can you complete this code? Input Format There will be three lines of input: - t

Question

can you complete this code?

Input Format
There will be three lines of input:

- the value that has to be searched.
- the size of the array.
- integers that make up the array .
Output Format
Output the index of 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

It is guaranteed that will occur in 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
. The value is the nd element in the array, but its index is since in this case, array indices start from (see array definition under Input Format.

Current Buffer (saved locally, editable)   



#include <bits/stdc++.h>

using namespace std;

vector<string> split_string(string);

// Complete the introTutorial function below.
int introTutorial(int V, vector<int> arr) {

}

int main()
{
ofstream fout(getenv("OUTPUT_PATH"));

int V;
cin >> V;
cin.ignore(numeric_limits<streamsize>::max(), ' ');

int n;
cin >> n;
cin.ignore(numeric_limits<streamsize>::max(), ' ');

string arr_temp_temp;
getline(cin, arr_temp_temp);

vector<string> arr_temp = split_string(arr_temp_temp);

vector<int> arr(n);

for (int i = 0; i < n; i++) {
int arr_item = stoi(arr_temp[i]);

arr[i] = arr_item;
}

int result = introTutorial(V, arr);

fout << result << " ";

fout.close();

return 0;
}

vector<string> split_string(string input_string) {
string::iterator new_end = unique(input_string.begin(), input_string.end(), [] (const char &x, const char &y) {
return x == y and x == ' ';
});

input_string.erase(new_end, input_string.end());

while (input_string[input_string.length() - 1] == ' ') {
input_string.pop_back();
}

vector<string> splits;
char delimiter = ' ';

size_t i = 0;
size_t pos = input_string.find(delimiter);

while (pos != string::npos) {
splits.push_back(input_string.substr(i, pos - i));

i = pos + 1;
pos = input_string.find(delimiter, i);
}

splits.push_back(input_string.substr(i, min(pos, input_string.length()) - i + 1));

return splits;
}

Explanation / Answer

Below is the code of the required function -

int introTutorial(int V, vector<int> arr) {

//use loop to traverse through all elements

for(int i=0;i<<arr.size();i++)
{

//Check whether the element at current index is required element
if(arr[i]==V)
return i; //return index at which element is found
}
}

I hope that i have provided sufficient comments for your better understanding Hope i have answered your question satisfactorily.Leave doubts in comment section if any

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