BINARY SEARCH USING C PROGRAM





1. Write a simple code for binary search in c programming language
2. Wap a c program to search an element in an array using binary search

#include<stdio.h>
int main(){

    int a[10],i,n,m,c=0,l,u,mid;

    printf("Enter the size of an array: ");
    scanf("%d",&n);

    printf("Enter the elements in ascending order: ");
    for(i=0;i<n;i++){
         scanf("%d",&a[i]);
    }

    printf("Enter the number to be search: ");
    scanf("%d",&m);

    l=0,u=n-1;
    while(l<=u){
         mid=(l+u)/2;
         if(m==a[mid]){
             c=1;
             break;
         }
         else if(m<a[mid]){
             u=mid-1;
         }
         else
             l=mid+1;
    }
    if(c==0)
         printf("The number is not found.");
    else
         printf("The number is found.");

    return 0;
}

Sample output:
Enter the size of an array: 5
Enter the elements in ascending order: 4 7 8 11 21
Enter the number to be search: 11
The number is found.





54 comments:

Anonymous said...

not working

Anonymous said...

Hello friend ur program is working sucessfully. Thanks for it.

Anonymous said...

cud u put some comments 2 make it easier 2 undrstand

Unknown said...

thanx bro!!! u got a good hand in C!!!!!!!!

Anonymous said...

please put some comments

Anonymous said...

thanks

priya said...

hey i need a program for graphical representation of binary search in c

Anonymous said...

hey can i get a program of queue in c using graphic.c

Unknown said...

Good Job

Anonymous said...

it saves our time ....... thanx

Anonymous said...

lucky guy

Anonymous said...

only works if input array is already sorted

Anonymous said...

seocndly ..for m<a[mid]){ u=mid-1; the loop will go infinte

Anonymous said...

sorry it wont go infinite..

Anonymous said...

I think l < u would be correct

Anonymous said...

It is not working......

Anonymous said...

can show us it's algorithm as well as flowchart.

Anonymous said...

good work

Anonymous said...

for searching,array must be sorted

rimmy said...

its realy gud thanx dear

Vijay said...

Could always the input be in ascending order?

Anonymous said...

nice work bro

ksp said...

thank you very much. i feel very easy in using 'c' in this blog. so simple and power

prasad..

Unknown said...

simple huffman algorithm
#include
#include
#include
void main()
{
struct huff
{
long int freq,code;
char data[10];
}h[50];
long int n,n1=100,i,j,n2=90,temp;
char tempc[10];
clrscr();
printf("\t\t\tHUFFMAN ALGORITHM\n\n");
printf("Enter the number of character");
scanf("%ld",&n);
for(i=0;i<n;i++)
{
printf("Enter the character::");
scanf("%s",h[i].data);
printf("Enter the frequency for character %s::", h[i].data);
scanf("%ld",&h[i].freq);
}
for(i=0;i<n-1;i++)
{
for(j=1;j<n;j++)
{
if(h[i].freq<h[j].freq)
{
temp=h[i].freq;
h[i].freq=h[j].freq;
h[j].freq=temp;
strcpy(tempc,h[i].data);
strcpy(h[i].data,h[j].data);
strcpy(h[j].data,tempc);
}
}
}
h[0].code=10;
h[1].code=11;
for(i=2;i<n;i++)
{
if(i%2==0)
{
h[i].code=h[i-2].code+n1;
n1=n1*10;
}
else
{
h[i].code=h[i-2].code+n2;
n2=n2*10;
}
}
printf("Huffman code");
printf("\nchar\tfreq\tcode");
for(i=0;i<n;i++)
{
printf("\n%s\t%ld\t%ld\n",h[i].data,h[i].freq,h[i].code);
}
getch();
}

Rohit said...

cud u pls do d programs fo findin hcf & lcm ?

debojyoti said...

good one.....

Anonymous said...

Yup it should always be in ascending order.

mahesh said...

thanx ....but pls put commands....sir

sandeep kumar said...

so many closed braces but only one open brace...how ca it work!!!

Antone said...

Thankzzzz....!!!!!
Itzz really very useful 2 meeee......!!!!!!

Anonymous said...

you try again

Anonymous said...

Thank you dude its working thanks a lot

Anonymous said...

good

anil chowdary...... said...

#include
int main(){

int a[10],i,n,m,c=0,l,u,mid;
int j;

printf("Enter the size of an array: ");
scanf("%d",&n);

printf("Enter the elements in ascending order: ");
for(i=0;i<n;i++){
scanf("%d",&a[i]);
}

printf("Enter the number to be search: ");
scanf("%d",&m);

for(j=0;j<n;j++)
{
if(m==a[j])
{
c=1;
}

}


if(c==0)
printf("The number is not found.");
else
printf("The number is found.");

return 0;
}

Anonymous said...

what is the diff between linear search and binary search

Anonymous said...

#include

int main()
{
int c, first, last, middle, n, search, array[100];

printf("Enter number of elements\n");
scanf("%d",&n);

printf("Enter %d integers\n", n);

for ( c = 0 ; c < n ; c++ )
scanf("%d",&array[c]);

printf("Enter value to find\n");
scanf("%d",&search);

first = 0;
last = n - 1;
middle = (first+last)/2;

while( first <= last )
{
if ( array[middle] < search )
first = middle + 1;
else if ( array[middle] == search )
{
printf("%d found at location %d.\n", search, middle+1);
break;
}
else
last = middle - 1;

middle = (first + last)/2;
}
if ( first > last )
printf("Not found! %d is not present in the list.\n", search);

return 0;
}

laxminarayana said...

its realy gud thanks

Unknown said...

i am getting it a postion before. ex 1 2 3 4 and i choose to search 3 it show the postion of 3 as 2. help :)

Unknown said...

can i get the logic of the program

Prashant said...

#include
#include
int key,A[5],m,i,lb,ub;
int binary(key,A[],lb,ub)
{
if(lb>ub)
{
printf("\n Search failed");
return;
}
m=lb+ub/2;
if(key==A[m])
{
printf("\n Match Found and value is %d",key);
return;
}
elseif(key<A[m])
{
ub=m-1;
binary(key,A[],lb,ub);
}
else
{
lb=m+1;
binary(key,A[],lb,ub);
}
}
void main()
{
clrscr();
for(i=0;i<5;i++)
{
printf("\n Enter the value in array:\n");
scanf("%d",&A[i])
}
printf("\n Enter the value of item to search:\n");
scanf("%d",&key);
lb=A[0];
ub=A[5];
binary(key,A[],lb,ub);
getch();
}

Unknown said...

plz give me the ans for this question


Given a number.You have to find the binary equivalent of that number first. Your task is to rearrange the 0's and 1's in the binary equivalent found to all the combinations and print all the numbers which form a co-prime with the digit entered.Print nil if no co-prime is found.

Unknown said...

This seems to go infinite unless specified differently, but I may very much be wrong, here is what I am thinking though,

if (m > n || m < 0 )
return 0;

Unless you want to allow the array to contain negative numbers, in that case I assume one would have to put in an algorithm to initiate the smallest number in the array to a variable, then insert that one instead of 0 in the above if statement, one way would be to insert below inside the for loop when scanning for a[i]:

// once there is exactly 2 elements in the array a[], then initialize min variable as the smallest number
if (i == 1)
min = a[0];
// Now one can compare each iteration to see if a number in the array a[] is smaller
if (a[i] < min)
min = a[i];

So after this, one can use the code above, now in this way:

if (m > n || m < min )
return 0;

But hey, I am actually very very new to this, so please let me know if this even works at all, I am currently learning how to implement binary search, so hopefully my thoughts here are in the ball park at least :)

Unknown said...

linear search starts from one value, like the smallest, then iterates through each value to see if it finds what it is searching for, so for example, if looking for the number 19 from 0-40, linear search would go:
1, 2, 3, 4, 5 etc til it reached 19

binary search instead, splits the value in 2, until it finds it, so it will do:
is the number in the middle of 0-40 ? (20, in other words)
No, then is the value more than, or less than 20?
If more, then check the middle between 20-40
If less, check 0-20
then repeat

So in this case, that would give us:
10 (0-20)
15 (10-20)
17(or 18) (15-20)
19, found it!

The reason this is efficient is that it makes it way faster to search through numbers
When the sample size is small, it might not make a difference, but when searching through really big numbers, it will show

For example, find 19000 out of 0-1000000 (one milion)
500000 (0-1000000)
250000 (0-500000)
125000 (0-250000)
62500 (0-125000)
31250 (0-62500)
15625 (0-31250)
23437 (15625-31250)
19531 (15625-23437)
17578 (15625-19531)
18555 (17578-19531)
19043 (18555-19531)
18799 (18555-19043)
18921 (18799-19043)
18982 (18921-19043)
19012 (18982-19043)
18997 (18982-19012)
19004 (18997-19012)
19000 (18997-19004), Found it!

So in this example it takes 19 steps out of a 1000000 (Milion) numbers
With linear search it would have taken 19000, in this case
I think that out of a 1000000000 (Bilion) numbers, it would only require 4-5 more steps to 19.

So basically, it stays pretty quick even if the numbers increase dramatically.

Unknown said...

not working for odd num of array element
ex a[15]={1,3,5,7,8,10,12,13,20,34,42,56,61,71,85}
for above array it not working

riya said...

Define a binary search.illustrate the binary search algorithm with graphical representation

riya said...

Please any body to can answer fast

Unknown said...
This comment has been removed by the author.
Unknown said...

any one can explain me this program please

Unknown said...

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along with theirs solution..https://github.com/harsit007/Programming_Made_Easy_With_Harsit

dodda venkatareddy said...

since the array a[i] is declared as integer, you need to give the array size with in its rage. That's it

dodda venkatareddy said...

since the array index starts with zero. so, a[o]=1, a[1]=2, a[2]=3, a[3]=4. If your printing 'i', it is an index number. then for element 3 the position or index number is 2

Javeria Fatima said...

ll the brackets ain't closed correctly!will this work?

Unknown said...

venkatareddy dodda how can i reach i want to talk

Unknown said...

u