Thread: Print 2nd largest value from array

Write a function to get the values in array but it souldnt get any negative values and print the second largest value from the values in the array?

Originally Posted by cssprasad

func()
{
int a[10];
int b;
int j;
clrscr();
for(int i=0;i<10;i++)
scanf("%d",&a[i]);
for(i=0;i<10;i++)
{
for(j=0;j<10;j++)
{
if(a[j]>a[j+1])
{
b=a[j];
a[j]=a[j+1];
a[j+1]=b;
}
}
printf("%d\t",a[i-2]);
}

sorry to say this i think its not right bcz its work as a swao my freind!

Originally Posted by cssprasad

func() { int a[10]; int b; int j; clrscr(); for(int i=0;i<10;i++) scanf("%d",&a[i]); for(i=0;i<10;i++) { for(j=0;j<10;j++) { if(a[j]>a[j+1]) { b=a[j]; a[j]=a[j+1]; a[j+1]=b; } } printf("%d\t",a[i-2]); }

//program to find second largest number from an array
#include<stdio.h>
#include<conio.h> void main() { int i,j,m,a[10]; clrscr(); printf("enter your ten numbers plz\n"); for(i=0;i<=9 && i>=0;i++) scanf("&#37;d",&a[i]); for(i=0;i<=9;i++) for(j=0;j<=9;j++) if(a[i]>a[j]) { m=a[i]; a[i]=a[j]; a[j]=m; } printf("\n\nthe second largest number in the array is=%d",a[1]);
getch(); }

Hai,
Here is the Query in SQL Server 2005, to get the Second highest Salary from a EMP Table.

select max(sal)<(select max(sal) from emp)


Regards,
Prashanth Chenna.

#include<stdio.h>
void main()
{
int i,j,n,a[20],first_max,second_max;
printf("Enter how many numbers");
scanf("%d",&n);
for(i=0;i<n;i++)
{
printf("Enter a positive number");
scanf("%d",&a[i]);
}

if(a[0]>a[1])
first_max=a[0];
second_max=a[1];
else
first_max=a[1];
second_max=a[0];

for(j=2;j<n;j++)
if(a[j]>second_max)
if(a[j]>first_max)
{
first_max=a[j];
second_max=first_max;
}
else
second_max=a[j];
printf("The second largest number is %d", second_max);
}

Originally Posted by sreenivasuluyc

#include<stdio.h>
void main()
{
int i,j,n,a[20],first_max,second_max;
printf("Enter how many numbers");
scanf("%d",&n);
for(i=0;i<n;i++)
{
printf("Enter a positive number");
scanf("%d",&a[i]);
}

if(a[0]>a[1])
first_max=a[0];
second_max=a[1];
else
first_max=a[1];
second_max=a[0];

for(j=2;j<n;j++)
if(a[j]>second_max)
if(a[j]>first_max)
{
first_max=a[j];
second_max=first_max;
}
else
second_max=a[j];
printf("The second largest number is %d", second_max);
}

this is definitely the best answer.. thank you

use simple concept......
swap the largest element at first position a[0]....second largest at second position at a[1].......smallest at last ...then print a[1]...
*******************************
#include<stdio.h>
#include<conio.h>
void main(){
int a[100],n,i,t;
clrscr();
printf("enter the value of n:\n");
scanf("%d",&n);
printf("enter the elements\n");
for(i=0;i<n;i++){
scanf("%d",&a[i]);}
for(i=0;i<n-1;i++){
if(a[i+1]>a[i]){t=a[i];
a[i]=a[i+1];
a[i+1]=t;}
}
printf("the second largest %d",a[1]);
getch();
}
*************************************
sumitsolution@gmail.com

sort the numbers in decending order.......so second element will be at a[1]......
*********************************************************
#include<stdio.h>
#include<conio.h>
void main(){
int a[100],n,i,t;
clrscr();
printf("enter the value of n:\n");
scanf("%d",&n);
printf("enter the elements\n");
for(i=0;i<n;i++){
scanf("%d",&a[i]);}
for(i=0;i<n-1;i++){
if(a[i+1]>a[i]){t=a[i];
a[i]=a[i+1];
a[i+1]=t;}
}
printf("the second largest %d",a[1]);
getch();
}

****************************************************
sumitsolution@gmail.com

#include<stdio.h>
#include<conio.h>
void main()
{
int a[15];
int i,j,n;
scanf("&#37;d",&n); /* size of the array */
for(i=0;i<n;i++)
scanf("%d",a[i]);
for(i=0;i<=-1;i++)
for(j=i;j<=n-1;j++)
if(a[i]>a[j])
a[i]^=a[j]^=a[i]^=a[j];

printf("second largest number: %d" ,a[n-2]);
getch();
}

Solution 1:

int arr[6] = {5,3,11,2,6,40};

int large,second_large,i;

large = second_large = arr[0];

for(i=1;i<6;i++)
{
if(arr[i] > second_large && arr[i] < large)
second_large = arr[i];

if(arr[i] > large)
{
second_large = large;
large = arr[i];
}
}
printf("\n Large = %d",large);
printf("\n Second Large = %d",second_large);


The time complexity is O(n).


Solution 2:
Do bubble sort; execute loop only two times. Bubble sort finds the Highest numbers first. If size of an array is n then index of second largest number is n-2 and Arr[n-2] will be the second largest number.

Time Complexity of this will be -
The first loop will be executed n times hence O(n)
Second loop will be executed n-1 times hence O(n-1)

Total complexity - n + n-1 = O(2n-1)


Solution 3:
1. Build a max heap - O(n)
2. Remove 2 elements - O(logn)

Total Complexity = O(nlogn)


Solution 3 is optimum.

here's a simple way to do it :P

Code:

int findmin(int array[], int max){
    int minNum = array[0];
    for(int i = 0; i < max; i++){
        if(array[i] < minNum){
            minNum = array[i];
        }
    }
    return minNum;
}

int findsecondhighest(int array[], int arraysize, int maxNum){
    int temp = findmin(array, arraysize);
    
    for(int i = 0; i < arraysize; i++){
        if(array[i] > temp && array[i] < maxNum){
            temp = array[i];
        }
    }
    return temp;
}

of course you'll need a findmax function too... so here's the code also :P

Code:

int findmax(int array[], int max){
    int maxNum = array[0];
    for(int i = 0; i < max; i++){
        if(array[i] > maxNum){
            maxNum = array[i];
        }
    }
    return maxNum;
}

this one is good!!