HackerRank Conditional Statements in c problem solution

In this HackerRank Conditional statements in c programming problem solution, 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 condition for each statement is only checked if all prior conditions in the chain are evaluated to false. Once a condition 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:

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

HackerRank Conditional Statements in c solution


HackerRank Conditional statements in c programming problem solution.

#include<stdio.h>
#include<string.h>
int main()
{
 int n;
 scanf("%d",&n);
  char *a[]={"zero","one","two","three","four","five","six","seven","eight","nine"};
  if(n<=9)
  {
     printf("%s",a[n]);
  }

 else
 {
     printf("Greater than 9");
 }
 return 0;
}


Second solution

#include <stdio.h>

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


Post a Comment

2 Comments

  1. Thank you for C solution but I have a question. how does the program know which one of the arrays to print? I mean, if the input is 5 then it returned "Five" but how does the program knows exactly?

    ReplyDelete
    Replies
    1. you just need to learn how if else statement exactly work

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