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;
+ }
+};