Count of smaller elements (GFG)

Question: 

Count of smaller elements

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Given an sorted array A of size N. Find number of elements which are less than or equal to given element X.

 

Example 1:

Input:
N = 6
A[] = {1, 2, 4, 5, 8, 10}
X = 9
Output:
5

 

Example 2:

Input:
N = 7
A[] = {1, 2, 2, 2, 5, 7, 9}
X = 2
Output:
4

 

Your Task:  
You don't need to read input or print anything. Your task is to complete the function countOfElements() which takes the array A[], its size N and an integer X as inputs and returns the number of elements which are less than or equal to given element.

 

Expected Time Complexity: O(N)
Expected Auxiliary Space: O(1)

 

Constraints:
1 <= N <= 105
1 <= Ai <= 105

0 <= X <= 105

Ans: 

Time Complexity of this solution: O(n);

public class Solution {

    public int countOfElements(int[] arr, int n, int x) {

        int count = 0;

        for (int i = 0; i < n; i++) {

            if (arr[i] <= x) { // If the current element is less than or equal to X, increment count

                count++;

            } else {

                break; // Since the array is sorted, we can break the loop if the current element exceeds X

            }

        }

        return count;

    }

}

In <= agr element x se chota hai ya barabar hai to conut ++ karega agr element x ke barabar to nhi hai lekin badha hai to else me jake ruk jayega break statement ki wjh se;

Means <= less than equals to operator ka use ham tab karte hai jab hame check krna hai ki element ya to x se chota ho ya barabar ho to true value dega otherwise false hoga.