First ask the user to enter the number to be searched. Store it inside a variable called key.
Now start comparing key value with a[i] using a for loop. If the key is found in the array, then assign 1 to flag orelse flag will be 0.

Depending upon the value of flag, print the message.

If flag is 1, search is successful.
If flag is 0, search failed. in the sense, the number is not present in the given array.

Video Tutorial: Linear Search: C

Full Source Code

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#include < stdio.h >
#include < conio.h >
 
void main()
{
	int a[20], key, N, flag = 0, i;
	clrscr();
 
	printf("Enter array limit\n");
	scanf("%d", &N);
 
	printf("Enter %d elements\n", N);
	for(i=0; i < N; i++)
	 scanf("%d",&a[i]);
 
	printf("Enter the number to be searched\n");
	scanf("%d", &key);
 
	for( i=0; i < N; i++)
	 if( a[i] == key )
	 {
		flag = 1;
		break;
	 }
 
	if(flag)
		printf("\nSearch Successful\n");
	else
		printf("\nSearch Failed\n");
	getch();
}

Output
Enter array limit
5
Enter 5 elements
77
0
13
9
55
Enter the number to be searched
9
Search Successful

Related posts:

1. Find Biggest In An Array, Without Sorting It: C
2. Bubble Sort: C
3. Find First and Second Biggest In An Array, Without Sorting It: C

Now check if this assumption is correct, if not, swap the values.

 
 if( sbig > fbig )
 {
    temp = sbig;
    sbig = fbig;
    fbig = temp;
 }

Now since we have already checked with a[0] and a[1], we start the comparison from a[2], till N.

for(i=2; i < n ; i++)
	 if(a[i] > fbig)
	 {
		sbig = fbig;
		fbig = a[i];
	 }
	 else if(a[i] > sbig)
		sbig = a[i];

Now, if a[i] contains a number which is bigger than fbig, we transfer the value of a[i] to fbig and the value present in fbig to sbig;
If the value of a[i] is not bigger than fbig, then we check it with sbig. If a[i] is bigger than sbig, then we assign the value of a[i] to sbig.

This way, at the end of the loop fbig will have the first biggest and sbig will have the second biggest element/number in the array.

Video Tutorial: Find First and Second Biggest In An Array, Without Sorting It: C

Full source code

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#include < stdio.h >
#include < conio.h >
 
void main()
{
	int a[20], N, i, fbig, sbig, temp;
	clrscr();
 
	printf("Enter array limit\n");
	scanf("%d", &N);
 
	printf("Enter %d array elements\n", N);
	for(i=0; i < n ; i++)
	 scanf("%d", &a[i]);
 
	fbig = a[0];
	sbig = a[1];
 
	if( sbig > fbig )
	{
	   temp = sbig;
	   sbig = fbig;
	   fbig = temp;
	}
 
	for(i=2; i < n ; i++)
	 if(a[i] > fbig)
	 {
		sbig = fbig;
		fbig = a[i];
	 }
	 else if(a[i] > sbig)
		sbig = a[i];
 
	 printf("First Big is %d and Second big is %d", fbig, sbig);
 
	 getch();
 
}

Output:
Enter array limit
6
Enter 6 array elements
99
108
777
723
786
999
First Big is 999 and Second big is 786

Related posts:

1. Find Biggest In An Array, Without Sorting It: C
2. Find Biggest of 2 Numbers: C++
3. Find Biggest of 3 Numbers: C++

Video Tutorial: Biggest In An Array, Without Sorting It: C

Note that, the previous value inside the variable big is discarded.

Full source code

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#include < stdio.h >
#include < conio.h >
 
void main()
{
	int big, a[20], N, pos, i;
	clrscr();
 
	printf("Enter the array size\n");
	scanf("%d", &N);
 
	printf("Enter %d elements\n", N);
	for(i=0; i<n ; i++)
	 scanf("%d", &a[i]);
 
	big = a[0];
	pos = 0;
 
	for(i=1; i<N; i++)
	 if(a[i] > big)
	 {
	  big = a[i];
	  pos = i+1;
 
	 }
 
	 printf("Big is %d and its position is %d", big, pos);
 
	 getch();
}
 
</n>

Output
Enter the array size
5
Enter 5 elements
2
0
100
108
55
Big is 108 and its position is 4

Related posts:

1. Find Biggest of 2 Numbers: C++
2. Find Biggest of 3 Numbers: C++
3. Biggest of Two Numbers Using Ternary Operator: C++

In bubble sort, a[0] is compared with a[1]. a[1] with a[2], a[2] with a[3] .. so on
if a[0] is greater than a[1], the values are swapped. This logic continues till the end of the array.

At the end of each iteration, 1 element gets sorted.

If the array size is 5, then we will have 4 iterations. i.e., if size is N, then we will have N-1 iterations.

Bubble Sort: Full source code for 5 elements

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#include < stdio.h >
#include < conio.h >
 
void main()
{
	int a[5] = { 15, 0, -2, 8, 3 };
	int i, j, temp;
	clrscr();
 
	printf("Array elements are..\n");
	for( i=0; i&lt;5; i++)
	 printf("\n%d", a[i]);
 
	for(j=3; j>=0; j--)
	 for(i=0; i< =j; i++)
	  if( a[i] > a[i+1] )
	  {
	    temp = a[i];
	    a[i] = a[i+1];
	    a[i+1] = temp;
	  }
 
	printf("\nafter sorting..\n");
	for(i=0; i&lt;5; i++)
	 printf("\n%d", a[i]);
 
	getch();
}

Output:
Array elements are..
15
0
-2
8
3

after sorting..
-2
0
3
8
15

Video Tutorial: Bubble Sort: C

Bubble Sort: Full source code for N elements

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#include < stdio.h >
#include < conio.h >
 
void main()
{
	int a[20];
	int i, j, temp, N;
	clrscr();
 
	printf("\nEnter array size\n");
	scanf("%d", &N);
 
	printf("Enter %d elements\n", N);
	for(i=0; i<n ; i++)
	 scanf("\n%d", &a[i]);
 
	printf("Array elements are..\n");
	for( i=0; i<N; i++)
	 printf("\n%d", a[i]);
 
	for(j=N-2; j>=0; j--)
	 for(i=0; i< =j; i++)
	  if( a[i] > a[i+1] )
	  {
	    temp = a[i];
	    a[i] = a[i+1];
	    a[i+1] = temp;
	  }
 
	printf("\nafter sorting..\n");
	for(i=0; i</n><n ; i++)
	 printf("\n%d", a[i]);
 
	getch();
}

Output:
Enter array size
10
Enter 10 elements
20
19
18
17
16
15
14
13
12
11

Array elements are..
20
19
18
17
16
15
14
13
12
11

after sorting..
11
12
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18
19
20

Related posts:

1. Selection Sort of N numbers: C
2. Array Basics in C++ : Find Sum
3. (Basic) Find Sum Using Dynamic Memory Allocation: C++

In selection sort, first we select one element and compare it with the rest of the elements. If the compared element is smaller than the selected element, then those numbers are sorted. This approach is continued till the last element in the array index.

The number of iterations is 1 less than the array size.
i.e., if the user enters array size as 5, then there will be 4 iterations.

Below is an example:
Lets assume, user has entered the array size as 5.
and entered 13, 0, -2, 8, 3 as array elements.


First Iteration:

In first iteration, a[0] is selected. and is compared with a[1], a[2], a[3] and a[4]. and is swapped with the least value in the array.

Second Iteration:

In the second iteration, we leave off a[0] and select a[1] and compare it with the rest of the elements in the array i.e., a[2], a[3] and a[4]. and swap the least value with a[1]

Third Iteration:

In third iteration, we select a[2] and compare it with a[3], a[4] and swap the least value to a[2]


Forth Iteration:

In forth iteration, we select a[3] and compare it with a[4] and swap the least value.

At the end of 4 iterations all array elements are sorted.

Video Tutorial: Selection Sort of N numbers: C

Since the array index starts from 0, and we need not compare the last element in the array with any other elements, we iterate the outer loop from 0 to N-2.
We iterate the inner loop from i+1 till N-1.

Full Source Code

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#include < stdio.h >
#include < conio.h >
 
void main()
{
	int a[20], i, j, N, temp;
	clrscr();
 
	printf("Enter the size of the array\n");
	scanf("%d", &N);
 
	printf("Enter %d elements\n", N);
	for(i=0; i<n ; i++)
	 scanf("%d", &a[i]);
 
	for( i=0; i<=N-2; i++ )
	 for( j=i+1; j<=N-1; j++)
	  if(a[i] > a[j])
	  {
	     temp = a[i];
	     a[i] = a[j];
	     a[j] = temp;
	  }
 
         printf("\nSorted Array:\n");
	  for(i=0; i</n><n ; i++)
	   printf("\n%d", a[i]);
 
	   getch();
 
}

Output:
Enter the size of the array
5
Enter 5 elements
13
0
-2
8
3
Sorted Array:
-2
0
3
8
13

Related posts:

1. Array Basics in C++ : Find Sum

In mathematics, the factorial of a non-negative integer n, denoted by n!, is the product of all positive integers less than or equal to n.

Ex:
if user enters 5, then the factorial is 1 * 2 * 3 * 4 * 5 i.e., factorial = 120
This logic must be handled with a c program.

Full Source Code

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#include < stdio.h >
#include < conio.h >
 
void main()
{
	int num, i, fact = 1;
	clrscr();
 
	printf("Enter a number\n");
	scanf("%d", &num);
 
	for( i = 1; i < = num; i++ )
	 fact = fact * i;                // fact *= i;
 
	printf("\nFactorial of %d is %d", num, fact);
 
	getch();
 
}

Here the for loop stars from 1 and not 0. As anything multiplied by 0 would also be zero.

Note: Factorial of 0 is 1. i.e., 0! = 1

fact *= i; is the compact representation of fact = fact * i;

Video Tutorial: Factorial in c

Output:
Enter a number
5
Factorial of 5 is 120

Also Watch/Read: Find the Factorial of a Number: C++

Related posts:

1. Find the Factorial of a Number: C++
2. Find Given Number Is Prime or Not: C
3. Find Sum of Digits In A Given Number: C

In this program we assign variable sum = 0 to avoid garbage values in sum before the calculation, which would result in wrong output.
We store the user entered value in num.

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 while( num )
  {
    rem = num % 10;
    sum = sum + rem;
    num = num / 10;
  }

Here the loop executes until the value of num is zero.

If user enters 123, we apply the modulus to get the individual values.
Ex:
123 % 10 = 3
12 % 10 = 2
1 % 10 = 1

We get 123, 12 and 1 by dividing the original value by 10.
Ex:
123 user entered value.
123 / 10 = 12
12 / 10 = 1

Now the sum.

sum = sum + rem;

3 = 0 + 3
5 = 3 + 2
6 = 5 + 1

So the sum of digits in the given integer number is 6.

Video Tutorial: Find Sum of Digits In A Given Number: C

Full Free Source code

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#include < stdio.h >
#include < conio.h >
 
void main()
{
	int num, rem, sum = 0;
	clrscr();
 
	printf("Enter an integer number\n");
	scanf("%d", &num);
 
	while(num)
	{
	  rem = num % 10;
	  sum = sum + rem;
	  num = num / 10;
	}
 
	printf("\nThe sum of digits is %d", sum);
	getch();
}

Output:
Enter an integer number
123
The sum of digits is 6

Related posts:

1. Reverse Given Number And Check For Palindrome: C++
2. Find Given Number Is Prime or Not: C
3. Array Basics in C++ : Find Sum

A natural number greater than 1 that is not a prime number is called a composite number. For example, 7 is prime, as only 1 and 7 divide it, whereas 10 is composite, since it has the divisors 2 and 5 in addition to 1 and 10.

In this video tutorial we show you 3 methods of finding whether the user entered number is a prime number or not.

Since every number is divisible by 1, we start the division(in for loop) from 2.

Note: A number can not be divided(to get a whole number) by another number which is greater than itself.

First Method:
Since every number is divisible by 1, we start the division(in for loop) from 2 till one number less than the user entered value.
Example: If user entered 10. For loop starts from 2 to 9.
for(i=2; i<10; i++) or for(i=2; i< =9; i++)

But the draw back: If user enters 500, the loop must get executed from 2 till 499. i.e., 497 times.

Second Method:
Here we reduce the number of iterations in for loop.
i.e., we divide the user entered value by 2 and divide the user entered value from 2 till num / 2;
Example: If user entered 500, we start the division from 2 till 500/2 i.e., till 250 times.

Third Method:
Here we still reduce the number of iterations in the for loop.
i.e., we take the square root of the user entered number and divide the user entered number from 2 till square root of number.
Example: If user entered 500, we start the division from 2 till sqrt(500) i.e., till 22 times.

So this third method is most optimum, as it involves least number of looping.

Video Tutorial: Find Given Number Is Prime or Not: C

YouTube Link: http://www.youtube.com/watch?v=3MYx9PrpXEg [Watch the Video In Full Screen.]

Full Free Source Code:

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#include< stdio .h> #include< conio .h> #include< math .h>   void main() { int num, i, fact, inum; clrscr();   printf("Enter a number\n"); scanf("%d", &num);   inum = sqrt(num);   for(i=2; i<inum ; i++) if( num % i == 0 ) { fact = 0; break; } else fact = 1;   if(fact) printf("%d is Prime\n", num); else printf("%d is NOT prime\n", num);   getch(); }

Table of all prime numbers up to 1,000:

Related posts:

1. Find the Factorial of a Number: C++
2. Reverse Given Number And Check For Palindrome: C++
3. Find Position of Left most Vowel: JavaScript

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