## INDEX

### 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)
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.

1. not working

1. you try again

2. 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

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

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

3. cud u put some comments 2 make it easier 2 undrstand

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

6. thanks

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

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

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

10. lucky guy

11. only works if input array is already sorted

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

13. sorry it wont go infinite..

14. I think l < u would be correct

15. It is not working......

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

1. This comment has been removed by the author.

17. good work

18. for searching,array must be sorted

19. its realy gud thanx dear

20. Could always the input be in ascending order?

1. Yup it should always be in ascending order.

21. nice work bro

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

23. 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();
}

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

25. good one.....

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

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

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

29. Thank you dude its working thanks a lot

30. good

31. #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)
else
printf("The number is found.");

return 0;
}

32. what is the diff between linear search and binary search

1. 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.

33. #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 )

return 0;
}

34. its realy gud thanks

35. 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 :)

1. 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

36. can i get the logic of the program

37. #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();
}

38. 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.

39. 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 :)

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