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

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	`1`	`+/*`
	`2`	`+ Approach:`
	`3`	`+ Keep the track of the ratios in a hash map`
	`4`	`+`
	`5`	`+ Time complexity : O(n)`
	`6`	`+ Space complexity: O(n)`
	`7`	`+`
	`8`	`+ n is number of rectangles`
	`9`	`+*/`
	`10`	`+`
	`11`	`+classSolution {`
	`12`	`+public:`
	`13`	`+longlonginterchangeableRectangles(vector<vector<int>>& rectangles) {`
	`14`	`+`
	`15`	`+ map<longdouble,int> hash;`
	`16`	`+longdouble ratio;`
	`17`	`+`
	`18`	`+longlong answer=0;`
	`19`	`+`
	`20`	`+for(int i=0;i<rectangles.size();i++){`
	`21`	`+ ratio = (longdouble)(rectangles[i][0])/`
	`22`	`+ (longdouble)(rectangles[i][1]);`
	`23`	`+`
	`24`	`+if(hash.find(ratio)!=hash.end()){`
	`25`	`+ hash[ratio]++;`
	`26`	`+ }`
	`27`	`+else{`
	`28`	`+ hash[ratio] =1;`
	`29`	`+ }`
	`30`	`+ }`
	`31`	`+`
	`32`	`+for(auto it:hash){`
	`33`	`+ answer+= (longlong)(it.second)*(longlong)(it.second-1)/2;`
	`34`	`+ }`
	`35`	`+`
	`36`	`+return answer;`
	`37`	`+ }`
	`38`	`+};`

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