Create 2001-number-of-pairs-of-interchangeable.cpp

cpp/2001-number-of-pairs-of-interchangeable-rectangles.cpp

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+/*
+    Approach: 
+    Keep the track of the ratios in a hash map
+
+    Time complexity : O(n)
+    Space complexity: O(n)
+
+    n is number of rectangles
+*/
+
+class Solution {
+public:
+    long long interchangeableRectangles(vector<vector<int>>& rectangles) {
+
+        map<long double,int> hash;
+        long double ratio;
+
+        long long answer=0;
+
+        for(int i=0;i<rectangles.size();i++){
+            ratio = (long double)(rectangles[i][0])/
+                    (long double)(rectangles[i][1]);
+
+            if(hash.find(ratio)!=hash.end()){
+                hash[ratio]++;
+            }
+            else{
+                hash[ratio] = 1;
+            }
+        }
+
+        for(auto it:hash){
+            answer+= (long long)(it.second)*(long long)(it.second-1)/2;
+        }
+
+        return answer;
+    }
+};