# Count the number of rectangles such that ratio of sides lies in the range [a,b]

Given the length and breadth of N rectangles and a range i.e. [a, b], the task is to count the number of rectangles whose sides(larger/smaller) ratio is in the range [a, b].

Examples:

Input: {{165, 100}, {180, 100}, {100, 170}}, a = 1.6, b = 1.7
Output: 2
165/100 = 1.65
170/100 = 1.7

Input: {{10, 12}, {26, 19}}, a = 0.8, b = 1.2
Output: 1

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

Approach: Iterate in the array of pairs, and increase the counter when max(a[i].first, a[i].second)/min(a[i].first, a[i].second) lies in the range a and b.

Below is the implementation of the above approach:

## C++

 `// C++ program to print the length of the shortest ` `// subarray with all elements greater than X ` `#include ` `using` `namespace` `std; ` ` `  `// Function to count the number of ratios ` `int` `countRatios(pair<``int``, ``int``> arr[], ``int` `n, ` `                           ``double` `a, ``double` `b) ` `{ ` `    ``int` `count = 0; ` ` `  `    ``// count the number of ratios ` `    ``// by iterating ` `    ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `        ``double` `large = max(arr[i].first, arr[i].second); ` `        ``double` `small = min(arr[i].first, arr[i].second); ` ` `  `        ``// find ratio ` `        ``double` `ratio = large / small; ` ` `  `        ``// check if lies in range ` `        ``if` `(ratio >= a && ratio <= b) ` `            ``count += 1; ` `    ``} ` ` `  `    ``return` `count; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``pair<``int``, ``int``> arr[] = { { 165, 100 }, ` `                             ``{ 180, 100 }, ` `                             ``{ 100, 170 } }; ` `    ``double` `a = 1.6, b = 1.7; ` `    ``int` `n = 3; ` ` `  `    ``cout << countRatios(arr, n, a, b); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to print the length of the shortest ` `// subarray with all elements greater than X ` `class` `GFG ` `{ ` `static` `int` `n = ``3``; ` `static` `class` `pair ` `{  ` `    ``int` `first, second;  ` `    ``public` `pair(``int` `first, ``int` `second)  ` `    ``{  ` `        ``this``.first = first;  ` `        ``this``.second = second;  ` `    ``}  ` `}  ` ` `  `// Function to count the number of ratios ` `static` `int` `countRatios(pair []arr, ``int` `n, ` `                       ``double` `a, ``double` `b) ` `{ ` `    ``int` `count = ``0``; ` ` `  `    ``// count the number of ratios ` `    ``// by iterating ` `    ``for` `(``int` `i = ``0``; i < n; i++)  ` `    ``{ ` `        ``double` `large = Math.max(arr[i].first,  ` `                                ``arr[i].second); ` `        ``double` `small = Math.min(arr[i].first, ` `                                ``arr[i].second); ` ` `  `        ``// find ratio ` `        ``double` `ratio = large / small; ` ` `  `        ``// check if lies in range ` `        ``if` `(ratio >= a && ratio <= b) ` `            ``count += ``1``; ` `    ``} ` `    ``return` `count; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String[] args)  ` `{ ` `    ``pair []arr = {``new` `pair(``165``, ``100``), ` `                  ``new` `pair(``180``, ``100``), ` `                  ``new` `pair(``100``, ``170``)}; ` `    ``double` `a = ``1.6``, b = ``1.7``; ` `    ``int` `n = ``3``; ` ` `  `    ``System.out.println(countRatios(arr, n, a, b)); ` `} ` `} ` ` `  `// This code is contributed by PrinciRaj1992 `

## C#

 `// C# program to print the length of the shortest ` `// subarray with all elements greater than X ` `using` `System; ` `     `  `class` `GFG ` `{ ` `static` `int` `n = 3; ` `class` `pair ` `{  ` `    ``public` `int` `first, second;  ` `    ``public` `pair(``int` `first, ``int` `second)  ` `    ``{  ` `        ``this``.first = first;  ` `        ``this``.second = second;  ` `    ``}  ` `}  ` ` `  `// Function to count the number of ratios ` `static` `int` `countRatios(pair []arr, ``int` `n, ` `                       ``double` `a, ``double` `b) ` `{ ` `    ``int` `count = 0; ` ` `  `    ``// count the number of ratios ` `    ``// by iterating ` `    ``for` `(``int` `i = 0; i < n; i++)  ` `    ``{ ` `        ``double` `large = Math.Max(arr[i].first,  ` `                                ``arr[i].second); ` `        ``double` `small = Math.Min(arr[i].first, ` `                                ``arr[i].second); ` ` `  `        ``// find ratio ` `        ``double` `ratio = large / small; ` ` `  `        ``// check if lies in range ` `        ``if` `(ratio >= a && ratio <= b) ` `            ``count += 1; ` `    ``} ` `    ``return` `count; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main(String[] args)  ` `{ ` `    ``pair []arr = {``new` `pair(165, 100), ` `                  ``new` `pair(180, 100), ` `                  ``new` `pair(100, 170)}; ` `    ``double` `a = 1.6, b = 1.7; ` `    ``int` `n = 3; ` ` `  `    ``Console.WriteLine(countRatios(arr, n, a, b)); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

Output:

```2
```

My Personal Notes arrow_drop_up Just another competitive programmer

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.

Improved By : princiraj1992, 29AjayKumar

Article Tags :

Be the First to upvote.

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.