#include <stdio.h>
#include<conio.h>
main()
{
int a[20], i,n;
printf("ent how many elements u want to enter");
scanf("%d",&n);
printf("enter the elements");
for(i=1;i<=n;i++)
scanf("%d",&a[i]));
s=a[1];
for(i=2;i<=n;i++)
{
if
{
s>a[i]
s=a[i]
}
}
printf("the smallest element is %d",s);
}

hope this correct if it is wrong plz let me know

first the given list must be sorted using any of the 
sorting tecniques and then taking the kth element in that 
resultent array will give the answer 
                or
     another tecnique is using quick sort it will be easier 
and time complexity is also very less.sorry i dont have 
enough time to write the program here dontmind

sorting is not required in this case.u can still use same 
technique(not exactly) as mentioned in Cormen.an algorithm 
named "QuickSelect" which has an average complexity of      
O(nlong(n)).
Anyways to reduce the number of comparisions one can use 
tournament algorithm and for kth element we have to 
recursively go through the loosers list.

#include<stdio.h>
#include<conio.h>
int main()
{
    int a[10],i,j,n,temp;
    printf("\nEnter the number of elements int he array ");
    scanf("%d",&n);
    printf("\nEnter the elements ");
    for(i=0;i<n;i++)
    {
    scanf("%d",&a[i]);
    }
    for(i=0;i<n;i++)
    {
                    for(j=i+1;j<n;j++)
                    {
                                      if(a[i]>a[j])
                                      {
                                                   temp=a[i];
                                                   a[i]=a[j];
                                                   a[j]=temp;
                                      }
                    }
    }
   printf("\nThe smallest element is "); 
    for(i=0;i<n;i++)
    {
                    printf("%d\n",a[i]);
                   break;            
    }  
                    getch();
}

int kmin(int a[],int n,int k)
{
int i;
int j;
int l;
int u;
int x;
int t;
l=1;
u=n;
	while(l<u)
	{
	i=l;
	j=u;
	x=a[k];
		while((i<=k)&&(j>=k))
		{
		while(a[i]<x) i++;
		while(a[j]>x) j--;
		swap(a[i],a[j]);
		i++;
		j--;
		}
	if(j<k) l=i;
	if(i>k) u=j;
	}
return(a[k]);
}