HackerRank For Loop in C programming problem solution

In this HackerRank For loop in c programming problem solution, In this challenge, you will learn the usage of the for loop, which is a programming language statement which allows code to be executed until a terminal condition is met. They can even repeat forever if the terminal condition is never met.

The syntax for the for loop is:

for ( <expression_1> ; <expression_2> ; <expression_3> )

    <statement>

expression_1 is used for intializing variables which are generally used for controlling the 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.

The following loop initializes i to 0, tests that i is less than 10, and increments i at every iteration. It will execute 10 times.

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

    ...

}

Task

For each integer [a,b] in the interval  (given as input) :

  1. If 1 <= n <= 9, then print the English representation of it in lowercase. That is "one" for 1, "two" for 2, and so on.
  2. Else if n > 9 and it is an even number, then print "even".
  3. Else if n > 9 and it is an odd number, then print "odd".

HackerRank For Loop in C programming solution


HackerRank For loop in c programming problem solution.

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

int main() 
{
    int a, b;
    scanf("%d\n%d", &a, &b);
    
    char labels[11][6] = {"one", "two", "three", "four", "five", "six", "seven", "eight", "nine", "even", "odd"};
    int labels_index;
      for (int i=a; i<=b; i++) {
        labels_index = i <= 9 ? i - 1 : 9 + i % 2;
        printf("%s\n", labels[labels_index]);
    }

    return 0;
}


Second solution

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

int main() 
{
    int a, b;
    scanf("%d\n%d", &a, &b);
     // Complete the code.
    for (int i=a; i<b+1; i++) {
        switch(i) {
            case 1: printf("one\n"); break;
            case 2: printf("two\n"); break;
            case 3: printf("three\n"); break;
            case 4: printf("four\n"); break;
            case 5: printf("five\n"); break;
            case 6: printf("six\n"); break;
            case 7: printf("seven\n"); break;
            case 8: printf("eight\n"); break;
            case 9: printf("nine\n"); break;
            default:
                if (i % 2)
                    printf("odd\n");
                else
                    printf("even\n");
        }
    }

    return 0;
}


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3 Comments

  1. anyone can explain this problem for me guys pls pls

    ReplyDelete
  2. what if the b value is smaller than value a ? the loop won't run isn't it?

    ReplyDelete