Hackerrank For Loop Solution

Hackerrank For Loop Solution

A for loop is a programming language statement which allows code to be repeatedly executed.

The syntax for this is

for ( <expression_1> ; <expression_2> ; <expression_3> )
    <statement>
  • expression_1 is used for intializing variables which are generally used for controlling terminating flag for the loop.
  • expression_2 is used to check for the terminating condition. If this evaluates to false, then the loop is terminated.
  • expression_3 is generally used to update the flags/variables.

A sample loop will be

for(int i = 0; i < 10; i++) {
    ...
}

Input Format

You will be given two positive integers,  and  (), separated by a newline.

Output Format.MathJax_SVG_LineBox {display: table!important} .MathJax_SVG_LineBox span {display: table-cell!important; width: 10000em!important; min-width: 0; max-width: none; padding: 0; border: 0; margin: 0}

For each integer  in the interval  :

  • If , then print the English representation of it in lowercase. That is "one" for , "two" for , and so on.
  • Else if  and it is an even number, then print "even".
  • Else if  and it is an odd number, then print "odd".

Note:

Sample Input

8
11

Sample Output

eight
nine
even
odd

Solution in cpp

Approach 1.

python
#include <iostream> #include <cstdio> using namespace std; const char *digits[] = {"one","two","three","four","five","six","seven","eight","nine"}; int main() { // Complete the code. int a; int b; cin >> a; cin >> b; while(b >= a ){ if((a >= 1) && (a <= 9)) cout << digits[a-1] << endl; else{ if( (a % 2) == 0) cout << "even" << endl; else cout << "odd" << endl; } ++a; } return 0; }

Approach 2.

python
#include <iostream> #include <cstdio> #include <string> #include <stdexcept> using namespace std; int main() { // Complete the code. int a, b; cin >> a; cin >> b; if (a < 1 || b < 1 || a > b) throw invalid_argument("invalid input"); string result[9] = {"one", "two", "three", "four", "five", "six", "seven", "eight", "nine"}; for (int i = a; i <= b; ++i) { if (1 <= i && i <= 9) cout << result[i-1] << endl; else cout << ((i & 1) ? "odd" : "even") << endl; } return 0; }

Approach 3.

python
#include <iostream> #include <cstdio> using namespace std; int main() { int a,b; cin>>a>>b; for(int n=a;n<=b;n++){ if(n==1) cout<<"one"<<endl; else if(n==2) cout<<"two"<<endl; else if(n==3) cout<<"three"<<endl; else if(n==4) cout<<"four"<<endl; else if(n==5) cout<<"five"<<endl; else if(n==6) cout<<"six"<<endl; else if(n==7) cout<<"seven"<<endl; else if(n==8) cout<<"eight"<<endl; else if(n==9) cout<<"nine"<<endl; else if(n%2==0) cout<<"even"<<endl; else cout<<"odd"<<endl; } return 0; }

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