Rank Solution Program In C++ For "Day 6: Let's Review " or Hacker rank 30 days of code Java Solution: Day 6: Let's Review solution or Hacker rank solution for 30 Days of Code Challenges.

Logic:- For this problem, we have to use a 'for loop', Let's take an example and with the help of example we will try to find out the solution suppose string is a "Hacker". for this there are two steps we follow.

Explanation:- we first divide the string into even and odd number as we can see that in the string "Hacker", H is in even place and a is in an odd place and so on. So our string is "Hacker" can be divided by even (H), odd (a), even (c), odd (k), even (e), odd (r).

Step 1:- Run the first loop up to the size of string and find the even character and print the even place character and print like below.

for(int i=0;i<s.size();i++)
{
if(i%2==0)
cout<<s[i];
}

after the first loop complete print space so we can print and separate the second loop.

cout<<" ";

Step 2:- Run the second loop up to the size of string and find the odd place character and print the odd place character and print like below.

for(int i=0;i<s.size();i++)
{
if(i%2!=0)
cout<<s[i];
}

At the end print the new line for next input to the early and next input print in the new line.

cout<<endl;

All solution provided here are in C++ (CPP) if you want these solutions in C, and Java comments below or sends a mail with your query like " day n solution in C / C++ / Java. If you want to 30 days solution (All previous solution from day 0 ) from Day 0 please check the below link. You can also find more program in below of this post. Check the below solution with the full explanation (explanation in simple language).

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Solution:-

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
#include<cstring>
using namespace std;

int main() 
{

    /* Enter your code here. Read input from STDIN. Print output to STDOUT */   
    int t;
    cin>>t;
    while(t--)
        {
        string s;
        cin>>s;
        for(int i=0;i<s.size();i++)
            {
            if(i%2==0)
            cout<<s[i];
            }
        cout<<" ";
        for(int i=0;i<s.size();i++)
            {
            if(i%2!=0)
            cout<<s[i];
            }
        cout<<endl;
    }

    return 0;
}


Output:-

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