# Lower Bound STL Hackerrank Solution in C++ | STL Solution

You are given N integers in sorted order. Also, you are given Q queries. In each query, you will be given an integer and you have to tell whether that integer is present in the array. If so, you have to tell at which index it is present and if it is not present, you have to tell the index at which the smallest integer that is just greater than the given number is present.

The lower bound is a function that can be used with a sorted vector. Learn how to use lower bound to solve this problem by clicking here.

Input Format

The first line of the input contains the number of integers N. The next line contains N integers in sorted order. The next line contains Q, the number of queries. Then Q lines follow each containing a single integer Y.

Note: If the same number is present multiple times, you have to print the first index at which it occurs. Also, the input is such that you always have an answer for each query.

Constraints

1 <= N <= 10^5
1 <= N <= 10^5, where X is the ith element in the array.
1 <= Q <= 10^5
1 <= Y <= 10^9

Output Format

For each query you have to print "Yes" (without the quotes) if the number is present and at which index(1-based) it is presently separated by a space.

If the number is not present you have to print "No" (without the quotes) followed by the index of the next smallest number just greater than that number.

You have to output each query in a new line.

Sample Input

8
1 1 2 2 6 9 9 15
4
1
4
9
15

Sample Output

Yes 1
No 5
Yes 6
Yes 8

So according to the first query 1 is present in an index 1 so our output is Yes 1, Now move on to the second query 4 is not present in an array that's why the output is No 5, according to the condition if the number is not present in an array then it should return a next greater number than query. Now next query is 9 it is present in an index 6 so the output is Yes 6. and the last query is 15 it is also present in an array that's why the output is Yes 8.

Note: Here an array index is starting with 1, which normally starts with 0.

Lower Bound STL Hackerrank Solution in C++

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``````#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

int main()
{
int m, num;
cin >> m;
vector<int> v;
for (int i = 0; i < m; i++)
{
cin >> num;
v.push_back(num);
}

int n, val;
cin >> n;
for (int i = 0; i < n; i++)
{
cin >> val;
vector<int>::iterator low = lower_bound(v.begin(), v.end(), val);
if (v[low - v.begin()] == val)
cout << "Yes " << (low - v.begin() + 1) << endl;
else
cout << "No " << (low - v.begin() + 1) << endl;
}

return 0;
}``````

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#### post written by: Ghanendra Yadav

Hi, I’m Ghanendra Yadav, SEO Expert, Professional Blogger, Programmer, and UI Developer. Get a Solution of More Than 500+ Programming Problems, and Practice All Programs in C, C++, and Java Languages. Get a Competitive Website Solution also Ie. Hackerrank Solutions and Geeksforgeeks Solutions. If You Are Interested to Learn a C Programming Language and You Don't Have Experience in Any Programming, You Should Start with a C Programming Language, Read: List of Format Specifiers in C.