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

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: 3
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++

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// 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;
}

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Java

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// 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..

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Python3

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# 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 =
        if rt[i] > rt[mxrt]: 
            mxrt =
  
    # 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

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C#

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// 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..

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PHP

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<?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
?>

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Output:

3

Time Complexity: O(N)



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Improved By : vt_m, inderDuMCA, rituraj_jain



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