Consider the following list, if user enters 5 then display “2 times” because “5” occurs 2 times in the list

Write pseudo code to count occurrence of a given number in the list. For example, consider the following list, if user enters 5 then display “2 times” because “5” occurs 2 times in the list.

Given Array

int A[] = {5, 3, 6, 8, 5, 1};

 Write pseudo code to count occurrence of a given number in the list. For example, consider the following list, if user enters 5 then display “2 times” because “5” occurs 2 times in the list.

Solution :

      By using C Language. 

#include <stdio.h>

int BinarySearch(int A[], int N, int x, int searchFirst)

{

    

    int low = 0, high = N - 1;

    

    int result = -1;

   

    while (low <= high)

    {

        

        int mid = (low + high)/2;

       

        if (x == A[mid])

        {

            result = mid;

            

            if (searchFirst)

                high = mid - 1;

            

            else

                low = mid + 1;

        }

        

        else if (x < A[mid])

            high = mid - 1;

      

        else

            low = mid + 1;

    }

    

    return result;

}

int main(void)

{

    int A[] = {5, 3, 6, 8, 5, 1};

    int target = 5;

    int n = sizeof(A)/sizeof(A[0]);

   

    int first = BinarySearch(A, n, target, 1);

    

    int last = BinarySearch(A, n, target, 0);

    int count = last - first + 1;

    if (first != -1)

        printf("Element %d occurs %d times.", target, count);

    else

        printf("Element not found in the array.");

    return 0;

}

Output : Element 5 occurs 2 times

Question :

Write pseudo code to count occurrence of a given number in the list. For example, consider the following list, if user enters 5 then display “2 times” because “5” occurs 2 times in the list.

Solution:

           As above mentioned. 

ATTENTION

Please let me know if you have any questions. I hope the above is useful to you.