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

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  • Difficulty Level : Easy
  • Last Updated : 29 Jun, 2022

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 implementation of above approach:
 

C++




// 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;
}


Java




// Java program to check if the
// number lies in given range
import java.io.*;
import java.util.*;
 
class GFG
{
 
    // Function that answers every query
    static void answerQueries(int[][] a, int n,
                               int[] queries, int q)
    {
 
        // container to store all range
        Vector<Integer> v = new Vector<>();
 
        // hash the L and R
        HashMap<Integer, Integer> mpp = new HashMap<>();
 
        // Push the element to container
        // and hash the L and R
        for (int i = 0; i < n; i++)
        {
            v.add(a[i][0]);
            mpp.put(a[i][0], 1);
            v.add(a[i][1]);
            mpp.put(a[i][1], 2);
        }
 
        // sort the elements in container
        Collections.sort(v);
        for (int i = 0; i < q; i++)
        {
 
            // each query
            int num = queries[i];
 
            // get the number same or greater than integer
            int ind = lowerBound(v, num);
 
            // if it lies
            if (ind >= 0 && v.elementAt(ind) == num)
                System.out.println("Yes");
 
            else
            {
 
                // check if greater is hashed as 2
                if (ind >= 0 && mpp.get(v.elementAt(ind)) == 2)
                    System.out.println("Yes");
 
                else // check if greater is hashed as 1
                    System.out.println("No");
            }
        }
    }
 
    // Lower bound implementation
    static int lowerBound(Vector<Integer> array, int value)
    {
        int low = 0;
        int high = array.size();
        while (low < high)
        {
            final int mid = (low + high) / 2;
            if (value <= array.elementAt(mid))
            {
                high = mid;
            }
            else
            {
                low = mid + 1;
            }
        }
        return low;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
 
        int[][] a = {{ 5, 6 }, { 1, 3 }, { 8, 10 }};
        int n = 3;
        int[] queries = {2, 3, 4, 7};
        int q = queries.length;
 
        // function call to answer queries
        answerQueries(a, n, queries, q);
    }
}
 
// This code is contributed by
// sanjeev2552


Python3




# Python program to check if the
# number lies in given range
from bisect import bisect_left as lower_bound
 
# Function that answers every query
def answerQueries(a: list, n, queries: list, q):
 
    # container to store all range
    v = list()
 
    # hash the L and R
    mpp = dict()
 
    # Push the element to container
    # and hash the L and R
    for i in range(n):
        v.append(a[i][0])
        mpp[a[i][0]] = 1
        v.append(a[i][1])
        mpp[a[i][1]] = 2
 
    # sort the elements in container
    v.sort()
    for i in range(q):
 
        # each query
        num = queries[i]
 
        # get the number same or greater than integer
        ind = lower_bound(v, num)
 
        # if it lies
        if v[ind] == num:
            print("Yes")
        else:
 
            # check if greater is hashed as 2
            if mpp[v[ind]] == 2:
                print("Yes")
 
            # check if greater is hashed as 1
            else:
                print("No")
 
# Driver Code
if __name__ == "__main__":
    a = [[5, 6], [1, 3], [8, 10]]
    n = 3
    queries = [2, 3, 4, 7]
    q = len(queries)
 
    # function call to answer queries
    answerQueries(a, n, queries, q)
 
# This code is contributed by
# sanjeev2552


C#




// C# program to check if the
// number lies in given range
using System;
using System.Collections.Generic;
 
class GFG
{
 
    // Function that answers every query
    static void answerQueries(int[,] a, int n,
                            int[] queries, int q)
    {
 
        // container to store all range
        List<int> v = new List<int>();
 
        // hash the L and R
        Dictionary<int, int> mpp = new Dictionary<int, int>();
 
        // Push the element to container
        // and hash the L and R
        for (int i = 0; i < n; i++)
        {
            v.Add(a[i, 0]);
            if(!mpp.ContainsKey(a[i, 0]))
                mpp.Add(a[i, 0], 1);
            v.Add(a[i, 1]);
            if(!mpp.ContainsKey(a[i, 1]))
                mpp.Add(a[i, 1], 2);
        }
 
        // sort the elements in container
        v.Sort();
        for (int i = 0; i < q; i++)
        {
 
            // each query
            int num = queries[i];
 
            // get the number same or greater than integer
            int ind = lowerBound(v, num);
 
            // if it lies
            if (ind >= 0 && v[ind] == num)
                Console.WriteLine("Yes");
 
            else
            {
 
                // check if greater is hashed as 2
                if (ind >= 0 && mpp[v[ind]] == 2)
                    Console.WriteLine("Yes");
 
                else // check if greater is hashed as 1
                    Console.WriteLine("No");
            }
        }
    }
 
    // Lower bound implementation
    static int lowerBound(List<int> array, int value)
    {
        int low = 0;
        int high = array.Count;
        while (low < high)
        {
            int mid = (low + high) / 2;
            if (value <= array[mid])
            {
                high = mid;
            }
            else
            {
                low = mid + 1;
            }
        }
        return low;
    }
 
    // Driver Code
    public static void Main(String[] args)
    {
 
        int[,] a = {{ 5, 6 }, { 1, 3 }, { 8, 10 }};
        int n = 3;
        int[] queries = {2, 3, 4, 7};
        int q = queries.Length;
 
        // function call to answer queries
        answerQueries(a, n, queries, q);
    }
}
 
// This code is contributed by 29AjayKumar


Javascript




<script>
// Javascript program to check if the
// number lies in given range
 
// Function that answers every query
function answerQueries(a, n, queries, q)
{
 
    // container to store all range
        let v = [];
  
        // hash the L and R
        let mpp = new Map();
  
        // Push the element to container
        // and hash the L and R
        for (let i = 0; i < n; i++)
        {
            v.push(a[i][0]);
            mpp.set(a[i][0], 1);
            v.push(a[i][1]);
            mpp.set(a[i][1], 2);
        }
  
        // sort the elements in container
        v.sort(function(a,b){return a-b;});
        for (let i = 0; i < q; i++)
        {
  
            // each query
            let num = queries[i];
  
            // get the number same or greater than integer
            let ind = lowerBound(v, num);
  
            // if it lies
            if (ind >= 0 && v[ind] == num)
                document.write("Yes<br>");
  
            else
            {
  
                // check if greater is hashed as 2
                if (ind >= 0 && mpp.get(v[ind]) == 2)
                    document.write("Yes<br>");
  
                else // check if greater is hashed as 1
                    document.write("No<br>");
            }
        }
}
 
// Lower bound implementation
function lowerBound(array,value)
{
    let low = 0;
        let high = array.length;
        while (low < high)
        {
            let mid = Math.floor((low + high) / 2);
            if (value <= array[mid])
            {
                high = mid;
            }
            else
            {
                low = mid + 1;
            }
        }
        return low;
}
 
// Driver Code
let a=[[ 5, 6 ], [ 1, 3 ], [ 8, 10 ]];
let n = 3;
let queries=[2, 3, 4, 7];
let q = queries.length;
 
// function call to answer queries
answerQueries(a, n, queries, q);
 
// This code is contributed by rag2127
</script>


Output: 

Yes
Yes
No
No

 

Time Complexity: O(n*log(n)+q*log(n))
Auxiliary Space: O(n)


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