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Create 2001-number-of-pairs-of-interchangeable.cpp
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/*
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Approach:
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Keep the track of the ratios in a hash map
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Time complexity : O(n)
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Space complexity: O(n)
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n is number of rectangles
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*/
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class Solution {
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public:
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long long interchangeableRectangles(vector<vector<int>>& rectangles) {
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map<long double,int> hash;
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long double ratio;
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long long answer=0;
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for(int i=0;i<rectangles.size();i++){
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ratio = (long double)(rectangles[i][0])/
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(long double)(rectangles[i][1]);
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if(hash.find(ratio)!=hash.end()){
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hash[ratio]++;
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}
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else{
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hash[ratio] = 1;
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}
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}
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for(auto it:hash){
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answer+= (long long)(it.second)*(long long)(it.second-1)/2;
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}
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return answer;
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}
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};

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