Created
November 4, 2018 22:36
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Implementing basic set operations - isSuperset, union, intersection and difference
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function isSuperset(set, subset) { | |
for (var elem of subset) { | |
if (!set.has(elem)) { | |
return false; | |
} | |
} | |
return true; | |
} | |
function union(setA, setB) { | |
var _union = new Set(setA); | |
for (var elem of setB) { | |
_union.add(elem); | |
} | |
return _union; | |
} | |
function intersection(setA, setB) { | |
var _intersection = new Set(); | |
for (var elem of setB) { | |
if (setA.has(elem)) { | |
_intersection.add(elem); | |
} | |
} | |
return _intersection; | |
} | |
function difference(setA, setB) { | |
var _difference = new Set(setA); | |
for (var elem of setB) { | |
if(setA.has(elem)) | |
_difference.delete(elem); | |
else | |
_difference.add(elem); | |
} | |
return _difference; | |
} | |
//Examples | |
var setA = new Set([1, 2, 3, 4]), | |
setB = new Set([2, 3]), | |
setC = new Set([3, 4, 5, 6]); | |
isSuperset(setA, setB); // => true | |
union(setA, setC); // => Set [1, 2, 3, 4, 5, 6] | |
intersection(setA, setC); // => Set [3, 4] | |
difference(setA, setC); // => Set [1, 2, 5, 6] |
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