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

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:

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// C++ program to print the length of the shortest
// subarray with all elements greater than X
#include <bits/stdc++.h>
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;
}

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Output:

2


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