Queries to check if a number lies in N ranges of L-R

Given N ranges and Q queries consisting of numbers. Every range consists of L and R. The task is to check if the given number lies in any of the given ranges or not for every query.

Note: There is no overlapping range.

Examples:



Input: range[] = { {5, 6}, {1, 3}, {8, 10}
Q = 4
1st query: 2
2nd query: 3
3rd query: 4
4th query: 7

Output:
Yes
Yes
No
No

1st query: 2 lies in a range 1-3
2nd query: 3 lies in a range 1-3
3rd query: 4 does not lie in any of the given range.
4th query: 7 does not lie in any of the given range.

Approach: Below is the step by step algorithm to solve this problem:

  1. Hash the L of every range as 1, and hash the R of every range as 2.
  2. Push the L and R separately into a container.
  3. Sort the elements in the container, all the range L and R will be adjacent to each other as do not overlap.
  4. For every query, use binary search to find the first occurrence of a number same or greater than it. This can be done using the lower_bound function.
  5. If there is any value which is equal to the query, then the number overlaps the range.
  6. If there is no value which is equal to the query, then check if the greater element is hashed as 1 or 2.
  7. If it is hashed as 1, then the number does not overlaps, else it does overlap.

Below is the implementaion of above approach:

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// C++ program to check if the
// number lies in given range
#include <bits/stdc++.h>
  
using namespace std;
  
// Function that answers every query
void answerQueries(int a[][2], int n, int queries[], int q)
{
  
    // container to store all range
    vector<int> v;
  
    // hash the L and R
    unordered_map<int, int> mpp;
  
    // Push the element to container
    // and hash the L and R
    for (int i = 0; i < n; i++) {
  
        v.push_back(a[i][0]);
        mpp[a[i][0]] = 1;
        v.push_back(a[i][1]);
        mpp[a[i][1]] = 2;
    }
  
    // sort the elements in container
    sort(v.begin(), v.end());
    for (int i = 0; i < q; i++) {
  
        // each query
        int num = queries[i];
  
        // get the number same or greater than integer
        int ind = lower_bound(v.begin(), v.end(), num) - v.begin();
  
        // if it lies
        if (v[ind] == num) {
            cout << "Yes\n";
        }
  
        else {
  
            // check if greater is hashed as 2
            if (mpp[v[ind]] == 2)
                cout << "Yes\n";
  
            else // check if greater is hashed as 1
                cout << "No\n";
        }
    }
}
  
// Driver code
int main()
{
    int a[][2] = { { 5, 6 }, { 1, 3 }, { 8, 10 } };
  
    int n = 3;
  
    int queries[] = { 2, 3, 4, 7 };
  
    int q = sizeof(queries) / sizeof(queries[0]);
  
    // function call to answer queries
    answerQueries(a, n, queries, q);
  
    return 0;
}

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Output:

Yes
Yes
No
No


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Striver(underscore)79 at Codechef and codeforces D

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