Printing consecutive entries as pairs from an array - c

Question

I have an array 'a' inserted by the user as:
printf("insert N");
scanf ("%d", &n);
printf ("insert group of numbers");
for(i=0; i<n; i++){
scanf("%d", &a[i]);
}
Then, from the original array, I have to make a sequence which will have all the entries of the original array but if an entry is coming multiple times consecutively, then instead of writing that number multiple times, only two numbers should be added to the sequence: first denoting the repeated number and the second denoting its frequency. Example:
Array : 1,3,4,1,1,1,1,6,7,1,1,1,1,1,1,9,3,2,5,6,1,2,1,1,1,1,1.
Sequence: 1,3,4,1,4,6,7,1,6,9,3,2,5,6,1,2,1,5
The first 3 numbers are written down as they are. Then '1' comes consecutively 4 times so we append 1,4 to the sequence and so on.
Actually i m confused how i can make i and j are on the same steps,
as far as i know i and j start from 0 then 1,2,3,4
but how can I make them start each in the same step so if a[i(that is 3 in the step)] = 10
so a[j( on step 3) is 10 too?
Thanks for the time! I appreciate!

Answer 1

Following code will do the job.
#include<stdio.h>
#include<stdlib.h>
void main()
{
int i,x,j=0,n,*a,*b;
printf("insert N");
scanf ("%d", &n);
a=(int *)malloc(n*sizeof(int));
b=(int *)malloc(n*sizeof(int));
printf ("insert group of numbers");
for(i=0; i<n; i++)
scanf("%d", &a[i]);
for( i=0;i<n;++i)
{
x=0;
while((i+1)<n&&a[i]==a[i+1])
{
++x;
++i;
}
if(x>1)
{
b[j++]=a[i];
b[j++]=x+1;
}
else b[j++]=a[i];
}
for(i=0;i<j;++i)
printf("%d ",b[i]);
}

Answer 2

Please check whether this code will help or not, insted of arr_1 elemets you could use the scanf and load the variables
#include<stdio.h> int main() {
int arr_1[]={1,3,4,1,1,1,1,6,7,1,1,1,1,1,1,9,3,2,5,6,1,2,1,1,1,1,1};
int arr_2[27];
int arr_len;
int i,j,cntr=0;
int flag = 0;
arr_len=sizeof(arr_1)/sizeof(int);
for(i=0,j=0;i<arr_len;i++)
{
if(arr_1[i] == arr_1[i+1])
{
if(flag == 0)
{
arr_2[j] = arr_1[i];
j++;
}
flag = 1;
cntr++;
}
else
{
if(cntr > 0)
{
arr_2[j] = cntr+1;
cntr=0;
flag = 0;
}
else
{
arr_2[j]=arr_1[i];
}
j++;
}
}
for(i=0;i<arr_len;i++)
printf("%d ",arr_1[i]);
printf("\n");
for(i=0;i<j;i++)
printf("%d ",arr_2[i]);
getch(); }

Answer 3

int count=0, last=0, j=0, n, i;
printf("insert N");
scanf ("%d", &n);
int b[n];
printf ("insert group of numbers");
for(i=0; i<n; i++){
scanf("%d", &a[i]);
if(a[i] == last){
count++;
}
else {
b[j] = last;
j++;
if(count != 0){
b[j] = count + 1;
j++;
count = 0;
}
last = a[i];
}
}
Rather than doing post processing as a few others have suggested you can process the input as it is received.
The if statement increments the count when a duplicate is encountered.
The else statement sets the second array to the last value (not current value) when the current value is different. If the last value was a duplicate, it adds the number of duplicates in the next if statement. After that, it updates the last variable.

Answer 4

Here is a demonstrative program that is based on using Variable Length Arrays (VLA). If you want you can change it such a way that instead of VLA(s) it used either arrays of fixed sizes or dynamically allocated arrays.
#include <stdio.h>
int main( void )
{
while ( 1 )
{
printf( "\nEnter array size: " );
size_t n = 0;
scanf( "%zu", &n );
if ( !n ) break;
int a[n];
printf( "Enter %zu elements of the array: ", n );
for ( size_t i = 0; i < n; i++ ) scanf( "%d", &a[i] );
printf( "\na[%zu] = { ", n );
for ( size_t i = 0; i < n; i++ ) printf( "%d, ", a[i] );
printf( "}\n" );
size_t m = 0;
size_t count;
for ( size_t i = 0; i < n; i += count )
{
count = 0;
while ( ++count + i < n && a[count + i] == a[i] );
m += count > 1 ? 2 : 1;
}
int b[m];
for ( size_t i = 0, j = 0; i < n; i += count )
{
count = 0;
while ( ++count + i < n && a[count + i] == a[i] );
b[j++] = a[i];
if ( count > 1 ) b[j++] = count;
}
printf( "b[%zu] = { ", m );
for ( size_t i = 0; i < m; i++ ) printf( "%d, ", b[i] );
printf( "}\n" );
}
}
The program might have the following output
Enter array size: 27
Enter 27 elements of the array: 1 3 4 1 1 1 1 6 7 1 1 1 1 1 1 9 3 2 5 6 1 2 1 1 1 1 1
a[27] = { 1, 3, 4, 1, 1, 1, 1, 6, 7, 1, 1, 1, 1, 1, 1, 9, 3, 2, 5, 6, 1, 2, 1, 1, 1, 1, 1, }
b[18] = { 1, 3, 4, 1, 4, 6, 7, 1, 6, 9, 3, 2, 5, 6, 1, 2, 1, 5, }
Enter array size: 6
Enter 6 elements of the array: 1 1 1 1 1 2
a[6] = { 1, 1, 1, 1, 1, 2, }
b[3] = { 1, 5, 2, }
Enter array size: 0