# TCS Coding Question 2022 Day 1 Slot 2

## Coding Question 2 for 2022 (September slot) ### TCS Coding Question Day 1 Slot 2 – Question 2

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

4 -> Value of Arr

8-> Value of Arr

2-> Value of Arr

9-> Value of Arr

Output :

3

Example 2:

5   -> Value of N, represents size of Arr

3  -> Value of Arr

4 -> Value of Arr

5 -> Value of Arr

8 -> Value of Arr

9 -> Value of Arr

Output :

5

Constraints

• 1<=N<=20
• 1<=Arr[i]<=10000
`#include <bits/stdc++.h>using namespace std; int main(){    int n,c=0,a,m=INT_MIN;    cin>>n;    while(n--)    {        cin>>a;        if(a>m)        {            m=a;            c++;        }    }    cout<<c;}`
`import java.util.*;class Solution{     public static void main(String[] args)     {                Scanner sc=new Scanner(System.in);                int n=sc.nextInt();                int arr[]=new int[n];                for(int i=0;i<n;i++)                        arr[i]=sc.nextInt();                           int max=Integer.MIN_VALUE;                int count=0;                for(int i=0;i<n;i++)                {                     if(arr[i]>max)                     {                                  max=arr[i];                                  count++;                      }                }                           System.out.println(count);    }}`
`import sysn=int(input())c=0m=-sys.maxsize-1while n:    n-=1    a=int(input())    if a>m:        m=a        c+=1print(c)`