Conditional Statements in C - Hackerrank step by step solution

Conditional Statements in C - Hackerrank solution

 
Welcome Back, Guys !!
Today in this post we will discuss Conditional statements in C  Hackerrank Practice problem in C programming language.
This problem is set for students to get better knowledge to handle conditional statement i.e if-else statement.

 Conditional Statements in C - Problem statement:

Objective

if and else are two of the most frequently used conditionals in C/C++, and they enable you to execute zero or one conditional statement among many such dependent conditional statements. We use them in the following ways:

if: This executes the body of bracketed code starting with statement1 if condition evaluates to true.

if (condition) {
    statement1;
    ...
}
if - else: This executes the body of bracketed code starting with statement1 if condition evaluates to true, or it executes the body of code starting with statement2 if condition evaluates to false. Note that only one of the bracketed code sections will ever be executed.

if (condition) {
    statement1;
    ...
}
else {
    statement2;
    ...
}
if - else if - else: In this structure, dependent statements are chained together and the  for each statement is only checked if all prior conditions in the chain are evaluated to false. Once a  evaluates to true, the bracketed code associated with that statement is executed and the program then skips to the end of the chain of statements and continues executing. If each condition in the chain evaluates to false, then the body of bracketed code in the else block at the end is executed.

if(first condition) {
    ...
}
else if(second condition) {
    ...
}
.
.
.
else if((n-1)'th condition) {
    ....
}
else {
    ...
}

Task

Given a positive integer denoting n, do the following:

If 1<=n<=9, then print the lowercase English word corresponding to the number (e.g., one for 1, two for 2, etc.).
If n>9, print Greater than 9.
Input Format

The first line contains a single integer denoting n.

Constraints
1<= n<= 10^9
Output Format

If 1<=n<=9, then print the lowercase English word corresponding to the number (e.g., one for 1, two for 2, etc.); otherwise, print Greater than 9 instead.

Sample Input

5
Sample Output

five
Sample Input #01

8
Sample Output #01

eight
Sample Input #02

44
Sample Output #02

Greater than 9

   Conditional Statements in C - solution code:

Given a number, if it is less than 10, you have to print it's English representation if it is less than 10 else print "Greater than 9". This can be solved using if-else loop.


#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {
    
    int n;
    scanf("%d", &n);
    if(n == 1) {
        printf("one");
    }
    else if(n == 2) {
        printf("two");
    }
    else if(n == 3) {
        printf("three");
    }
    else if(n == 4) {
        printf("four");
    }
    else if(n == 5) {
        printf("five");
    }
    else if(n == 6) {
        printf("six");
    }
    else if(n == 7) {
        printf("seven");
    }
    else if(n == 8) {
        printf("eight");
    }
    else if(n == 9) {
        printf("nine");
    }
    else {
        printf("Greater than 9");
    }
    
    return 0;

}


The code can be made short by using arrays let me show you how,

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {

    int n;
    char* a[10] = {"Greater than 9", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine"};

    scanf("%d", &n);

    if(n > 9) 
       puts(a[0]);
  
    else 
        puts(a[n]);
    

    return 0;

}
I hope it added valuable knowledge to you.
Feel free to share your doubts in the comment section below.
See you next time.