Count Repeated Elements in an Array in C Programming

Count Repeated Elements in an Array in C Programming is generally asked in interviews.

This problem can be solved in any language.

Here we will discuss algorithms to count repeated elements and implement them in C programming.

**Fig: Count Repeated Elements in an Array in C Programming**

What is Repeated Elements

Repeated elements are elements that occur more than one time in a list or in a sequence.

Consider the following list

1 2 3 4 5 6 7 7 2 3 3 3

Here element 2 is repeating 2 times 3 is repeating 4 times and 7 is repeating 2 times.

This thing we have to implement in programming.

So our input is a list or sequence of elements.

and output is repeated details and repetition numbers.

So how will we turn our input into output?

Algorithm to Count Repeated Elements

Let’s define our Algorithm that will get input and give us the desired output.

Create Array arr[] and repeated[][]
Take input from the user and store it to arr
For each element of the arr check is it in the repeated array or not
If an element is found then increase the element count
Else add an element to the repeated array
The print repeated array that has count 2 or more

Above is our simple step to developing our program

Program to Count Repeated Elements in C Programming

#include<stdio.h>
int main() {
  int arr[20];
  int repeated[20][2];
  int elementCount = 0;
  int count = 0;
  int found = 0;
  int position = 0;
  int i = 0, j = 0;
  printf("\nEnter size of an array\n");
  scanf("%d", & count);
  for (i = 0; i < count; i++) {
    printf("Enter element\n");
    scanf("%d", & arr[i]);
  }
  for (i = 0; i < count; i++) {
    for (int j = 0; j <= elementCount; j++) {
      if (arr[i] == repeated[j][0]) {
        found = 1;
        position = j;
        break;
      }
    }
    if (found == 1) {
      repeated[position][1] += 1;
    } else {
      repeated[elementCount][0] = arr[i];
      repeated[elementCount][1] = 1;
      elementCount++;
    }
  }
  printf("\nRepeated Elements are \n ");
  printf("\nElement     No of Repeatatione\n ");
  for (i = 0; i < elementCount; i++) {
    if (repeated[i][1] > 1) {
      printf("\n%d\t\t\t  %d", repeated[i][0], repeated[i][1]);
    }
  }
}

#include<stdio.h>

int main() {

int arr[20];

int repeated[20][2];

int elementCount = 0;

int count = 0;

int found = 0;

int position = 0;

int i = 0, j = 0;

printf("\nEnter size of an array\n");

scanf("%d", & count);

for (i = 0; i < count; i++) {

printf("Enter element\n");

scanf("%d", & arr[i]);

}

for (i = 0; i < count; i++) {

for (int j = 0; j <= elementCount; j++) {

if (arr[i] == repeated[j][0]) {

found = 1;

position = j;

break;

}

if (found == 1) {

repeated[position][1] += 1;

} else {

repeated[elementCount][0] = arr[i];

repeated[elementCount][1] = 1;

elementCount++;

}

printf("\nRepeated Elements are \n ");

printf("\nElement No of Repeatatione\n ");

for (i = 0; i < elementCount; i++) {

if (repeated[i][1] > 1) {

printf("\n%d\t\t\t %d", repeated[i][0], repeated[i][1]);

}

Based on the above algorithm we develop this program.

Result

Enter size of an array
15   
Enter element
1
Enter element
2
Enter element
3
Enter element
4
Enter element
5    
Enter element
6
Enter element
4
Enter element
4
Enter element
3
Enter element
1
Enter element
1
Enter element    
3
Enter element
1
Enter element
5
Enter element
6

Repeated Elements are 
Element     No of Repetition
 
1			  4
3			  3
4			  3
5			  2
6			  2

Enter size of an array

Enter element

Repeated Elements are

Element No of Repetition

1 4

3 3

4 3

5 2

6 2

Hope you learn from this example.