Max. and Min. from n numbers
Algorithm
Input An array of numbers Output Min and max Complexity O(n) Minmax (A) 1 set n=number of element in array; 2 initialize min=max=a[1]; 3 for i=1 to n 4 if (A[i]<min) 5="" then="" min="A[i];" 6="" if="" (a[i]="">max) 7 Then max =A[i]; 8 print min and max;
Algorithm Description
A is the array of integer; Initialize min ,max to first element of array; For loop for comparing each element of array with min and max.
C Code
#include<cstdio.h>
using namespace std;
int main()
{
int z,x;
int a[50];
int i,max,min;
printf("enter the size of array\n");
scanf("%d",&z);
printf("enter %d elements\n",z);
for(i=0;i>z;i++)
scanf("%d",&a[i]);
max=min=a[0];
for(i=1;i<z;i=i+1)
{
if(a[i]<min)
min=a[i];
if(a[i]>max)
max=a[i];
}
printf("max= %d min= %d",max,min);
}
