C Program To Insert New Element At Specified Position of An Array

Write a C program to insert new element/number at specified position of an array. The array elements need not be in sorted order.

Related Read:
C Program To Shift Elements of An Array by n Position

Example: Expected Input/Output

Enter 10 integer numbers
1
2
3
4
5
6
8
9
10
11
Enter the position where new number has to be inserted
6
Enter a new number to be inserted at position 6
7
Array after inserting 7 at position 6
1
2
3
4
5
6
7
8
9
10
11

Visual Representation

insert element at specified position of an array

Video Tutorial: C Program To Insert New Element At Specified Position of An Array


[youtube https://www.youtube.com/watch?v=oT31byliQPE]

YouTube Link: https://www.youtube.com/watch?v=oT31byliQPE [Watch the Video In Full Screen.]

Source Code: C Program To Insert New Element At Specified Position of An Array

#include<stdio.h>
#define N 11

int main()
{
    int a[N], i, pos, num;

    printf("Enter %d integer numbers\n", (N - 1));
    for(i = 0; i < (N -1); i++)
        scanf("%d", &a[i]);

    printf("Enter the position where new number has to be inserted\n");
    scanf("%d", &pos);

    if(pos < N)
    {
        printf("Enter a new number to be inserted at position %d\n", pos);
        scanf("%d", &num);
        for(i = N - 1; i > pos; i--)
                a[i] = a[i - 1];

        a[pos] = num;

        printf("Array after inserting %d at position %d\n", num, pos);
        for(i = 0; i < N; i++)
            printf("%d\n", a[i]);
    }
    else
    {
        printf("Please enter a position within the range/size of the array!\n");
    }

    printf("\n");

    return 0;
}

Output:
Enter 10 integer numbers
11
25
52
36
98
92
45
69
59
2
Enter the position where new number has to be inserted
5
Enter a new number to be inserted at position 5
100
Array after inserting 100 at position 5
11
25
52
36
98
100
92
45
69
59
2

Logic To Insert New Element At Specified Position of An Array

We ask the user to enter (N – 1) number of elements and store it inside array variable a[N]. We leave the last index position empty(technically it’ll automatically have zero as the element). Now we ask the user to enter the position where he or she wants to insert new number/element. Next we ask the user to input the new number to be inserted at the specified position. Once we get the position and the new number, we start the “for loop”.

We initialize i with last index of the array, which is N – 1. We iterate the array until i is greater than the user entered position.

Note: We iterate the for loop only until i is greater than user specified position because, we need to shift the elements to right by 1 position only after the position/index indicated by the user. We won’t move any elements above the user specified position.

Inside for loop
Inside for loop we move the elements by 1 position to the bottom or right. Ex: if i value is 5, then a[5] will be assigned whatever the value is present in it’s previous index a[4].

        for(i = N - 1; i > pos; i--)
                a[i] = a[i - 1];

        a[pos] = num;

Once all the elements from user input position move 1 position right/down, we insert the new element/number at user specified position.

Explanation With Example

If a[6] = {5, 7, 3, 2, 1};
Position to insert new element/number: 2
New number to be inserted: 9

        for(i = N - 1; i > pos; i--)
                a[i] = a[i - 1];

        a[pos] = num;

We initialize i to last index of the array, which is N – 1. Since N is 6 in this case, 6 -1 is 5. So i is initialized to 5. We’ve not assigned any element/value to a[5]. Technically it’ll have 0.

iposa[i]a[i – 1]a[i] = a[i – 1]
52a[5] = 0a[4] = 1a[5] = 1
42a[4] = 1a[3] = 2a[4] = 2
32a[3] = 2a[2] = 3a[3] = 3
22

Now that index i is 2 and user input position is also 2. So 2 > 2 returns false, so the control exits the for loop. Outside for loop we’ve a[pos] = num. So the position is 2 and the new number to be inserted is 9. i.e., a[2] = 9.

So the array elements after execution of above logic:
a[6] = {5, 7, 9, 3, 2, 1};

That’s how we successfully inserted new element/number 9 at position 2.

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C Program To Count Prime Numbers Between Range

Lets write a C program to count prime numbers between user entered/input range of numbers, using function/method.

Prime Number: is a natural number greater than 1, which has no positive divisors other than 1 and itself.

Related Read:
C Program To Find Prime Number or Not using For Loop

Video Tutorial: C Program To Count Prime Numbers Between Range


[youtube https://www.youtube.com/watch?v=vY6HFMblBJY]

YouTube Link: https://www.youtube.com/watch?v=vY6HFMblBJY [Watch the Video In Full Screen.]

Source Code: C Program To Count Prime Numbers Between Range

#include<stdio.h>
#include<math.h>

int isPrime(int num)
{
    int inum = sqrt(num), prime = 1, count;

    for(count = 2; count <= inum; count++)
    {
         if(num % count == 0)
         {
            prime = 0;
            break;
         }
    }

    return(prime);
}

int main()
{
    int start, end, temp, num, slno = 0, on_off;

    printf("Enter start and end value\n");
    scanf("%d%d", &start, &end);

    printf("Do you want to print prime numbers? (yes = 1, no = 0)\n");
    scanf("%d", &on_off);

    if(start > end)
    {
        temp  = start;
        start = end;
        end   = temp;
    }

    if(on_off)
        printf("\nPrime numbers between %d and %d are:\n\n", start, end);

    for(num = start; num <= end; num++)
    {
        if(num == 1)
        {
            continue;
        }

        if( isPrime(num) )
        {
            slno++;

            if(on_off)
            {
                printf("%d. %d is prime number.\n", slno, num);
            }
         }
    }

    printf("\nThere are %d prime numbers between %d and %d.\n", 
             slno, start, end);

    return 0;
}

Output 1:
Enter start and end value
10
30
Do you want to print prime numbers? (yes = 1, no = 0)
1

Prime numbers between 10 and 30 are:

1. 11 is prime number.
2. 13 is prime number.
3. 17 is prime number.
4. 19 is prime number.
5. 23 is prime number.
6. 29 is prime number.

There are 6 prime numbers between 10 and 30.

Output 2:
Enter start and end value
10
30
Do you want to print prime numbers? (yes = 1, no = 0)
0

There are 6 prime numbers between 10 and 30.

Output 3:
Enter start and end value
1
10
Do you want to print prime numbers? (yes = 1, no = 0)
1

Prime numbers between 1 and 10 are:

1. 2 is prime number.
2. 3 is prime number.
3. 5 is prime number.
4. 7 is prime number.

There are 4 prime numbers between 1 and 10.

Output 4:
Enter start and end value
1
300
Do you want to print prime numbers? (yes = 1, no = 0)
0

There are 62 prime numbers between 1 and 300.

Output 5:
Enter start and end value
1
1000
Do you want to print prime numbers? (yes = 1, no = 0)
0

There are 168 prime numbers between 1 and 1000.

Logic To Count Prime Numbers Between Range

First we ask the user to enter the range and store it inside address of variables start and end. We make sure that start value is less than end value. If start value is greater than end value, we use a temporary variable to swap the values of start and end.

Swap 2 Numbers Using a Temporary Variable: C

Next we ask the user if he / she wants to display the prime numbers between the range or just want to know the count of prime numbers between the entered range. We store the user answer in a variable called on_off.

We start the for loop: We initialize the loop counter variable num to start and iterate through the loop until num is less than or equal to end. For each iteration of the for loop we increment the value of num by 1. This for loop selects number one by one from start to end. And this selected number, which is present inside variable num is checked for prime or not. If its prime we display it to the console window and keep track of the count of prime numbers, if not we simply ignore that non-prime number.

We use a function to check if the selected number present in variable num is a prime number or not. We call the function/method isPrime() inside if condition and pass the value of num to it. isPrime() returns 1 or 0 value. 1 means true and 0 means false. So if isPrime() returns 1, then the if condition becomes true and we increment the value of slno by 1 and optionally printout the prime number. If isPrime() returns 0, then we simply ignore it and go to the next number.

isPrime() function has the logic to determine if the given number is prime or not. We have explained the complete logic of this in a separate video tutorial, link to which is present below.

C Program To Find Prime Number or Not using For Loop

Note:
1. We are including math.h header file or library file since we are using sqrt() builtin method. sqrt() method is present inside math.h file.

2. We are also using continue and break keywords in our program and we’ve explained about it in detail in separate videos. Please watch them for more clarity about the topic.

Continue Statement In C Programming Language
break Statement In C Programming Language

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C Program To Find Armstrong Numbers Between Range using Function

In today’s video tutorial lets find all the Armstrong numbers or Narcissistic numbers between user entered range, using function / method.

An Armstrong number or Narcissistic number is an n-digit base b number such that the sum of its (base b) digits raised to the power n is the number itself.

Example 1:
If number = 370
It has 3 digits: 3, 7 and 0. So n = 3.
result = 33 + 73 + 03 = 27 + 343 + 0 = 370.
So the original number 370 is equal to the result. So it’s an Armstrong Number.

Example 2:
If number = 8208
It has 4 digits: 8, 2, 0 and 8. So n = 4.
result = 84 + 24 + 04 + 84 = ‭4096‬ + 16 + 0 + ‭4096‬ = 8208.
So the original number 8208 is equal to the result. So it’s an Armstrong Number.

Related Read:
Swap 2 Numbers Using a Temporary Variable: C
Function / Methods In C Programming Language
C Program to Check Armstrong Number
C Program to print Armstrong Numbers Between Two Integers

Video Tutorial: C Program To Find Armstrong Numbers Between Range using Function


[youtube https://www.youtube.com/watch?v=nsVcZxEvIc8]

YouTube Link: https://www.youtube.com/watch?v=nsVcZxEvIc8 [Watch the Video In Full Screen.]


To count number of digits in a number

    int n = 0, temp;

    temp = num;

    while(temp)
    {
        temp = temp / 10;
        n++;
    }

To Iterate through the digits in the number

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

To know above code logic, please visit and watch the video tutorial present at C Program to Check Armstrong Number.

Full Source Code: C Program To Find Armstrong Numbers Between Range using Function

#include<stdio.h>
#include<math.h>

float armstrong(int);

int main()
{
    int count, start, end, temp;

    printf("Enter start and end values\n");
    scanf("%d%d", &start, &end);

    if(start > end)
    {
        temp = start;
        start= end;
        end  = temp;
    }

    printf("Armstrong numbers between %d and %d are\n", start, end);

    for(count = start; count <= end; count++)
    {
        if(count == armstrong(count))
        {
            printf("%d is an Armstrong number\n", count);
        }
    }

    return 0;
}

float armstrong(int num)
{
    int rem, n = 0, temp;
    float sum = 0.0;

    temp = num;

    while(temp)
    {
        temp = temp / 10;
        n++;
    }

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

    return(sum);
}

Output 1:
Enter start and end values
1
999
Armstrong numbers between 1 and 999 are
1 is an Armstrong number
2 is an Armstrong number
3 is an Armstrong number
4 is an Armstrong number
5 is an Armstrong number
6 is an Armstrong number
7 is an Armstrong number
8 is an Armstrong number
9 is an Armstrong number
153 is an Armstrong number
370 is an Armstrong number
371 is an Armstrong number
407 is an Armstrong number

Output 2:
Enter start and end values
500
99999
Armstrong numbers between 500 and 99999 are
1634 is an Armstrong number
8208 is an Armstrong number
9474 is an Armstrong number
54748 is an Armstrong number
92727 is an Armstrong number
93084 is an Armstrong number

Output 3:
Enter start and end values
1
99999
Armstrong numbers between 1 and 99999 are
1 is an Armstrong number
2 is an Armstrong number
3 is an Armstrong number
4 is an Armstrong number
5 is an Armstrong number
6 is an Armstrong number
7 is an Armstrong number
8 is an Armstrong number
9 is an Armstrong number
153 is an Armstrong number
370 is an Armstrong number
371 is an Armstrong number
407 is an Armstrong number
1634 is an Armstrong number
8208 is an Armstrong number
9474 is an Armstrong number
54748 is an Armstrong number
92727 is an Armstrong number
93084 is an Armstrong number

Output 4:
Enter start and end values
1
500
Armstrong numbers between 1 and 500 are
1 is an Armstrong number
2 is an Armstrong number
3 is an Armstrong number
4 is an Armstrong number
5 is an Armstrong number
6 is an Armstrong number
7 is an Armstrong number
8 is an Armstrong number
9 is an Armstrong number
153 is an Armstrong number
370 is an Armstrong number
371 is an Armstrong number
407 is an Armstrong number

Logic To Find Armstrong Numbers Between Range using Function

We ask the user to enter start and end value i.e., the range. Using for loop we iterate through all the numbers between start and end. For each iteration of for loop, variable count holds the number which we need to check if its Armstrong number or not.

Value of count is passed to function armstrong. Function armstrong counts the number of digits present in the integer number(we store it inside variable n) and then multiplies all the individual digits of the number by n and adds them. The sum is returned back to the calling function.

If value present in count and the value returned by armstrong method are equal, then the value present in count is an Armstrong number and will be displayed on the console window.

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C Program To Find Sum of All Even Numbers Between Range, using For loop

Lets write a C program to find sum of all the even numbers between range or between 2 integers input by the user, using for loop.

Even Number: An even number is an integer that is exactly divisible by 2.

For Example: 14 % 2 == 0. When we divide 14 by 2, it gives a reminder of 0. So number 14 is an even number.

Note: In this C program we ask the user to input start and end value. We assume that the user enters bigger value for variable end and smaller value for variable start. i.e., start < end If start value is greater than value present in end, we swap the values of variable start and end.

If user enters start = 5 and end = 14. C program finds all the even numbers between 5 and 14, including 5 and 14. So the even numbers are 6, 8, 10, 12, 14. We add all these even numbers and output the sum to the console window. i.e., 6 + 8 + 10 + 12 + 14 = 50. We out put the value 50 as result.

Related Read:
Decision Control Instruction In C: IF
For Loop In C Programming Language
Even or Odd Number: C Program
C Program to Generate Even Numbers Between Two Integers

You can also watch C Program To Find Sum of All Even Numbers Between Two Integers, using While loop

Video Tutorial: C Program To Find Sum of All Even Numbers Between Range, using For loop


[youtube https://www.youtube.com/watch?v=9nKzyNnM2CY]

YouTube Link: https://www.youtube.com/watch?v=9nKzyNnM2CY [Watch the Video In Full Screen.]

Logic To Find Sum of All Even Numbers Between Range, using For loop

Step 1: We ask the user to enter start and end value.

Step 2: We initialize count to start and iterate through the for loop until value of count is less than or equal to value of variable end. For each iteration of for loop count value increments by 1.

Step 3: For every iteration we check if value present in variable count is a even number. i.e., count % 2 == 0. If this condition is true, then we add the value present in variable count to the previous value of variable sum.

Step 4: Once the control exits for loop, we print the value present in variable sum – which has the sum of all the even numbers between the range entered by the user.

Source Code: C Program To Find Sum of All Even Numbers Between Range, using For loop

#include<stdio.h>

int main()
{
    int start, end, temp, count, sum = 0;

    printf("Enter start and end values\n");
    scanf("%d%d", &start, &end);

    if(start > end)
    {
        temp  = start;
        start = end;
        end   = temp;
    }

    printf("Even numbers between %d and %d are:\n", start, end);
    for(count = start; count <= end; count++)
    {
        if(count % 2 == 0)
        {
            printf("%d\n", count);
            sum = sum + count;
        }
    }

    printf("Sum of all the even numbers from %d to %d is %d\n", start, end, sum);

    return 0;
}

Output 1:
Enter start and end values
5
14
Even numbers between 5 and 14 are:
6
8
10
12
14
Sum of all the even numbers from 5 to 14 is 50

Output 2:
Enter start and end values
23
14
Even numbers between 14 and 23 are:
14
16
18
20
22
Sum of all the even numbers from 14 to 23 is 90

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C Program To Find Sum of All Odd Numbers Between Range, using For loop

Lets write a C program to find sum of all odd numbers between range or between 2 integers input by the user, using For loop.

Odd Number: An odd number is an integer that is not exactly divisible by 2.

For Example: 15 % 2 != 0. When we divide 15 by 2, it does not give a reminder of 0. So number 15 is an odd number.

Note: In this C program we ask the user to input start and end value. We assume that the user enters bigger value for variable end and smaller value for variable start. i.e., start < end If not, we swap the values of variable start and end.

If user enters start = 14 and end = 23. C program finds all the odd numbers between 14 and 23, including 14 and 23. So the odd numbers are 15, 17, 19, 21, 23. We add all these odd numbers and output the sum to the console window. i.e., 15 + 17 + 19 + 21 + 23 = 95. We out put the value 95 as result.

Related Read:
Decision Control Instruction In C: IF
For Loop In C Programming Language
Even or Odd Number: C Program
C Program to Generate Odd Numbers Between Two Integers

Video Tutorial: C Program To Find Sum of All Odd Numbers Between Range, using For loop


[youtube https://www.youtube.com/watch?v=tUudL6AKPOg]

YouTube Link: https://www.youtube.com/watch?v=tUudL6AKPOg [Watch the Video In Full Screen.]

Logic To Find Sum of All Odd Numbers Between Range, using For loop

Step 1: We ask the user to enter start and end value.

Step 2: Variable count is initialized to start and for loop executes until count is less than or equal to end. For each iteration of the for loop, value of count increments by 1.

Step 3: For every iteration of the for loop, we check if value present in variable count is an odd number. i.e., count % 2 != 0. If this condition is true, then we add the value present in variable count to the previous value of variable sum.

Step 4: Once the control exits for loop, we print the value present in variable sum – which has the sum of all the odd numbers between the range entered by the user.

Source Code: C Program To Find Sum of All Odd Numbers Between Range, using For loop

#include<stdio.h>

int main()
{
    int start, end, temp, count, sum = 0;

    printf("Enter start and end value\n");
    scanf("%d%d", &start, &end);

    if(start > end)
    {
        temp  = start;
        start = end;
        end   = temp;
    }

    printf("Odd numbers from %d to %d are\n", start, end);
    for(count = start; count <= end; count++)
    {
        if(count % 2 != 0)
        {
            printf("%d\n", count);
            sum = sum + count;
        }
    }

    printf("Sum of all the Odd numbers from %d to %d is %d\n", start, end, sum);

    return 0;
}

Output 1:
Enter start and end value
5
14
Odd numbers from 5 to 14 are
5
7
9
11
13
Sum of all the Odd numbers from 5 to 14 is 45

Output 2:
Enter start and end value
14
23
Odd numbers from 14 to 23 are
15
17
19
21
23
Sum of all the Odd numbers from 14 to 23 is 95

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