Lower Bound – STL

Earlier this week, I received an email message from HackerRank suggesting a C++ challenge. Finally I was able to work on it. This past week I was busy with work.

If interested, please take a look at the requirements for the Lower Bound – STL challenge and solve it. When done continue reading to see if our solutions are quite different. If so, please leave me a message in the area provided towards the end of this post.

Following is a screen capture of a console using the sample input for the challenge:

8
 1 1 2 2 6 9 9 15
 4
 1
Yes 1
 4
No 5
 9
Yes 6
 15
Yes 8

Following is the C++ code I generated using Visual Studio 2017 Enterprise Edition IDE:

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>

using namespace std;

/*
*/
int main() {

	// **** get tshe number of elements in the vector ****

	int N = 0;
	cin >> N;
	//cout << "main << N: " << N << endl;

	// **** declare the vector of integers ****

	vector<int>	vv(N);
	
	// **** populate the vector ****

	for (int n = 0; n < N; n++) { cin >> vv.at(n);
		//cout << "main <<< vv[" << n << "]: " << vv.at(n) << endl; } // **** get the number of queries **** int Q = 0; cin >> Q;
	//cout << "main <<< Q: " << Q << endl;

	// **** read and process queries ****

	int query = 0;
	for (int q = 0; q < Q; q++) { // **** read query **** cin >> query;
		//cout << "main <<< query: " << query << endl;

		// **** process query ****

		//vector<int>::iterator low, up;
		vector<int>::iterator low;
		low = lower_bound(vv.begin(), vv.end(), query);
		//up	= upper_bound(vv.begin(), vv.end(), query);

		// **** display result of query ****

		//cout << "main <<< low: " << low - vv.begin() << endl;
		//cout << "main <<<  up: " << up - vv.begin() << endl;

		if (vv.at(low - vv.begin()) == query)
		{
			cout << "Yes ";
		}
		else
		{
			cout << "No ";
		}
		cout << low - vv.begin() + 1 << endl;
	}

	// **** ****

	return 0;
}

The code passed all 17 test cases.

If you have comments or questions please let me know.

Enjoy;

John

Follow me on Twitter:  @ john_canessa

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