# Search an element in given N ranges

• Difficulty Level : Medium
• Last Updated : 07 Jun, 2021

Given an array of N sorted ranges and a number K. The task is to find the index of the range in which K lies. If K does not lie in any of the given ranges then print -1
Note: None of the given ranges coincide.
Examples:

Input: arr[] = { { 1, 3 }, { 4, 7 }, { 8, 11 } }, K = 6
Output:
6 lies in the range {4, 7} with index = 1
Input: arr[] = { { 1, 3 }, { 4, 7 }, { 9, 11 } }, K = 8
Output: -1

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Naive approach: The following steps can be followed to solve the above problem.

• Traverse all the ranges.
• Check if the condition K >= arr[i].first && K <= arr[i].second holds in any of the iterations.
• If the number K does not lie in any of the given range then print -1.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to return the index of the range``// in which K lies and uses linear search``int` `findNumber(pair<``int``, ``int``> a[], ``int` `n, ``int` `K)``{` `    ``// Iterate and find the element``    ``for` `(``int` `i = 0; i < n; i++) {` `        ``// If K lies in the current range``        ``if` `(K >= a[i].first && K <= a[i].second)``            ``return` `i;``    ``}` `    ``// K doesn't lie in any of the given ranges``    ``return` `-1;``}` `// Driver code``int` `main()``{``    ``pair<``int``, ``int``> a[] = { { 1, 3 }, { 4, 7 }, { 8, 11 } };``    ``int` `n = ``sizeof``(a) / ``sizeof``(a);``    ``int` `k = 6;``    ``int` `index = findNumber(a, n, k);``    ``if` `(index != -1)``        ``cout << index;``    ``else``        ``cout << -1;` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``class` `GFG``{``static` `class` `pair``{``    ``int` `first, second;``    ``public` `pair(``int` `first, ``int` `second)``    ``{``        ``this``.first = first;``        ``this``.second = second;``    ``}``}` `// Function to return the index``// of the range in which K lies``// and uses linear search``static` `int` `findNumber(pair a[],``                      ``int` `n, ``int` `K)``{` `    ``// Iterate and find the element``    ``for` `(``int` `i = ``0``; i < n; i++)``    ``{` `        ``// If K lies in the current range``        ``if` `(K >= a[i].first &&``            ``K <= a[i].second)``            ``return` `i;``    ``}` `    ``// K doesn't lie in any``    ``// of the given ranges``    ``return` `-``1``;``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``pair a[] = {``new` `pair(``1``, ``3` `),``                ``new` `pair(``4``, ``7` `),``                ``new` `pair(``8``, ``11` `)};``    ``int` `n = a.length;``    ``int` `k = ``6``;``    ``int` `index = findNumber(a, n, k);``    ``if` `(index != -``1``)``        ``System.out.println(index);``    ``else``        ``System.out.println(-``1``);``}``}` `// This code is contributed by Rajput-Ji`

## Python3

 `# Python 3 implementation of the approach` `# Function to return the index of the range``# in which K lies and uses linear search``def` `findNumber(a, n, K):``    ` `    ``# Iterate and find the element``    ``for` `i ``in` `range``(``0``, n, ``1``):``        ` `        ``# If K lies in the current range``        ``if` `(K >``=` `a[i][``0``] ``and` `K <``=` `a[i][``1``]):``            ``return` `i` `    ``# K doesn't lie in any of the``    ``# given ranges``    ``return` `-``1` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ``a ``=` `[[``1``, ``3``], [``4``, ``7``], [``8``, ``11``]]``    ``n ``=` `len``(a)``    ``k ``=` `6``    ``index ``=` `findNumber(a, n, k)``    ``if` `(index !``=` `-``1``):``        ``print``(index, end ``=` `"")``    ``else``:``        ``print``(``-``1``, end ``=` `"")``        ` `# This code is contributed by``# Surendra_Gangwar`

## C#

 `// C# implementation of the approach``using` `System;` `class` `GFG``{``    ` `class` `pair``{``    ``public` `int` `first, second;``    ``public` `pair(``int` `first, ``int` `second)``    ``{``        ``this``.first = first;``        ``this``.second = second;``    ``}``}` `// Function to return the index``// of the range in which K lies``// and uses linear search``static` `int` `findNumber(pair []a,``                    ``int` `n, ``int` `K)``{` `    ``// Iterate and find the element``    ``for` `(``int` `i = 0; i < n; i++)``    ``{` `        ``// If K lies in the current range``        ``if` `(K >= a[i].first &&``            ``K <= a[i].second)``            ``return` `i;``    ``}` `    ``// K doesn't lie in any``    ``// of the given ranges``    ``return` `-1;``}` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``pair []a = {``new` `pair(1, 3 ),``                ``new` `pair(4, 7 ),``                ``new` `pair(8, 11 )};``    ``int` `n = a.Length;``    ``int` `k = 6;``    ``int` `index = findNumber(a, n, k);``    ``if` `(index != -1)``        ``Console.WriteLine(index);``    ``else``        ``Console.WriteLine(-1);``}``}` `// This code is contributed by 29AjayKumar`

## Javascript

 ``
Output:
`1`

Time Complexity: O(N)
Efficient Approach: Binary Search can be used to find the element in O(log N).
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to return the index of the range``// in which K lies and uses binary search``int` `findNumber(pair<``int``, ``int``> a[], ``int` `n, ``int` `K)``{` `    ``int` `low = 0, high = n - 1;` `    ``// Binary search``    ``while` `(low <= high) {` `        ``// Find the mid element``        ``int` `mid = (low + high) >> 1;` `        ``// If element is found``        ``if` `(K >= a[mid].first && K <= a[mid].second)``            ``return` `mid;` `        ``// Check in first half``        ``else` `if` `(K < a[mid].first)``            ``high = mid - 1;` `        ``// Check in second half``        ``else``            ``low = mid + 1;``    ``}` `    ``// Not found``    ``return` `-1;``}` `// Driver code``int` `main()``{``    ``pair<``int``, ``int``> a[] = { { 1, 3 }, { 4, 7 }, { 8, 11 } };``    ``int` `n = ``sizeof``(a) / ``sizeof``(a);``    ``int` `k = 6;``    ``int` `index = findNumber(a, n, k);``    ``if` `(index != -1)``        ``cout << index;``    ``else``        ``cout << -1;` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``class` `GFG``{``static` `class` `pair``{``    ``int` `first, second;``    ``public` `pair(``int` `first, ``int` `second)``    ``{``        ``this``.first = first;``        ``this``.second = second;``    ``}``}` `// Function to return the index of the range``// in which K lies and uses binary search``static` `int` `findNumber(pair a[], ``int` `n, ``int` `K)``{``    ``int` `low = ``0``, high = n - ``1``;` `    ``// Binary search``    ``while` `(low <= high)``    ``{` `        ``// Find the mid element``        ``int` `mid = (low + high) >> ``1``;` `        ``// If element is found``        ``if` `(K >= a[mid].first &&``            ``K <= a[mid].second)``            ``return` `mid;` `        ``// Check in first half``        ``else` `if` `(K < a[mid].first)``            ``high = mid - ``1``;` `        ``// Check in second half``        ``else``            ``low = mid + ``1``;``    ``}` `    ``// Not found``    ``return` `-``1``;``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``pair a[] = { ``new` `pair(``1``, ``3``),``                 ``new` `pair(``4``, ``7``),``                 ``new` `pair(``8``, ``11``) };``    ``int` `n = a.length;``    ``int` `k = ``6``;``    ``int` `index = findNumber(a, n, k);``    ``if` `(index != -``1``)``        ``System.out.println(index);``    ``else``        ``System.out.println(-``1``);``    ``}``}` `// This code is contributed by Princi Singh`

## Python3

 `# Python3 implementation of the approach` `# Function to return the index of the range``# in which K lies and uses binary search``def` `findNumber(a, n, K):` `    ``low ``=` `0``    ``high ``=` `n ``-` `1` `    ``# Binary search``    ``while` `(low <``=` `high):` `        ``# Find the mid element``        ``mid ``=` `(low ``+` `high) >> ``1` `        ``# If element is found``        ``if` `(K >``=` `a[mid][``0``] ``and` `K <``=` `a[mid][``1``]):``            ``return` `mid` `        ``# Check in first half``        ``elif` `(K < a[mid][``0``]):``            ``high ``=` `mid ``-` `1` `        ``# Check in second half``        ``else``:``            ``low ``=` `mid ``+` `1` `    ``# Not found``    ``return` `-``1` `# Driver code``a``=` `[ [ ``1``, ``3` `], [ ``4``, ``7` `], [ ``8``, ``11` `] ]``n ``=` `len``(a)``k ``=` `6``index ``=` `findNumber(a, n, k)``if` `(index !``=` `-``1``):``    ``print``(index)``else``:``    ``print``(``-``1``)` `# This code is contributed by mohit kumar`

## C#

 `// C# implementation of the above approach``using` `System;``class` `GFG``{``public` `class` `pair``{``    ``public` `int` `first, second;``    ``public` `pair(``int` `first, ``int` `second)``    ``{``        ``this``.first = first;``        ``this``.second = second;``    ``}``}` `// Function to return the index of the range``// in which K lies and uses binary search``static` `int` `findNumber(pair []a, ``int` `n, ``int` `K)``{``    ``int` `low = 0, high = n - 1;` `    ``// Binary search``    ``while` `(low <= high)``    ``{` `        ``// Find the mid element``        ``int` `mid = (low + high) >> 1;` `        ``// If element is found``        ``if` `(K >= a[mid].first &&``            ``K <= a[mid].second)``            ``return` `mid;` `        ``// Check in first half``        ``else` `if` `(K < a[mid].first)``            ``high = mid - 1;` `        ``// Check in second half``        ``else``            ``low = mid + 1;``    ``}` `    ``// Not found``    ``return` `-1;``}` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``pair []a = {``new` `pair(1, 3),``                ``new` `pair(4, 7),``                ``new` `pair(8, 11)};``    ``int` `n = a.Length;``    ``int` `k = 6;``    ``int` `index = findNumber(a, n, k);``    ``if` `(index != -1)``        ``Console.WriteLine(index);``    ``else``        ``Console.WriteLine(-1);``    ``}``}` `// This code is contributed by Rajput-Ji`

## Javascript

 ``
Output:
`1`

Time Complexity: O(log N)

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