// this program takes a sorted array as input and prints the unique numbers in that array as the output
// time complexity is of the order n  and also checks whether the array //is sorted or not
#include<stdio.h>
 
void main(char* argv)
{
	int a[100],b[100];
	int i=0,j=0,k=0,n=0;
	int flag=0;
	printf("Enter number of elements in the array \n");
	scanf("%d",&n);
	printf("Enter all the %d elements \n",n);
	for(i=0;i<n;i++)
		scanf("%d",&a[i]);
	// to check whether if the array is sorted or not
	for(i=0;i<n;i++)
		if(!(a[i]<=a[i+1]))
			flag=1;
	if(flag==1)
		printf("The given array is not sorted \n");
	else
	{
		for(i=0,j=i+1;j<=n;j++)
		{
			if(a[i]==a[j])	// we compare if a[i] and a[j] are equal or not if equal then contiue
				continue;
			b[k++]=a[i];// if they are not equal then b[k++] is equal to the repeating value 
			i=j;		// assign i==j to start from the end of the repeating elemnet
		}
		printf("The Unique numbers in the array are \n");
		for(i=0;i<k;i++)
			printf("%d \n",b[i]);	
	}
}