Required knowledge

Logic to reverse the order of words in a given string

1. Read a sentence from user in any variable say in string.
2. Find a word from the end of string. Learn how to find words in a string.
3. Append this word to reverseString (where reverseString holds the reverse order of words of the original string).
4. Repeat step 2-3 till the beginning of string.

Program to reverse the order of words in a given string

/**
 * C program to reverse order of words in a string
 */

#include <stdio.h>
#include <string.h>

int main()
{
    char string[100], reverse[100];
    int len, i, index, wordStart, wordEnd;

    printf("Enter any string: ");
    gets(string);

    len   = strlen(string);
    index = 0;

    // Start checking of words from the end of string
    wordStart = len - 1;
    wordEnd  = len - 1;

    while(wordStart > 0)
    {
        // If a word is found
        if(string[wordStart] == '')
        {
            // Add the word to the reverse string
            i = wordStart + 1;
            while(i <= wordEnd)
            {
                reverse[index] = string[i];

                i++;
                index++;
            }
            reverse[index++] = '';

            wordEnd = wordStart - 1;
        }

        wordStart--;
    }

    // Finally add the last word
    for(i=0; i<=wordEnd; i++)
    {
        reverse[index] = string[i];
        index++;
    }
    reverse[index] = '\0'; // Adds a NULL character at the end of string

    printf("Original string \n%s\n\n", string);
    printf("Reverse ordered words \n%s", reverse);


    return 0;
}

12345
23454
34543
45432
54321

Required knowledge

Logic to print the given square number pattern

12345
23454
34543
45432
54321

PART 1

12345
2345-
345--
45---
5----

PART 2

-----
----4
---43
--432
-4321

1. The pattern consists of N rows (where N is the number of rows to be printed). Hence the outer loop formation is for(i=1; i<=N; i++).
2. To print the first part of the pattern run an inner loop from i to N (where i is the current row number). Hence the loop formation will be for(j=i; j<=N; j++). Inside this loop print the current value j.
3. To print the second part of the pattern. Initialize a variable k = j - 2. After that run another inner loop from 1 to i-1. Loop formation for this will be for(j=1; j<i; j++, k--). Inside this loop print the current value of k.

Program to print the given square number pattern

/**
 * C program to print given number pattern
 */

#include <stdio.h>

int main()
{
    int i, j, N;
    int k = 1;

    printf("Enter N: ");
    scanf("%d", &N);

    for(i=1; i<=N; i++)
    {
        // Prints first part of the pattern
        for(j=i; j<=N; j++)
        {
            printf("%d ", j);
        }

        // Prints second part of the pattern
        k = j - 2;
        for(j=1; j<i; j++, k--)
        {
            printf("%d ", k);
        }

        printf("\n");
    }

    return 0;
}

Screenshot

12345
2   4
3   3
4   2
54321

Required knowledge

Logic to print the given number pattern

12345
2   4
3   3
4   2
54321

12345
2---
3--
4-
5

-----
----4
---43
--432
-4321

1. The pattern consists of N rows (where N is the number of rows to be printed). Hence the outer loop formation is for(i=1; i<=N; i++).
2. To print the first part of the pattern run an inner loop from i to N (where i is the current row number). Hence the loop formation will be for(j=i; j<=N; j++). Inside this loop print the current value j.
3. To print the second part of the pattern. Initialize a variable k = j - 2. After that run another inner loop from 1 to i-1. Loop formation for this will be for(j=1; j<i; j++, k--). Inside this loop print the current value of k.
4. NOTE: Numbers only gets printed for first and last row or column otherwise spaces gets printed.

Program to print hollow square number pattern

/**
 * C program to print given number pattern
 */

#include <stdio.h>

int main()
{
    int i, j, N;
    int k = 1;

    printf("Enter N: ");
    scanf("%d", &N);

    for(i=1; i<=N; i++)
    {
        // Prints first part of the pattern
        for(j=i; j<=N; j++)
        {
            if(i==1 || i==N || j==i)
                printf("%d", j);
            else
                printf("");
        }

        // Prints second part of the pattern
        k = j - 2;
        for(j=1; j<i; j++, k--)
        {
            if(i==1 || i==N || j==i-1)
                printf("%d", k);
            else
                printf("");
        }

        printf("\n");
    }

    return 0;
}

12345
2   4
3   3
4   2
54321

Screenshot

1       1
 2     2
  3   3
   4 4
    5
   4 4
  3   3
 2     2
1       1

Required knowledge

Logic to print X number pattern

1       1
 2     2
  3   3
   4 4
    5
   4 4
  3   3
 2     2
1       1

  PART 1

1       1
 2     2
  3   3
   4 4
    5

  PART 2

   4 4
  3   3
 2     2
1       1

1       1
 2     2
  3   3
   4 4
    5

1. The pattern consists of total N rows (where N is the total number of rows * 2 - 1). Hence the first outer loop formation to iterate through rows will be for(i=1; i<=N; i++)
2. Now notice each row in the first part of the pattern. First spaces gets printed then an integer then some more spaces finally the integer. Hence we will use two inner loops to print spaces. You can hover on the above pattern to count the total number of spaces gets printed.
3. Notice the trailing spaces in the first part of pattern. You might have noticed that for each row total number of trailing spaces is i - 1 (Where i is the current row number). Hence to iterate through the trailing spaces the loop formation will be for(j=1; j<i; j++). Inside this loop print a single space.
4. After printing the trailing space an integer which is the current row number gets printed. Hence print the current value of i.
5. Now the center spaces. Center spaces are arranged in tricky fashion. Each row contains exactly (N - i) * 2 - 1 spaces. Hence the second inner loop formation to iterate through spaces is for(j=1; j<= (N - i) * 2 - 1; j++). Inside this loop print single spaces.
6. After the above loop print the value of i.

   4 4
  3   3
 2     2
1       1

1. The second part of the pattern consists of N - 1 rows. Hence the outer loop formation to iterate through rows is for(i=N-1; i>=1; i--). Now, I have used a descending order loop, since the numbers printed are in descending order.
2. Like first part of the pattern. Here also first trailing spaces gets printed then an integer then the central spaces and finally the same integer gets printed.
3. Have a close look to the trailing spaces. Each row contains exactly i - 1 spaces i.e. the first row contains 4 - 1 => 3 spaces, second contains 3 - 1 => 2 spaces and so on. Hence the first inner loop formation will be for(j=1; j<i; j++). Inside this loop print the single space.
4. After printing tailing spaces print the current value of i.
5. Now notice the central spaces carefully. Each row contains exactly (N - i) * 2 - 1 central spaces. Hence the loop formation for central spaces will be for(j=1; j<=((N - i) * 2 - 1); j++). Inside this loop print single space.
6. Again after central loop print the current value of i.

Program to print X number pattern

/**
 * C program to print X number pattern
 */

#include <stdio.h>

int main()
{
    int i, j, N;

    printf("Enter N: ");
    scanf("%d", &N);

    //First part of the pattern
    for(i=1; i<=N; i++)
    {
        //Prints trailing spaces
        for(j=1; j<i; j++)
        {
            printf("");
        }

        printf("%d", i);

        //Prints central spacces
        for(j=1; j<=((N - i) * 2 - 1); j++)
        {
            printf("");
        }

        //Don't print for last row
        if(i != N)
            printf("%d", i);

        //Moves on to the next row
        printf("\n");
    }

    //Second part of the pattern
    for(i=N-1; i>=1; i--)
    {
        //Prints trailing spaces
        for(j=1; j<i; j++)
        {
            printf("");
        }

        printf("%d", i);

        //Prints central spaces
        for(j=1; j<=((N - i ) * 2 - 1); j++)
        {
            printf("");
        }

        printf("%d", i);

        //Moves on to the next line
        printf("\n");
    }

    return 0;
}

1       1
 2     2
  3   3
   4 4
    5
   4 4
  3   3
 2     2
1       1

Screenshot

1
121
12321
1234321
123454321
1234321
12321
121
1

Required knowledge

Logic to print the given half diamond number pattern

1
121
12321
1234321
123454321
1234321
12321
121
1

1
121
12321
1234321
123454321

1234321
12321
121
1

1
121
12321
1234321
123454321

1. The given pattern consists of total N rows. Hence the outer loop formation to iterate through rows will be for(i=0; i<=N; i++).
2. Now, when you look carefully at the series in which each row gets printed. You will find that it contains two series.

   
   1
   12 
   123  
   1234   
   12345    
   

   
   
     1
      21
       321
        4231
   

   We will print these two pattern in two separate inner loops. These patterns are also explained separately in my two earlier posts
   • Number pattern 23
   • Number pattern 25
   1. The first part contains total i number of columns (where i is the current row number). Hence the first inner loop formation will be for(j=1; j<=i; j++). Inside this loop print the current value of j.
   2. The second part contains total i-1 number of columns each row. Hence the second inner loop formation will be for(j=i; j>=1; j--). Inside this loop print the current value of j.
   And you are done with the first upper half of the half diamond number pattern.

1234321
12321
121
1

1. The given pattern consists of total N - 1 number of rows. Hence the outer loop formation to iterate through rows will be for(i=N-1; i>=1; i--). I have used a decreasing order loop as the pattern is in decreasing order.
2. Now, as the upper part of the pattern. This can also be divided again in two parts.

   
   1234   
   123  
   12 
   1
   

   
       321
      21
     1
   
   

   We need two separate inner loops to print these.
   1. The first part contains total i columns (where i is the current row number). Hence the first inner loop formation will be for(j=1; j&lt=i; j++). Inside this loop print the current value of j. This part is also explained separately in my previous post.
   2. The second part contains total i - 1 columns. Hence the second inner loop formation will be for(j=i; j>=i; j--). Inside this loop print the current value of j. This pattern is also explained in detail in my previous number pattern post.

Program to print the given half diamond number pattern

/**
 * C program to print half diamond number pattern series
 */

#include <stdio.h>

int main()
{
    int i, j, N;

    printf("Enter rows: ");
    scanf("%d", &N);

    // Print the first upper half
    for(i=1; i<=N; i++)
    {
        for(j=1; j<=i; j++)
        {
            printf("%d", j);
        }
        for(j=i-1; j>=1; j--)
        {
            printf("%d", j);
        }

        printf("\n");
    }

    // Print the lower half of the pattern
    for(i=N-1; i>=1; i--)
    {
        for(j=1; j<=i; j++)
        {
            printf("%d", j);
        }
        for(j=i-1; j>=1; j--)
        {
            printf("%d", j);
        }

        printf("\n");
    }

    return 0;
}

Screenshot of half diamond number pattern

*
*1*
*121*
*12321*
*1234321*
*123454321*
*1234321*
*12321*
*121*
*1*
*

Required knowledge

Logic to print the given half diamond number pattern with star border

1
121
12321
1234321
123454321
1234321
12321
121
1

Program to print the given half diamond number pattern with star border

/**
 * C program to print the half diamond number pattern with star border
 */

#include <stdio.h>

int main()
{
    int i, j, N;

    printf("Enter rows: ");
    scanf("%d", &N);

    printf("*\n");
    // Print the first upper half
    for(i=1; i<=N; i++)
    {
        printf("*");
        for(j=1; j<=i; j++)
        {
            printf("%d", j);
        }

        for(j=i-1; j>=1; j--)
        {
            printf("%d", j);
        }
        printf("*");

        printf("\n");
    }

    // Print the lower half of the pattern
    for(i=N-1; i>=1; i--)
    {
        printf("*");
        for(j=1; j<=i; j++)
        {
            printf("%d", j);
        }

        for(j=i-1; j>=1; j--)
        {
            printf("%d", j);
        }
        printf("*");

        printf("\n");
    }
    printf("*");

    return 0;
}

Screenshot to print half diamond number pattern series with star border

Write a C program to create, initialize and demonstrate the use of a pointer variable. How to access values and addresses using a pointer variable in C programming.

Required knowledge

Basic C programming, Pointers

Basic of pointers

C programming is extremely powerful at low level. Pointer is one of its tool that provide such low level memory handling. Data in memory is organized as a sequence of bytes. Where each byte is accessed through its unique address. These addresses ranges from zero (0) to some positive integer. Pointers in C programming provides an efficient way to handle the low level memory activities.

Like other variables, pointers also allows to declare variables of pointer types. Now, what does that mean? Like other variables we can declare a pointer variable. But what type of data does pointer variable contains? Unlike other variables pointer holds memory address of some other variable. We will later see the use of these memory addresses and how pointers are a magic wand to the C programmers. Let us now get familiar with some pointer concepts.

Reading memory address of any variable

We know that each and every declared variable has a name, memory location and value. Name is the identifier name which we give during declaration of the variable. Values are the constants that the variable contains. For example - int num = 10;num is the variable name and 10 is the value stored in that variable. But what about the memory address?

In C programming the unary & (Address of) operator is used to get memory address of any variable. Address of operator when prefixed with any variable returns the actual memory address of that variable. Let us see a program to get actual memory address of variables.

Program to get memory address using address of operator

/**
 * C program to get memory address using address of operator
 */

#include <stdio.h>

int main()
{
    /* Simple declarations */
    char character = 'C';
    int integer = 1;
    float real = 10.4f;
    long long biginteger = 989898989ll;

    /* Print variable value with their memory address */
    printf("Value of character = %c, Address of character = %u\n", character, &character);
    printf("Value of integer = %d, Address of integer = %u\n", integer, &integer);
    printf("Value of real = %f, Address of real = %u\n", real, &real);
    printf("Value of biginteger = %lld, Address of biginteger = %u", biginteger, &biginteger);

    return 0;
}

Value of character = C, Address of character = 6356751
Value of integer = 1, Address of integer = 6356744
Value of real = 10.400000, Address of real = 6356740
Value of biginteger = 989898989, Address of biginteger = 6356728

Creating, initializing and using pointer variables

Pointers can handle many low level memory operations (including dynamic memory allocation). But before we get in deep of pointers let us first learn to declare a pointer variable. Like other variable declarations pointer also follow the same syntax -

Syntax to declare pointer variable

<data-type> * <variable-name>

Example of pointer declaration

int * integer_pointer;
float * float_ptr
char * charPtr;
long * lptr;

Program to create, initialize and use pointer variable

/**
 * C program to create, initialize and use pointers
 */

#include <stdio.h>

int main()
{
    int num = 10;
    int * ptr;

    /* Stores the address of num to pointer type */
    ptr = &num;

    printf("Address of num = %d\n", &num);
    printf("Value of num = %d\n", num);

    printf("Address of ptr = %d\n", &ptr);
    printf("Value of ptr = %d\n", ptr);
    printf("Value pointed by ptr = %d\n", *ptr);

    return 0;
}

Address of num = 6356748.
Value of num = 10
Address of ptr = 6356744
Value of ptr = 6356748
Value pointed by ptr = 10

Happy coding ;)

Write a C program to read two numbers from user and add them using pointers. How to find sum of two number using pointers in C programming. Program to perform arithmetic operations on number using pointers.

Input num1: 10
Input num2: 20

Sum = 30
Difference = -10
Product = 200
Quotient = 0

Required knowledge

Basic C programming, Pointers

Since the first days of writing programs in C. We have performed arithmetic operations so many times. In case you want a quick recap check out these links without pointers -

In previous post I explained how to store address of a variable in pointer variable. Let us now learn to work with the value stored in the pointer variable.

There are two common operators used with pointers -

1. & (Address of) operator - When prefixed with any variable returns the actual memory address of that variable.
2. * (Dereference) operator - When prefixed with any pointer variable it evaluates to the value stored at the address of a pointer variable.

Program to add two numbers using pointers

/**
 * C program to add two number using pointers
 */

#include <stdio.h>

int main()
{
    int num1, num2, sum;
    int *ptr1, *ptr2;

    ptr1 = &num1; // ptr1 stores the address of num1
    ptr2 = &num2; // ptr2 stores the address of num2

    printf("Enter any two numbers: ");
    scanf("%d%d", ptr1, ptr2);

    sum = *ptr1 + *ptr2;

    printf("Sum = %d", sum);

    return 0;
}

You have noticed that, I haven't used any & (address of) operator in the scanf() function. scanf() takes the actual memory address where the user input is to be stored. The pointer variable ptr1 and ptr2 contains the actual memory addresses of num1 and num2. Therefore we need not to prefix the & operator in scanf() function in case of pointers.

Now using this, it should be an easy workaround to compute all arithmetic operations using pointers. Let us write another program that performs all arithmetic operations using pointers.

Program to perform all arithmetic operations using pointers

/**
 * C program to perform all arithmetic operations using pointers
 */

#include <stdio.h>

int main()
{
    float num1, num2; // Normal variables
    float *ptr1, *ptr2; // Pointer variables

    float sum, diff, mult, div;

    ptr1 = &num1; // ptr1 stores the address of num1
    ptr2 = &num2; // ptr2 stores the address of num2

    /* User input through pointer */
    printf("Enter any two real numbers: ");
    scanf("%f%f", ptr1, ptr2);

    /* Perform arithmetic operation */
    sum  = (*ptr1) + (*ptr2);
    diff = (*ptr1) - (*ptr2);
    mult = (*ptr1) * (*ptr2);
    div  = (*ptr1) / (*ptr2);

    /* Print the results */
    printf("Sum = %.2f\n", sum);
    printf("Difference = %.2f\n", diff);
    printf("Product = %.2f\n", mult);
    printf("Quotient = %.2f\n", div);

    return 0;
}

Enter any two real numbers: 20
5
Sum = 25.00
Difference = 15.00
Product = 100.00
Quotient = 4.00

Happy coding ;)

Write a C program to swap two numbers using pointers and functions. How to swap two numbers using call by reference method. Logic to swap two number using pointers in C program.

Input num1: 10
Input num2: 20

Num1 = 20
Num2 = 10

Required knowledge

Basic C programming, Functions, Pointers

Logic to swap two numbers using call by reference

Swapping two numbers is simple and a fundamental thing. You need not to know any rocket science for swapping two numbers. Simple swapping can be achieved in three steps -

1. Copy the value of first number say num1 to some temporary variable say temp.
2. Copy the value of second number say num2 to the first number. Which is num1 = num2.
3. Copy back the value of first number stored in temp to second number. Which is num2 = temp.

Let us implement this logic using call by reference concept in functions.

Program to swap two numbers using call by reference

/**
 * C program to swap two number using call by reference
 */

#include <stdio.h>

/* Swap function declaration */
void swap(int * num1, int * num2);

int main()
{
    int num1, num2;

    /* Input numbers */
    printf("Enter two numbers: ");
    scanf("%d%d", &num1, &num2);

    /* Print original values of num1 and num2 */
    printf("Before swapping in main \n");
    printf("Value of num1 = %d \n", num1);
    printf("Value of num2 = %d \n\n", num2);

    /* Pass the addresses of num1 and num2 */
    swap(&num1, &num2);

    /* Print the swapped values of num1 and num2 */
    printf("After swapping in main \n");
    printf("Value of num1 = %d \n", num1);
    printf("Value of num2 = %d \n\n", num2);

    return 0;
}


/**
 * Function to swap two numbers
 */
void swap(int * num1, int * num2)
{
    int temp;

    // Copy the value of num1 to some temp variable
    temp = *num1;

    // Copy the value of num2 to num1
    *num1= *num2;

    // Copy the value of num1 stored in temp to num2
    *num2= temp;

    printf("After swapping in swap function \n");
    printf("Value of num1 = %d \n", *num1);
    printf("Value of num2 = %d \n\n", *num2);
}

Before moving on to other post, learn more ways to play with swapping numbers.

Enter two numbers: 10 20
Before swapping in main
Value of num1 = 10
Value of num2 = 20

After swapping in swap function
Value of num1 = 20
Value of num2 = 10

After swapping in main
Value of num1 = 20
Value of num2 = 10

Happy coding ;)

Write a C program to print 8 star pattern series using loop. How to print 8 star pattern series using for loop in C programming. Logic to print 8 star pattern series of N columns in C program.

Input N: 5

 ***
*   *
*   *
*   *
 ***
*   *
*   *
*   *
 ***

Required knowledge

Basic C programming, If else, Loop

Must have programming knowledge for this program.

Logic to print 8 star pattern

To simply things for beginners I have divided entire logic in three main sub tasks.

• Print a hollow square star pattern with N columns and (N*2) - 1 rows.
• Modify above hollow square pattern logic. Instead of star(*), print space at top and bottom corners and in the center intersection edge.
• Finally modify hollow square logic to print star instead of space for Nth row.

Below is the step by step descriptive logic to print 8 star pattern.

1. Read number of columns to print from user. Store it in some variable say N.
2. Iterate through (N*2)-1 rows. Run an outer loop with structure for(i=1; i<N*2; i++).
3. Iterate through N columns. Run an inner loop with structure for(j=1; j<=N; j++).
4. Inside inner loop first check conditions for corner and center intersection spaces. Print space for
   1st row and 1st or Nth column = if(i==1 && (j==1 || j==N))
   Nth row and 1st or Nth column = if(i==N && (j==1 || j==N))
   N*2-1 row and 1st or Nth column = if(i==N*2-1 && (j==1 || j==N))
   
5. Print stars for
   1st row or 1st column = if(i==1 || j==1)
   Nth row or Nth column = if(i==N || j==N)
   N*2-1th row = if(i==N*2-1)
   
6. Otherwise print spaces if above conditions does not matches.

Program to print 8 star pattern

/**
 * C program to print 8 star pattern series
 */
#include <stdio.h>

int main()
{
    int i, j, size;

    printf("Enter size: ");
    scanf("%d", &size);

    for(i=1; i<size*2; i++)
    {
        for(j=1; j<=size; j++)
        {
            // Condition for corner and center intersection space
            if((i==1 && (j==1 || j==size)) || (i==size && (j==1 || j==size)) || (i==size*2-1 && (j==1 || j==size)))
                printf("");
            else if(i==1 || i==size || i==(size*2)-1 || j==1 || j==size)
                printf("*");
            else
                printf("");
        }

        printf("\n");
    }

    return 0;
}

Enter size: 6
 ****
*    *
*    *
*    *
*    *
 ****
*    *
*    *
*    *
*    *
 ****

Screenshot

Happy coding ;)

Write a C program to left rotate an array by n position. Logic to rotate an array to left by n position in C program.

Input 10 elements in array: 1 2 3 4 5 6 7 8 9 10
Input number of times to rotate: 3

Array after left rotation: 4 5 6 7 8 9 10 1 2 3

Required knowledge

Basic C programming, Loop, Array, Function

Logic to left rotate an array

Below is the step by step descriptive logic to left rotate an array.

1. Read elements in an array say arr.
2. Read number of times to rotate in some variable say N.
3. Left Rotate the given array by 1 for N times. In real left rotation is shifting of array elements to one position left and copying first element to last.

Algorithm to left rotate an array
Begin:
read(arr)
read(n)
Fori←1 to ndo
rotateArrayByOne(arr)
End for
End
rotateArrayByOne(arr[], SIZE)
Begin:
first← arr[0]
Fori←1 to SIZE - 1do
arr[i] ← arr[i + 1]
End for
arr[SIZE - 1] ← first
End

Program to left rotate an array

/**
 * C program to left rotate an array
 */

#include <stdio.h>
#define SIZE 10 /* Size of the array */

void printArray(int arr[]);
void rotateByOne(int arr[]);


int main()
{
    int i, N;
    int arr[SIZE];

    printf("Enter 10 elements array: ");
    for(i=0; i<SIZE; i++)
    {
        scanf("%d", &arr[i]);
    }
    printf("Enter number of times to left rotate: ");
    scanf("%d", &N);

    /* Actual rotation */
    N = N % SIZE;

    /* Print array before rotation */
    printf("Array before rotation\n");
    printArray(arr);

    /* Rotate array n times */
    for(i=1; i<=N; i++)
    {
        rotateByOne(arr);
    }

    /* Print array after rotation */
    printf("\n\nArray after rotation\n");
    printArray(arr);

    return 0;
}


void rotateByOne(int arr[])
{
    int i, first;

    /* Store first element of array */
    first = arr[0];

    for(i=0; i<SIZE-1; i++)
    {
        /* Move each array element to its left */
        arr[i] = arr[i + 1];
    }

    /* Copies the first element of array to last */
    arr[SIZE-1] = first;
}


/**
 * Print the given array
 */
void printArray(int arr[])
{
    int i;

    for(i=0; i<SIZE; i++)
    {
        printf("%d ", arr[i]);
    }
}

Enter 10 elements array: 1 2 3 4 5 6 7 8 9 10
Enter number of times to left rotate: 3
Array before rotation
1 2 3 4 5 6 7 8 9 10

Array after rotation
4 5 6 7 8 9 10 1 2 3

Happy coding ;)

Write a C program to right rotate an array by n position. Logic to rotate an array to right by n position in C program.

Input 10 elements in array: 1 2 3 4 5 6 7 8 9 10
Input number of times to rotate: 3

Array after right rotation: 8 9 10 1 2 3 4 5 6 7

Required knowledge

Basic C programming, Loop, Array, Function

Logic to right rotate an array

Below is the step by step descriptive logic to rotate an array to right by n positions.

1. Read elements in an array say arr.
2. Read number of times to rotate in some variable say N.
3. Right rotate the given array by 1 for N times. In real right rotation is shifting of array elements to one position right and copying last element to first.

Algorithm to right rotate an array
Begin:
read(arr)
read(n)
Fori←1 to ndo
rotateArrayByOne(arr)
End for
End
rotateArrayByOne(arr[], SIZE)
Begin:
last← arr[SIZE - 1]
Fori← SIZE-1 to 0 do
arr[i] ← arr[i - 1]
End for
arr[0] ← last
End

Program to right rotate an array

/**
 * C program to right rotate an array
 */

#include <stdio.h>
#define SIZE 10 /* Size of the array */

void printArray(int arr[]);
void rotateByOne(int arr[]);


int main()
{
    int i, N;
    int arr[SIZE];

    printf("Enter 10 elements array: ");
    for(i=0; i<SIZE; i++)
    {
        scanf("%d", &arr[i]);
    }
    printf("Enter number of times to right rotate: ");
    scanf("%d", &N);

    /* Actual rotation */
    N = N % SIZE;

    /* Print array before rotation */
    printf("Array before rotation\n");
    printArray(arr);

    /* Rotate array n times */
    for(i=1; i<=N; i++)
    {
        rotateByOne(arr);
    }

    /* Print array after rotation */
    printf("\n\nArray after rotation\n");
    printArray(arr);

    return 0;
}


void rotateByOne(int arr[])
{
    int i, last;

    /* Store last element of array */
    last = arr[SIZE - 1];

    for(i=SIZE-1; i>0; i--)
    {
        /* Move each array element to its right */
        arr[i] = arr[i - 1];
    }

    /* Copy last element of array to first */
    arr[0] = last;
}


/**
 * Print the given array
 */
void printArray(int arr[])
{
    int i;

    for(i=0; i<SIZE; i++)
    {
        printf("%d ", arr[i]);
    }
}

Enter 10 elements array: 1 2 3 4 5 6 7 8 9 10
Enter number of times to right rotate: 3
Array before rotation
1 2 3 4 5 6 7 8 9 10

Array after rotation
8 9 10 1 2 3 4 5 6 7

Happy coding ;)