# 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.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++ implementation of the above approach ` `#include ` `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 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 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# 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.. `

 ` ``\$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 ` `?> `

Output:
```3
```

Time Complexity: O(N)

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Improved By : vt_m, inderDuMCA, rituraj_jain

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