Skip to content
Related Articles

Related Articles

Improve Article

Find a range that covers all the elements of given N ranges

  • Difficulty Level : Easy
  • Last Updated : 25 May, 2021

Given N ranges containing L and R. The task is to check or find the index(0-based) of the range which covers all the other given N-1 ranges. If there is no such range, print -1.
Note: All L and R points are distinct.
Examples: 
 

Input: L[] = {1, 2}, R[] = {1, 2} 
Output: -1
Input: L[] = {2, 4, 3, 1}, R[] = {4, 6, 7, 9} 
Output:
Range at 3rd index i.e. 1 to 9 covers 
all the elements of other N-1 ranges.

 

Approach: Since all L and R points are distinct, find the range with the smallest L point and the range with the largest R point, if both are in the same Range, it would mean that all other ranges lie within it, otherwise it is not possible.
Below is the implementation of the above approach: 
 

C++




// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to check range
int findRange(int n, int lf[], int rt[])
{
 
    // Index of smallest L and largest R
    int mnlf = 0, mxrt = 0;
    for (int i = 1; i < n; i++) {
        if (lf[i] < lf[mnlf])
            mnlf = i;
        if (rt[i] > rt[mxrt])
            mxrt = i;
    }
 
    // If the same range has smallest L
    // and largest R
    if (mnlf == mxrt)
        return mnlf;
    else
        return -1;
}
 
// Driver Code
int main()
{
    int N = 4;
    int L[] = { 2, 4, 3, 1 };
    int R[] = { 4, 6, 7, 9 };
 
    cout << findRange(N, L, R);
 
    return 0;
}

Java




// Java implementation of the above approach
 
import java.io.*;
 
class GFG {
// Function to check range
static int  findRange(int n, int lf[], int rt[])
{
 
    // Index of smallest L and largest R
    int mnlf = 0, mxrt = 0;
    for (int i = 1; i < n; i++) {
        if (lf[i] < lf[mnlf])
            mnlf = i;
        if (rt[i] > rt[mxrt])
            mxrt = i;
    }
 
    // If the same range has smallest L
    // and largest R
    if (mnlf == mxrt)
        return mnlf;
    else
        return -1;
}
 
// Driver Code
 
 
    public static void main (String[] args) {
            int N = 4;
    int[] L = { 2, 4, 3, 1 };
    int []R = { 4, 6, 7, 9 };
 
    System.out.println( findRange(N, L, R));
    }
}
// This code is contributed by anuj_67..

Python3




# Python3 implementation of the
# above approach
 
# Function to check range
def findRange(n, lf, rt):
 
    # Index of smallest L and
    # largest R
    mnlf, mxrt = 0, 0
    for i in range(1, n):
        if lf[i] < lf[mnlf]:
            mnlf = i
        if rt[i] > rt[mxrt]:
            mxrt = i
 
    # If the same range has smallest
    # L and largest R
    if mnlf == mxrt:
        return mnlf
    else:
        return -1
 
# Driver Code
if __name__ == "__main__":
 
    N = 4
    L = [2, 4, 3, 1]
    R = [4, 6, 7, 9]
 
    print(findRange(N, L, R))
 
# This code is contributed
# by Rituraj Jain

C#




// C# implementation of the
// above approach
using System;
 
class GFG
{
// Function to check range
static int findRange(int n, int []lf,
                            int []rt)
{
 
    // Index of smallest L and largest R
    int mnlf = 0, mxrt = 0;
    for (int i = 1; i < n; i++)
    {
        if (lf[i] < lf[mnlf])
            mnlf = i;
        if (rt[i] > rt[mxrt])
            mxrt = i;
    }
 
    // If the same range has smallest L
    // and largest R
    if (mnlf == mxrt)
        return mnlf;
    else
        return -1;
}
 
// Driver Code
public static void Main ()
{
    int N = 4;
    int[] L = { 2, 4, 3, 1 };
    int []R = { 4, 6, 7, 9 };
     
    Console.WriteLine(findRange(N, L, R));
}
}
 
// This code is contributed by anuj_67..

PHP




<?php
// PHP implementation of the above approach
 
// Function to check range
function findRange($n, $lf, $rt)
{
 
    // Index of smallest L and largest R
    $mnlf = 0; $mxrt = 0;
    for ($i = 1; $i <$n; $i++)
    {
        if ($lf[$i] < $lf[$mnlf])
            $mnlf = $i;
        if ($rt[$i] > $rt[$mxrt])
            $mxrt = $i;
    }
 
    // If the same range has smallest
    // L and largest R
    if ($mnlf == $mxrt)
        return $mnlf;
    else
        return -1;
}
 
// Driver Code
$N = 4;
$L = array( 2, 4, 3, 1 );
$R = array( 4, 6, 7, 9 );
 
echo findRange($N, $L, $R);
 
// This code is contributed
// by inder_verma
?>

Javascript




<script>
 
// JavaScript implementation of the above approach
 
// Function to check range
function findRange(n, lf, rt)
{
 
    // Index of smallest L and largest R
    let mnlf = 0, mxrt = 0;
    for (let i = 1; i < n; i++) {
        if (lf[i] < lf[mnlf])
            mnlf = i;
        if (rt[i] > rt[mxrt])
            mxrt = i;
    }
 
    // If the same range has smallest L
    // and largest R
    if (mnlf == mxrt)
        return mnlf;
    else
        return -1;
}
 
// Driver Code
    let N = 4;
    let L = [ 2, 4, 3, 1 ];
    let R = [ 4, 6, 7, 9 ];
 
    document.write(findRange(N, L, R));
 
 
// This code is contributed by Surbhi Tyagi.
 
</script>
Output: 



3

 

Time Complexity: O(N)
 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :