C Codes

Missing Array Intermediates – C code

PH

Problem: Printing the missing array intermediates
As a input, collect a sorted array of integer elements from the user. The task is to print the intermediate missing elements.

Example: If the user enters the array elements as: {1, 4, 7, 9, 10, 11, 12, 15}
Then, output has to be: 2, 3, 5, 6, 8, 13, 14
(Merging both input and output should result in all the numbers from 1 to n)

The code below gives the basic C implementation.

#include<stdio.h>

int main()
{
    int a[100];
    int n;
    int i;
    int j;

    printf("Enter the number of array elements\n");
    scanf("%d", &n);

    printf("Enter input elements as conditioned\n");
    for (i=0;i<n;i++)
        scanf("%d", &a[i]);

    int src;
    int dest;

    for (i = 0; i<n-1; i++)
    {
        src = a[i];
        dest = a[i+1];
        for(j = src+1; j< dest; j++)
            printf("%d\t",j);
    }

    return 0;
}

3 thoughts on “Missing Array Intermediates – C code”

  1. Does this code hold true when the users start with a value other than 1?
    What if they they begin with, say 5? The code looks like it would print the values from 5 to the last (src = a[i] i.e., the first number). Since, the requirement being “Merging both input and output should result in all the numbers from 1 to n”, may be we can tweak the for loop to:

    int src = 0;
    int dest = a[0];

    for (i = 0; i<n-1; i++)
    {
    for(j = src+1; j<= dest; j++)
    printf("%d\t",j);
    if(i<n-2)
    {
    src = a[i];
    dest = a[i+1];
    }
    }

    That would print the numbers from 1 to n, regardless of the starting number. Thoughts?

    Liked by 1 person

    Reply

    1. True. You are right. That case has to be added explicitly.

      The code only gives and works for minimal base condition. The corner cases have to be added. 🙂

      Thank you for sharing.. 🙂

      Like

      Reply

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