##### Intro to Tutorial Challenges Hackerrank

Sample Challenge

This is a simple challenge to get things started. Given a sorted array (ar) and a number (V), can you print the index location of V in the array?

The next section describes the input format. You can often skip it, if you are using included methods.

**Input Format **

There will be three lines of input:

V – the value that has to be searched.

n – the size of the array.

ar – n numbers that make up the array.

**Output Format **

Output the index of V in the array.

The next section describes the constraints and ranges of the input. You should check this section to know the range of the input.

Constraints

1<=N<=1000
-1000<=V<=1000 V{ar}
It is guaranteed that V will occur in ar exactly once.
This "sample" shows the first input test case. It is often useful to go through the sample to understand a challenge.
Sample Input

4

6

1 4 5 7 9 12

Sample Output

1

Explanation

V=4. The value 4 is the 2^{nd} element in the array, but its index is 1 since array indices start from 0, so the answer is 1.

##### Solution

```
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main() {
int v, n, a[1000],i;
scanf("%d",&v);
scanf("%d",&n);
for(i=0; i<n;i++){
scanf("%d",&a[i]);
}
for(i=0; i<n;i++){
if(a[i]==v)
printf("%d",i);
}
/* Enter your code here. Read input from STDIN. Print output to STDOUT */
return 0;
}
```

```
V=input();
N=int(input());
A=input();
A=A.split();
print(A.index(V))
```

```
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
public static void main(String[] args) {
/* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */
Scanner in = new Scanner(System.in);
int v = in.nextInt();
int n = in.nextInt();
int[] ar = new int[n];
for(int i=0;i<n;i++){
ar[i]=in.nextInt();
}
int i=0;
while (ar[i]<v){
i++;
}
System.out.println(i);
}
}
```

#include<string> #include<iostream> #include<iterator> #include<algorithm> #include<vector> using namespace std; class Binary { public: int Search(vector<int>V, int Value) { int low = 0; int high = V.size(); while (low <= high) { int mid = low + (high - low) / 2; if (Value == V[mid]) { return mid; } else if (V[mid] < Value) { low = mid + 1; } else { high = mid - 1; } } cout << "Target wasn't found" << endl; } }; //Binary Search Implementation vector<int>Ans; Binary B; auto main()->int { int n; cin >> n; int l; cin>>l; for (int i = 0; i != l; i++) { int p; cin >> p; Ans.push_back(p); } cout<<B.Search(Ans,n); return 0; }