✅ Count Elements Greater Than All Prior Elements

Given an integer array Arr of size N the task is to find the count of elements whose value is greater than all of its prior elements.

Note : 1st element of the array should be considered in the count of the result.

For example,

Arr[]={7,4,8,2,9}

As 7 is the first element, it will consider in the result.

8 and 9 are also the elements that are greater than all of its previous elements.

Since total of  3 elements is present in the array that meets the condition.

Hence the output = 3.
  

Example 1:

Input 

5 -> Value of N, represents size of Arr

7-> Value of Arr[0]

4 -> Value of Arr[1]

8-> Value of Arr[2]

2-> Value of Arr[3]

9-> Value of Arr[4]

Output :

3
  

Example 2:

5   -> Value of N, represents size of Arr

3  -> Value of Arr[0]

4 -> Value of Arr[1]

5 -> Value of Arr[2]

8 -> Value of Arr[3]

9 -> Value of Arr[4]

Output : 

5
  

Constraints

1<=N<=20

1<=Arr[i]<=10000
  

#include <stdio.h>

int main()
{
    int n;
    printf("Enter the size of array: ");
    scanf("%d", &n);

    int arr[n];

    printf("Enter array elements:\n");
    for (int i = 0; i < n; i++)
    {
        scanf("%d", &arr[i]);
    }

    int count = 1;           // First element is always counted
    int max = arr[0];

    for (int i = 1; i < n; i++)
    {
        if (arr[i] > max)
        {
            count++;
            max = arr[i];
        }
    }

    printf("Output: %d\n", count);

    return 0;
}
  

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