**PROBLEM LINK**

Author : Hatim Tai

Tester : Deepanshu Bhatia

Editorialist : Hatim Tai

**DIFFICULTY :**

Simple

**PREREQUISITES:**

Linear Search

**PROBLEM :**

Given a number **O** and an array **A** having **N** numbers, If the first element of the array is less than **O** output ‘-1’ otherwise find the count of numbers which are greater than or equal to **O**.and this count can go upto **M**.

**EXPLANATION :**

To find the count of numbers greater than or equal to **O** first initialise cnt=0 and simply traverse the whole array, compare each number if it is greater than or equal to **O** ? If it is, increment ‘cnt’ by ‘1’.

Finally see if this ‘cnt’ is greater than **M** or not ? If it is then set cnt = M, since the value of ‘cnt’ can go upto M.

```
cnt = 0
if(A[0] < O)
cnt = -1
else
{
for(i = 0; i < N ; i++)
{
if(A[i] >= cnt)
cnt = cnt + 1
}
}
if(cnt != -1)
cnt = min(cnt,M)
print cnt
```