# Search an element in given N ranges

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: 1
6 lies in the range {4, 7} with index = 1

Input: arr[] = { { 1, 3 }, { 4, 7 }, { 9, 11 } }, K = 8
Output: -1

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++ 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 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 `

 `# 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# 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 `

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++ 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 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 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# 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 `

Output:
```1
```

Time Complexity: O(log 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.

Striver(underscore)79 at Codechef and codeforces D

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Article Tags :