Pairs of similar rectangles

matchingRects.js

/*
Given a 2D array A[][2] of size N (1 ≤ N ≤ 103), where A[i][0] and A[i][1] denotes the length and breadth of rectangle i respectively.

Two rectangle i and j where (i < j) are similar if the ratio of their length and breadth is equal

A[i][0] / A[i][1] = A[j][0] / A[j][1]

Input : A[][2] = {{4, 8}, {15, 30}, {3, 6}, {10, 20}}
Output: 6
Explanation: Pairs of similar rectangles are (0, 1), {0, 2), (0, 3), (1, 2), (1, 3), (2, 3). For every rectangle, ratio of length : breadth is 1 : 2

Input : A[][2] = {{2, 3}, {4, 5}, {7, 8}}
Output: 0
Explanation: No pair of similar rectangles exists.

Approach: Follow the steps to solve the problem 

Traverse the array.
For every pair such that (i < j), check whether the rectangles are similar or not by checking if the condition A[i][0] / A[i][1] = A[j][0] / A[j][1] is satisfied or not.
If found to be true, increment the count.
Finally, print the count obtained.

*/
function getMatchingPairs(arr){
  let count = 0;
  for (var i = 0; i<arr.length; i++){
    for (var j = i+1; j<arr.length; j++){
      // let prod1 = arr[i][0] * arr[j][1]
      // let prod2 = arr[i][1] * arr[j][0]
      // if (prod1 === prod2) count += 1

      let div1 = arr[i][0] / arr[j][0]
      let div2 = arr[i][1] / arr[j][1]
      if (div1.toFixed(2) === div2.toFixed(2)) count += 1
    }
  }
  return count
}

// test cases
// [4, 8], [15, 30], [3, 6], [10, 20]] => 6
// [[2, 3], [4, 5], [7, 8]] => 0
// [[2, 3], [4, 6], [7, 8]] => 1

var rectsArr = [[4, 8], [15, 30], [3, 6], [10, 20]]
getMatchingPairs(rectsArr)

Saw this kind of problem and the solution of looping inside of a loop was what I could think of but Hackerank complained of time complexity. Is there another way this can be achieved without looping inside of a for loop

You can try to use Dictionary