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# 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 using namespace std; // Function to check rangeint 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 Codeint 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 rangestatic 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 rangedef 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 Codeif __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 approachusing System; class GFG{// Function to check rangestatic 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 Codepublic 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

 \$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


Output:

3

Time Complexity: O(N)

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