Skip to main content

Largest and smallest number from array in java

//File name--> MinMax.java
import java.io.*;
class MinMax{
 public static void main(String args[])throws Exception
 {
  BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
  int n,min=0,max=0;
  
  System.out.println("Enter number of element you want in array.");
  n=Integer.parseInt(br.readLine());
  int[] a=new int[n];
  
  for(int i=0;i < n;i++)
  {
   System.out.println("Enter elementa["+i+"]:");
   a[i]=Integer.parseInt(br.readLine());
  }
  min=max=a[0];
  System.out.print("Content of Array:");
  for(int i=0;i < n;i++)
  {
   System.out.print("\t"+a[i]);
   
   if(min>a[i])
    min=a[i];
    
   if(max < a[i])
    max=a[i];
  }
  System.out.println("\nLargest element from array is:"+max);
  System.out.println("Smallest element from array is:"+min);
 }
}

OUTPUT

javac MinMax.java

java MinMax
Enter number of element you want in array.
10
Enter elementa[0]:
-5
Enter elementa[1]:
11
Enter elementa[2]:
88
Enter elementa[3]:
77
Enter elementa[4]:
66
Enter elementa[5]:
55
Enter elementa[6]:
44
Enter elementa[7]:
33
Enter elementa[8]:
22
Enter elementa[9]:
11
Content of Array:       -5      11      88      77      66      55      44
33      22      11
Largest element from array is:88
Smallest element from array is:-5


Comments

  1. How to play on the sands casino - Shootercasino.com
    The main difference between a 샌즈카지노 slot machine and a video poker is that most of the slot machines, except 온카지노 for a few, 12bet have games like

    ReplyDelete

Post a Comment

Popular posts from this blog

MVT (Multiprogramming Variable Task) in C Programming

#include #include void main() { int i,os_m,nPage,total,pg[25]; clrscr(); printf("\nEnter total memory size:"); scanf("%d",&total); printf("\nEnter memory for OS:"); scanf("%d",&os_m); printf("\nEnter no. of pages:"); scanf("%d",&nPage); for(i=0;i =pg[i]) { printf("\n Allocate page %d",i+1); total=total-pg[i]; } else printf("\n page %d is not allocated due to insufficient memory.",i+1); } printf("\n External Fragmentation is:%d",total); getch(); } OUTPUT Enter total memory size:1024 Enter memory for OS:256 Enter no. of pages:4 Enter size of page[1]:128 Enter size of page[2]:512 Enter size of page[3]:64 Enter size of page[4]:512 Allocate page 1 Allocate page 2 Allocate page 3 page 4 is not allocated due to insufficient memory. External Fragmentation is:64

First Come First Serve (FCFS) Page replacement algorithm in C Programming

#include #include int fsize; int frm[15]; void display(); void main() { int pg[100],nPage,i,j,pf=0,top=-1,temp,flag=0; clrscr(); printf("\n Enter frame size:"); scanf("%d",&fsize); printf("\n Enter number of pages:"); scanf("%d",&nPage); for(i=0;i OUTPUT Enter frame size:3 Enter number of pages:12 Enter page[1]:1 Enter page[2]:2 Enter page[3]:3 Enter page[4]:4 Enter page[5]:1 Enter page[6]:2 Enter page[7]:5 Enter page[8]:1 Enter page[9]:2 Enter page[10]:3 Enter page[11]:4 Enter page[12]:5 page | Frame content -------------------------------------- 1 | 1 -1 -1 2 | 1 2 -1 3 | 1 2 3 4 | 4 2 3 1 | 4 1 3 2 | 4 1 2 5 | 5 1 2 1 | 5 1 2 2 | 5 1 2 3 | 5 3 2 4 | 5 3 4 5 | 5 3 4 ---------------------------...

Deadlock Prevention using Banker’s Algorithm in C Programming

#include #include void main() { int allocated[15][15],max[15][15],need[15][15],avail[15],tres[15],work[15],flag[15]; int pno,rno,i,j,prc,count,t,total; count=0; clrscr(); printf("\n Enter number of process:"); scanf("%d",&pno); printf("\n Enter number of resources:"); scanf("%d",&rno); for(i=1;i OUTPUT Enter number of process:5 Enter number of resources:3 Enter total numbers of each resources:10 5 7 Enter Max resources for each process: for process 1:7 5 3 for process 2:3 2 2 for process 3:9 0 2 for process 4:2 2 2 for process 5:4 3 3 Enter allocated resources for each process: for process 1:0 1 0 for process 2:3 0 2 for process 3:3 0 2 for process 4:2 1 1 for process 5:0 0 2 available resources: 2 3 0 Allocated matrix Max need 0 1 0| 7 5 3| 7 4 3 3 0 2| 3 2 2| 0 2 0 3 0 2| 9 0 2| 6 0 0 2 1 1| ...